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125 is multiplied by either 10 or 15. The resultant number is again multiplied by either 10 or 15. This process continues.
Which of the following CANNOT be a resultant number at any point in time?
125 = $$5^3$$
10 = $$2\times\ 5$$
15 = $$3\times\ 5$$
In each step, we are multiplying 125 by either 10 or 15. So, the power of 5 and the power of either 2 or 3 will only increase by one in every step. So at any point of time, the difference between the power of 5 and the power of 2 and 3 combined will remain constant (which is 3).
Option A : Power of 5 = 691
Power of 2 = 235
Power of 3 = 453
Difference of powers = 691 - (235+453) = 3
Option B : Power of 5 = 1080
Power of 2 = 253
Power of 3 = 824
Difference of powers = 1080 - (235+824) = 3
Option C : Power of 5 = 1604
Power of 2 = 689
Power of 3 = 912
Difference of powers = 1604 - (689+912) = 3
Option D : Power of 5 = 1034
Power of 2 = 476
Power of 3 = 455
Difference of powers = 1034 - (476+455) = 103
Option E : Power of 5 = 1145
Power of 2 = 689
Power of 3 = 453
Difference of powers = 1145 - (689+453) = 3
Only in option D the difference of power is different than 3.
$$\therefore\ $$ The answer is D.
Consider a seven-digit number 735x6y4, divisible by 44, where the two digits x and y are unknown.
Consider the following two additional pieces of information:
I. x and y are even numbers
II. x and y are equal
To determine the values of x and y UNIQUELY, which of the above pieces of information is/are MINIMALLY SUFFICIENT?
A seven-digit number 735x6y4 is divisible by 44. It means that the number is also divisible by 4 and 11.
Divisibility rule of 4 = Last two digits of the number is divisible by 4.
Divisibility rule of 11 = Sum of odd digits starting from left - Sum of even digits starting from left
For the given number to be divisible by 4, y4 should be divisible by 4.
So, the possible values of y = 0,2,4,6,8
For the number to be divisible by 11, (7+5+6+4)-(3+x+y) = 19-(x+y) should be divisible by 11.
Case 1 : Both x and y are even numbers.
If y = 0, x = 8
If y = 2, x = 6
If y = 4, x = 4
If y = 6, x = 2
If y = 8, x = 0
In this case, we got 5 different combinations of values.
Case 2 : Both x and y are equal.
If y = 4, x = 4
In this case, we got a unique solution.
Hence, we only need condition 2 to uniquely determine the values of x and y.
$$\therefore\ $$ The required answer is C.
Arun selected an integer $$x$$ between 2 and 40, both inclusive. He noticed that the greatest common divisor of the selected integer $$x$$ and any other integer between 2 and 40, both inclusive, is 1.
How many different choices for such an $$x$$ are possible?
We know that if the greatest common divisor of two numbers is 1, they are co-prime to each other.
Given that the greatest common divisor of the selected integer $$x$$ and any other integer between 2 and 40, both inclusive, is 1.
So, the selected number has to be a prime number, and that too greater than $$\dfrac{40}{2}$$ or $$20$$ because prime numbers less than 20, like 17, will have the greatest common divisor = 17 with their multiple, like 34 which is also in the range of selected integers.
Hence, the only possible values of $$x$$ are 23, 29, 31 and 37 i.e. a total of 4 values.
If a, b, and c are all positive integers, with 4a > b, then which of the following conditions is BOTH NECESSARY AND SUFFICIENT for the expression $$\sqrt[3]{(3)^a(21)^{(3a - b)}(49)^{(2b + c)}}$$ to be a positive integer?
The given expression is, $$\sqrt[3]{(3)^a(21)^{(3a - b)}(49)^{(2b + c)}}$$ and it can also be written as,
$$\sqrt[3]{(3)^a\ \left(3\right)^{3a\ -\ b}\ \ \left(7\right)^{3a\ -\ b}\ (7)^{4b+2c}}$$ $$=\ \sqrt[3]{\ \left(3\right)^{4a\ -\ b}\ \ (7)^{3a\ +\ 3b+2c}}$$
For the above expression to be a positive integer, the power of the expression must be an integer after applying the cube root for the expression inside. This means that the power of the expression inside must be a multiple of 3.
The expression is,
$$\sqrt[3]{\ \left(3\right)^{4a\ -\ b}\ \ (7)^{3a\ +\ 3b+2c}}\ =\ 3^{\frac{\left(4a\ -\ b\right)}{3}}\ \times\ 7^{\frac{\left(3a\ +\ 3b\ +\ 2c\right)}{3}}$$
The value $$\dfrac{4a\ -\ b}{3}=\ \dfrac{3a\ +\ a\ -\ b}{3}\ =\ a\ +\dfrac{a\ -\ b}{3}$$ must be an integer which means that a - b must be a multiple of 3.
Similarly, the value $$\dfrac{3a\ +\ 3b\ +\ 2c}{3}=\ a\ +\ b\ +\ \dfrac{2c}{3}$$ must also be an integer which gives us the condition that c must be a multiple of 3.
So, the necessary and sufficient conditions are a - b and c must be a multiple of 3.
Hence, the correct answer is option E.
Consider a 4-digit number of the form abbb, i.e., the first digit is a (a > 0) and the last three digits are all b.
Which of the following conditions is both NECESSARY and SUFFICIENT to ensure that the 4-digit number is divisible by a?
abbb can be written as (1000*a+100*b+10*b+1*b) = 1000a+111b
Now, we need to check to the necessary and sufficient condition for which (1000a+111b) is divisible by a.
We know that 1000a is always divisible by 'a', hence, we need to check for which condition, 111b is always divisible by a.
111b can be written as (3*37*b) => (3b*37) must be divisible by a.
a can't be a factor of 37, which implies 'a' is a factor of 3b => 3b is divisible by 'a'
The correct option is E
The least common multiple of a number and 990 is 6930. The greatest common divisor of that number and 550 is 110.
What is the sum of the digits of the least possible value of that number?
Let's say the unknown number is x—the LCM of x, 990= 6930.
LCM is the greatest power of a prime number in those two numbers.
Now let's prime factorize 990 and 6930.
990=2*5*11*$$3^2$$
6930=2*$$3^2$$*5*7*11
The number 990 has all the powers of prime numbers, which are also present in 6930, except 7. So, x definitely has to have 7.
Then, it is given that the GCD of x and 550 is 110. So, x has to be a factor of 110. 110 is not a multiple of 7. So, we have to multiply the 110 by 7. That is 770, which is the minimum value of x.
The sum of digits of x is 7+7+0=14
Amit has forgotten his 4-digit locker key. He remembers that all the digits are positive integers and are different from each other. Moreover, the fourth digit is the smallest and the maximum value of the first digit is 3. Also, he recalls that if he divides the second digit by the third digit, he gets the first digit.
How many different combinations does Amit have to try for unlocking the locker?
Let the 4-digit locker key be $$abcd$$, where $$9\ge\ a,b,c,d>0$$, and they are all unique.
It is given that $$a\le\ 3\ \&\ \frac{b}{c}=a$$
It is also given that 'd' is the smallest number.
Case (i): $$a=1$$
Not possible, as 'd' is the smallest number, 'a' cannot be 1.
Case (ii): $$a=2$$
Then 'd' can take only one value, i.e. $$d=1$$
$$b=2c$$
$$if\ c=3,\ then\ b=6$$
$$if\ c=4,\ then\ b=8$$
for $$c\ge\ 5$$, 'b' will not be a single-digit number.
Hence, two cases are possible, i.e. (2631, 2841).
Case (iii): $$a=3$$
'd' can take two values, i.e. $$d=1\ or\ d=2\ $$
& $$b=3c$$
If d = 1, then 'c' can take only one value. i.e. $$\ c=2,\ then\ b=6$$
If d = 2, then 'c' cannot take any value.
Hence one case is possible, i.e. (3621)
Amit has to try 3 different combinations. Option (E) is correct.
Suppose Haruka has a special key $$\triangle$$ in her caculator called delta key:
Rule 1: If the display shows a one-digit number, pressing delta key $$\triangle$$ replace the displayed number with twice its value.
Rule 2: If the display shows a two-digits number, pressing delta key $$\triangle$$ replace the displayed number with the number sum of two digits.
Suppose Haruka enters the value 1 and then presses delta key $$\triangle$$ repeated.
After pressing the key for 68 times, what will be the displayed number?
The first number Haruka entered is 1.
From Rule-1, the series will be 2, 4, 8, 16
From Rule 2, 16 will be followed by 7
Like-wise, the series will be 1, 2, 4, 8, 16, 7, 14, 5, 10, 1, 2, 4,......
The series is repeating itself for every 9 iterations.
Initially, input is 1 and after first iteration, result is 2.
This implies,
9n = 1
9n + 1 = 2
9n + 2 = 4
.
.
.
.
9n + 7 = 5
9n + 8 = 10
In the question, it is given that key is pressed 68 times.
68 = 9(7) + 5
68 is in the form of 9n + 5
This implies, 7 is the output.
The answer is option A.
The addition of 7 distinct positive integers is 1740. What is the largest possible “greatest common divisor” of these 7 distinct positive integers?
Let 'm' be the Greatest common divisor and A, B, C, D, E, F & G be the 7 unique numbers.
When 'm' divides A, B, ......& G, we get seven unique quotients. Let the quotients be a, b, c, d, e, f, and g.
It is given,
A + B + C + D + E + F + G = 1740
m(a + b + c + d + e + f + g) = 1740
$$\ \ 1740=2^2\times3\times5\times29$$
m(a + b + c + d + e + f + g) = $$\ \ 1740=2^2\times3\times5\times29$$
29 is a prime number and cannot be factorised further.
Therefore, largest possible value m can take is 60.
Least possible sum of 7 unique numbers is 1 + 2 + 3 +...+ 7 = 28
Replacing 7 with 8, we will get 29.
This implies 60(1 + 2 + 3 .....+ 6 + 8) = 1740
Largest possible G.C.D is 60.
The answer is option B.
Raju and Sarita play a number game. First, each one of them chooses a positive integer independently. Separately, they both multiply their chosen integers by 2, and then subtract 20 from their resultant numbers. Now, each of them has a new number. Then, they divide their respective new numbers by 5. Finally, they added their results and found that the sum is 16. What can be the maximum possible difference between the positive integers chosen by Raju and Sarita?
Let the positive integers Raju & Sarita chose be 'r' & 's'.
After doing the given operations, the final numbers they get are $$\frac{2r-20}{5}\ \&\frac{2s-20}{5}$$
Adding results we get, $$\frac{2r-20}{5}\ +\frac{2s-20}{5}\ =\ 16$$
$$2r-20\ +2s-20\ =\ 80$$
$$2r+2s\ =\ 80+40+40=120$$
$$r+s\ =60$$
For maximum value of |r-s|, one of 'r' & 's' has to be maximum and other has to be minimum.
As r & s are positive integers minimum value they can take is "1".
If $$r=1,\ then\ s=59$$
$$\left|r-s\right|=\left|1-59\right|=58$$.
If $$s=1,\ then\ r=59$$
$$\left|r-s\right|=\left|59-1\right|=58$$
Hence maximum value of $$\left|r-s\right|$$ is 58. Option B is correct.
There are three sections in a question paper and each section has 10 questions. First section only has multiple-choice questions, and 2 marks will be awarded for each correct answer. For each wrong answer, 0.5 marks will be deducted. Any unattempted question in this section will be treated as a wrong answer. Each question in the second section carries 3 marks, whereas each question in the third section carries 5 marks. For any wrong answer or un-attempted question in the second and third sections, no marks will be deducted. A student’s score is the addition of marks obtained in all the three sections. What is the sixth highest possible score?
From the given information we get the following table,
The possible maximum marks are shown below.
Therefore, sixth highest possible score is 94.
Option B is correct.
A supplier receives orders from 5 different buyers. Each buyer places their order only on a Monday. The first buyer places the order after every 2 weeks, the second buyer, after every 6 weeks, the third buyer, after every 8 weeks, the fourth buyer, every 4 weeks, and the fifth buyer, after every 3 weeks. It is known that on January 1st, which was a Monday, each of these five buyers placed an order with the supplier.
On how many occasions, in the same year, will these buyers place their orders together excluding the order placed on January 1st?
The supplier receives his orders from the five buyers once every 2 weeks, once every 6 weeks, once every 8 weeks, once every 4 weeks, and once every 3 weeks.
The number of occasions where all the five buyers place the order on the same day is :
The LCM of the 5-time frames during which the 5 buyers place their orders :
Hence the LCM is :
(2, 6, 8, 4, 3).
= 24 weeks.
Once every 24 weeks, all five of them place the order simultaneously.
A year has 53 weeks in total :
Hence all five of them place the orders after 24 weeks, 48 weeks.
Wilma, Xavier, Yaska and Zakir are four young friends, who have a passion for integers. One day, each of them selects one integer and writes it on a wall. The writing on the wall shows that Xavier and Zakir picked positive integers, Yaska picked a negative one, while Wilma’s integer is either negative, zero or positive. If their integers are denoted by the first letters of their respective names, the following is true:
$$W^{4}+X^{3}+Y^{2}+Z\leq4$$
$$X^{3}+Z\geq2$$
$$W^{4}+Y^{2}\leq2$$
$$Y^{2}+Z\geq3$$
Given the above, which of these can $$W^{2}+X^{2}+Y^{2}+Z^{2}$$ possibly evaluate to?
Given that X, Z are positive Y is negative and W can be either positive or zero or negative.
The given conditions are :
$$W^{4}+X^{3}+Y^{2}+Z\leq4$$
$$X^{3}+Z\geq2$$
$$W^{4}+Y^{2}\leq2$$
$$Y^{2}+Z\geq3$$
For $$W^4+\ Y^2\ \le\ 2$$. Since Y is negative $$but\ Y^2$$ is always positive and must be less than 2 because $$W^4$$ is a nonnegative value. Hence Y = -1 is the only possibility. For W this can take any value among -1, 0, 1.
$$Y^2+Z\ \ge\ 3$$. Since Y = -1, Z must be at least equal to 2 so the value of $$Y^2+Z\ \ge\ 3$$ is greater than 2.
X is a positive value and must at least be equal to 1.
The condition: $$W^{2}+X^{2}+Y^{2}+Z^{2}$$ here has all the independent values: $$X^2,\ Y^2,\ Z^2,\ W^2$$are nonnegative.
$$W^{4}+X^{3}+Y^{2}+Z\leq4$$ :
Since the value of Z is at least equal to 2 the value of $$Y^2$$ is equal to 1.
Since X is a positive number in order to have the condition of $$W^{4}+X^{3}+Y^{2}+Z\leq4$$ satisfied. The value of Z must be the minimum possible so that $$X^3+Y^2+Z$$ to have a value equal to 4 when X takes the minimum possible positive value equal to 1.
Hence X must be 1. W must be equal to 0 so that :
$$W^{4}+X^{3}+Y^{2}+Z\leq4$$. = The sum = (0+1+1+2) = 4. The only possible case.
The value of $$W^{2}+X^{2}+Y^{2}+Z^{2}$$ = (0+1+1+4) = 6.
Fatima found that the profit earned by the Bala dosa stall today is a three-digit number. She also noticed that the middle digit is half of the leftmost digit, while the rightmost digit is three times the middle digit. She then randomly interchanged the digits and obtained a different number. This number was more than the original number by 198.
What was the middle digit of the profit amount?
From the given conditions :
Considering the three-digit number to be a b c.
With the given conditions :
a = 2b, c = 3b.
Hence the number is of the form : 2b b 3b.
Since all three of the values must be less than 10 and non-negative:
This takes values : b = 1, b = 2, b = 3.
Hence the possible numbers are : (213, 426, 639) :
The interchanged number must be greater than the original by 198.
Hence the different rearrangements for the three numbers are :
213 : (312, 321, 132, 123, 231).
426 : (462, 624, 642, 246, 264)
639 : ( 693, 963, 936, 396, 369)
The only possible value which is higher than the original by 198 is :
(426, 624).
The middle digit is 2.
The Madhura Fruits Company is packing four types of fruits into boxes. There are 126 oranges, 162 apples, 198 guavas and 306 pears. The fruits must be packed in such a way that a given box must have only one type of fruit and must contain the same number of fruit units as any other box.
What is the minimum number of boxes that must be used?
The number of oranges, apples, guavas, and pears = 126, 162, 198, and 306.
Each box must contain an equal number of fruits with only one type of fruit. The additional condition provided is that there should be a minimum number of boxes in total.
The distribution is possible in multiple ways in such a way that distribution in each box is placed in such that each box contains a certain number of fruits n which is a factor for all the four given number of fruits :
Arrangement of 1 fruit of one kind in a basket.
2 is a factor of 126, 162, 198, and 306. So we can place 2 fruits of a particular kind in a basket.
Since we were asked for the minimum number of boxes this is possible when a maximum number of fruits of a kind are placed in a box.
Hence each box must contain the Highest common factor for the four numbers :
The prime factorization for the four numbers :
126 : $$2\cdot7\cdot9$$, 162 : $$2\cdot9\cdot9$$, 198 : $$2\cdot9\cdot11$$, $$306\ =\ 2\cdot9\cdot17$$
The HCF is 18.
The number of boxes required for each :
$$\frac{126}{18},\ \frac{162}{18},\ \frac{198}{18},\ \frac{306}{18}$$
7+9+11+17 = 44.
An encryption system operates as follows:
Step 1. Fix a number k $$(k \leq 26)$$.
Step 2. For each word, swap the first k letters from the front with the last k letters from the end in reverse order. If a word contains less than 2k letters, write the entire word in reverse order.
Step 3. Replace each letter by a letter k spaces ahead in the alphabet. If you cross Z in the process to move k steps ahead, start again from A.
Example: k = 2: zebra --> arbez --> ctdgb.
If the word “flight” becomes “znmorl” after encryption, then the value of k:
Flight become znmorl
Let's assume $$k>3$$
So flight will become thgilf -> znmrol. Hence the value of k will be 6
Zahir and Raman are at the entrance of a dark cave. To enter this cave, they need to open a number lock. Raman sees a note on a rock: “ ... chest of pure diamonds kept for the smart one ... number has six digits ... second last digit is 2, third last is 4 ... divisible by all prime numbers less than 15 ...”. Excited, Zahir and Raman seek your help: which of these can be the first digit of the six-digit number that will help them open the lock?
Let the 6 digit number be _ _ _ 42_
It is divisible by 2,3,5,7,11,13
Since the number is divisible by both 2 and 5 the last digit of the number must be 0.
The number is also divisible by 3, 7, 11, and 13.
Hence the number must also be divisible by 7*11*13.
=7*11*13 = 1001.
A number which is a multiple of 1001 is of the form abcabc.
This is because abc*(1001) = abc*(1000+1) = abc000 + abc = abcabc.
Hence the number is 420420.
The first digit is 4.
Given that :
(A O B) = (A + B) * B.
For (5 O 2) = ( 5+2 )*2 = 14.
((14) O 5 )= ( 14 + 5 )*5 = 19*5 = 95.
When expressed in a decimal form, which of the following numbers will be non - terminating as well as non-repeating?
Option A: $$\left(\frac{\pi}{2}\right) \left[\left(\frac{1}{\pi}\right) + 1\right] - \frac{\pi}{2}$$ =1/2
Option B: $$\sin^21^{\circ}+\sin^22^{\circ}+....+\sin^289^{\circ}=44+1/2\ $$ because $$\sin^2\left(89\right)=\cos^2\left(1\right)\ \&\ \sin^2\left(1\right)+\cos^2\left(1\right)=1$$
Option C: $$\sqrt{2}\left(3\sqrt{2} - \frac{4}{\sqrt{2}}\right) + \sqrt{3}$$ = $$6-4+\sqrt{3}$$=$$2+\sqrt{3}$$ which is non-terminating and non repeating.
Option D: $$\frac{\left(\sqrt[3]{729}\right)}{3} + \frac{22}{7}$$ = 3+22/7 = 6.142857142857....
Option E: $$\left(\frac{\pi}{4}\right) + \left(\frac{\pi}{4}\right)^2 + \left(\frac{\pi}{4}\right)^3 + ...$$ (infinite terms)= $$\frac{1}{1-\frac{\pi\ }{4}}=\frac{4}{4\ -\pi\ }\ $$ => $$(4 - \pi)[1 + \left(\frac{\pi}{4}\right) + \left(\frac{\pi}{4}\right)^2 + \left(\frac{\pi}{4}\right)^3 + ...$$ (infinite terms)] =4
What is the remainder if $$19^{20} - 20^{19}$$ is divided by 7?
Using Fermat's theorem :
If p is a prime number and a, p are co primes $$\left(a^{p-1}\right)\ mod\ p=1$$
Remainder when $$19^{20}$$ is divided by 7 = $$19^2$$ mod 7 =4. ( Here $$19^{20\ }=\ \left(\left(19\right)^6\right)^3\cdot\left(19\right)^2$$
Since the remainder for $$19^6$$ is 1 the remainder for $$\ 19^{20}$$ is equivalent to the $$\frac{19^2}{7}$$ = 4.
Remainder when $$20^{19}$$ is divided by 7 = $$20^1$$ mod 7 =6.( Here $$\frac{20^6}{7}$$ the remainder is 1 and since
$$20^{19}=\ \left(20^6\right)^3\cdot\left(20\right)^1\ =\ \frac{\left(1\cdot20\right)}{7}$$. The remainder is 6.
Remainder when $$19^{20} - 20^{19}$$ is divided by 7=4-6=-2=> 5.
If $$\sqrt[3]{7^a\times 35^{b+1} \times 20^{c+2}}$$ is a whole number then which one of the statements below is consistent with it?
In the given statement, the expression becomes a whole number only when the powers of all the prime numbers are also whole numbers.
Let us first simplify the expression a bit by expressing all terms in terms of prime numbers.
$$\sqrt[3]{7^a\times 35^{b+1} \times 20^{c+2}}$$
$$\Rightarrow \sqrt[3]{7^a\times 5^{b+1} \times 7^{b+1} \times 2^{2(c+2)} \times 5^{c+2}}$$
$$\Rightarrow \sqrt[3]{2^{2c+4} \times 5^{b+c+3} \times 7^{a+b+1}}$$
$$\Rightarrow 2^{\frac{2c+4}{3}} 5^{\frac{b+c+3}{3}} 7^{\frac{a+b+1}{3}} $$
Now, from the given options, we can put in values of the variables and check the exponents of all the numbers.
Option A : a = 2, b = 1, c = 1 :
In this case, we can see that exponent of 5 ie $$\frac{b+c+3}{3} = \frac{5}{3} $$ is not a whole number.
Option B : a = 1, b = 2, c = 2
In this case, we can see that exponent of 2 ie $$\frac{2c+4}{3} = \frac{8}{3} $$ is not a whole number.
Option C : a = 2, b = 1, c = 2
In this case, we can see that exponent of 2 ie $$\frac{2c+4}{3} = \frac{8}{3} $$ is not a whole number.
Option D : a = 3, b = 1, c = 1
In this case, we can see that exponent of 5 ie $$\frac{b+c+3}{3} = \frac{5}{3} $$ is not a whole number.
Option E : a = 3, b = 2, c = 1
In this case, we can see that all exponents are whole numbers.
Thus, option E is the correct option.
Find the value of the expression: $$10 + 10^3 + 10^6 + 10^9$$
$$10 + 10^3 + 10^6 + 10^9 = 10 + 1000 + 1000000 + 1000000000$$
=$$1001001010$$
Therefore, option D is the right answer.
An antique store has a collection of eight clocks. At a particular moment, the displayed times on seven of the eight clocks were as follows: 1:55 pm, 2:03 pm, 2:11 pm, 2:24 pm, 2:45 pm, 3:19 pm and 4:14 pm. If the displayed times of all eight clocks form a mathematical series, then what was the displayed time on the remaining clock?
Let us find out the difference between the times given to figure out the pattern.
The times given are 1:55 pm, 2:03 pm, 2:11 pm, 2:24 pm, 2:45 pm, 3:19 pm and 4:14 pm.
The difference between 2 consecutive times given are 8 minutes, 8 minutes, 13 minutes, 21 minutes, 34 minutes, and 55 minutes.
We can observe that the difference between the times are in the Fibonacci series.
8 + 13 = 21
21 + 13 = 34
34 + 21 = 55
The Fibonacci series is as follows:
1,1,2,3,5,8,13,21,34,55.
But the first difference in the times given is 8.
Therefore, the missing time must be such that it divides the interval of 8 minutes into 3 minutes and 5 minutes.
The missing time should be 1:58 pm and hence, option B is the right answer.
X and Y are the digits at the unit's place of the numbers (408X) and (789Y) where X ≠ Y. However, the digits at the unit's place of the numbers $$(408X)^{63}$$ and $$(789Y)^{85}$$ are the same. What will be the possible value(s) of (X + Y)?
All numbers from 1 to 9 repeat their last digits over a cycle of 4.
63 can be written as 4k+3.
85 can be written as 4k+1.
Some number's third power should yield the first digit as some number's first power.
$$2^3$$ will yield $$8$$ as the last digit (2 and 8 is a possible solution).X+Y = 10
$$3^3$$ will yield $$7$$ as the last digit (3 and 7 is a possible solution).X + Y = 10
$$4^3$$ will yield $$4$$ as the last digit and hence, can be eliminated.
$$5$$ and $$6$$ yield $$5$$ and $$6$$ respectively as the last digit for any power and hence, can be eliminated.
$$7^3$$ will yield $$3$$ as the last digit (7 and 3 is a possible solution). X+Y=10.
$$8^3$$ will yield $$2$$ as the last digit. 8+2 =10.
As we can see, X+Y = 10 in all the cases. Therefore, option B is the right answer.
David has an interesting habit of spending money. He spends exactly £X on the Xth day of a month. For example, he spends exactly £5 on the 5th of any month. On a few days in a year, David noticed that his cumulative spending during the last 'four consecutive days' can be expressed as $$2^N$$ where N is a natural number. What can be the possible value(s) of N?
The sum of 4 consecutive numbers should be a power of 2.
The powers of 2 are 1, 2, 4, 8, 16, 32, 64, 128. The maximum possible value that 4 consecutive days can take is 28 + 29 + 30 + 31 = 118.
We can eliminate 1, 2, 4, and 8 since sum of 4 consecutive integers is always greater than 9.
Let the first day be x.
x+x+1+x+2+x+3 = 16
=> 4x = 10
x = 2.5
We can eliminate 16.
Let us check for 32.
x+x+1+x+2+x+3 = 32
4x = 26.
x is not an integer.
Let us check the case in which x is the last day of the month.
Even in the month of February, the least value that x can take is 28.
28+1+2+3 = 34 > 32.
We can eliminate 32 as well.
Let us check for 64.
4x+6 = 64
4x = 58
The sum of no 4 consecutive days in the same month can be expressed as 4k+6.
Let us check the cases in which 2 months are involved.
29+30+1+2 = 62 < 64.
30 + 31 + 1 + 2 = 64. This is a possible combination.
There are 7 months with 31 days in a year. We have to eliminate December since 1 will spill over to the 1st of January of the next year.
Therefore, in a year, there will be 6 such instances. Therefore, option B is the right answer.
If $$N = (11^{p + 7})(7^{q - 2})(5^{r + 1})(3^{s})$$ is a perfect cube, where $$p, q, r$$ and $$s$$ are positive integers, then the smallest value of $$p + q + r + s$$ is :
It has been given that $$N = (11^{p + 7})(7^{q - 2})(5^{r + 1})(3^{s})$$ is a perfect cube. All the factors given are prime. Therefore, the power of each number should be a multiple of 3 or 0.
$$p,q,r$$ and $$s$$ are positive integers. Therefore, only the power of the expressions in which some number is subtracted from these variables or these variables are subtracted from some number can be made $$0$$.
$$11^{p + 7}$$:
This expression must be made a perfect cube. The nearest perfect cube is $$11^9$$. Therefore, the least value that $$p$$ can take is $$9-7=2$$.
$$7^{q - 2}$$
The least value that $$q$$ can take is 2. If $$q=2$$, then the value of the expression $$7^{q-2}$$ will become $$7^0=1$$, without preventing the product from becoming a perfect cube.
$$5^{r+1}$$:
The least value that $$r$$ can take is $$2$$.
$$3^{s})$$:
The least value that $$s$$ can take is $$3$$.
Therefore, the least value of the expression $$p + q + r + s$$ is $$2+2+2+3=9$$.
Therefore, option E is the right answer.
Two numbers in the base system B are 2061$$_{B}$$ and 601$$_{B}$$. The sum of these two numbers in decimal system is 432. Find the value of 1010$$_B$$ in decimal system.
Converting numbers in decimal system
=> $$2061_B = 2B^3 + 6B + 1$$
and $$601_B = 6B^2 + 1$$
Acc. to ques, => $$2B^3 + 6B + 1 + 6B^2 + 1 = 432$$
Solving above equation, we get : $$B = 5$$
$$\therefore 1010_5$$ in decimal system = $$1(5)^3 + 1(5) = 130$$
Note :- This question is technically wrong as we cannot have digit 6 in Base 5.
For two positive integers a and b, if $$(a + b)^{(a + b)}$$ is divisible by 500, then the least possible value of a $$\times$$ b is:
If $$(a + b)^{(a + b)}$$ is divisible by 500,
$$500 = 2^2 \times 5^3$$
=> Least value of $$a + b = 2 \times 5 = 10$$
For least $$a$$ and $$b$$, let $$a = 1$$
=> $$b = 10 - 1 = 9$$
$$\therefore$$ Min $$(a \times b) = 1 \times 9 = 9$$
If a, b and c are 3 consecutive integers between -10 to +10 (both inclusive), how many integer values are possible for the expression?
$$\frac{a^3+b^3+c^3+3abc}{(a+b+c)^2}$$=?
Since a,b,c are consecutive integers
=> $$a = b-1$$ and $$c = b+1$$
Expression : $$\frac{a^3+b^3+c^3+3abc}{(a+b+c)^2}$$
= $$\frac{(b - 1)^3 + b^3 + (b + 1)^3 + 3 (b - 1) b (b + 1)}{(b - 1 + b + b + 1)^2}$$
= $$\frac{b^3 + 3b + b^3 + b^3 + 3b + 3b^3 - 3b}{9 b^2}$$
= $$\frac{6 b^3 + 3 b}{9 b^2} = \frac{2 b^2 + 1}{3 b}$$
Putting different values of b from - 10 to 10, we can verify that only - 1 and 1 satisfies to get integer values for the expression.
Ans - (C)
If a, b, c and d are four different positive integers selected from 1 to 25, then the highest possible value of ((a + b) + (c +d ))/((a + b) + (c - d)) would be:
Expression : $$\frac{a + b + c + d}{a + b + c - d}$$
To maximize the above expression, we have to minimize the denominator
Minimum value of the denominator = 1
So we can make $$a + b + c = 26$$ and $$d = 25$$ (as maximizing d will give denominator the least value).
So required maximum value = $$\frac{a + b + c + d}{a + b + c - d}$$
= $$\frac{26 + 25}{26 - 25} = 51$$
An ascending series of numbers satisfies the following conditions:
i. When divided by 3, 4, 5 or 6, the numbers leave a remainder of 2.
Ii. When divided by 11, the numbers leave no remainder.
The 6th number in this series will be:
L.C.M. of 3,4,5,6 = 60
Number is of the form = $$60 k_1 + 2$$ -------------(i)
When divided by 11, it leaves 0 remainder so number will also be of the form = $$11 k_2$$ ---------(ii)
Hence equating (i) and (ii), we get,
$$60k_1 + 2 = 11k_2$$
$$60k_1 - 11k_2 = -2$$ or $$11k_2 - 60k_1 = 2$$ -----------(iii)
It means $$60k_1$$ will leave remainder 9 when divide by 11.
Lets consider values for 60k1, if k1=1, 60k1=60, reminder is 60mod11=5
120mod11 will be 5+5=10, 180mod11 will be 5+5+5=15, since 15>11, reminder will be 15-11=4
240mod11 reminder will be 4+5=9
$$\therefore$$ By remainder root $$\frac{4 k_1}{11}$$ should leave remainder as 9 or -2
=> Possible values of $$K_1 = 4, 15, 26, 37, 48, 59$$ (As 11 and 60 are co-prime)
$$\therefore$$ Required value = $$60 \times 59 + 2 = 3540 + 2 = 3542$$
Alternatively,
L.C.M. of 3,4,5,6 = 60
As the number 60k+2 is divisible by 11, 60k leaves a reminder of 9
60mod11=5, 120mod11=10, 180mod11=4, 240mod11=9
Hence the first number where both conditions are satisfied as 242.
As 60 and 11 are co-prime, the next number where this is true is 242+60*11
Hence, the numbers are in the form 242+660k
For 6th number, k=5 => 3300+242=3542
In an examination, two types of questions are asked: one mark questions and two marks questions. For each wrong answer, of one mark question, the deduction is 1/4 of a mark and for each wrong answer, of two marks question, the deduction is 1/3 of a mark.
Moreover, 1/2 of a mark is deducted for any unanswered question. The question paper has 10 one mark questions and 10 two marks questions. In the examination, students got all possible marks between 25 and 30 and every student had different marks. What would be the rank of a student, who scores a total of 27.5 marks?
It is evident that, 1 wrong 2 marks question would result in 2.33 deduction from the total (As negative in 2 marks question is 1/3 of a mark)
1 wrong of 1 mark question lead to deduction of 1.25 marks
1 unattempted of 1 mark question lead to deduction of 1.5 marks
1 unattempted of 2 marks question lead to deduction of 3 marks
$$\therefore$$ Rank of student who scores 27.5 = 5
A three - digit number has digits in strictly descending order and divisible by 10. By changing the places of the digits a new three - digit number is constructed in such a way that the new number is divisible by 10. The difference between the original number and the new number is divisible by 40. How many numbers will satisfy all these conditions?
Since the three digit number is divisible by 10, then the unit's digit is 0
Let the three digit number = $$a b 0$$
After the digits are interchanged, the new number is also divisible by 10, thus only a and b are interchanged.
=> New number = $$b a 0$$
Difference between number is divisible by 40
=> $$(100a + 10b) - (100b + 10a) = 40 k$$ (k is constant)
=> $$90a - 90b = 90 (a - b) = 40 k$$
=> k = $$\frac{9\left(a-b\right)}{4}$$
Since k is a natural number (a-b) should be a multiple of 4
If a = 9 , the values of b that satisfies the given equation are 1,5
If a = 8 , the value of b that satisfies the given equation is 4
If a = 7 , the values of b that satisfies the given equation is 3
If a = 6 , the values of b that satisfies the given equation is 2
If a = 5 , the values of b that satisfies the given equation is 1
The number could be = 510,620,730,840,950, 910
Thus, there are 6 numbers that satisfy these conditions.
If the last 6 digits of [(M)! - (N)!] are 999000, which of the following option is not possible for (M) × (M - N)? Both (M) and (N) are positive integers and M > N. (M)! is factorial M.
None of the answers given are correct. The reasoning is as given below.
999000 is a multiple of 8 but not of 16. If N! is a multiple of 16, M! would also be a multiple of 16 and hence M!-N! would be a multiple of 16.
Hence, as M!-N! = 999000, it would imply that N! is a multiple of 8 and not of 16. Therefore, N is either 4 or 5. So, N! is either 24 or 120. So, it would imply that M! is either 999024 or 999120. Both of which are not factorials for any natural number.
Hence, the given question is wrong.
$$x, 17, 3x - y^{2} - 2$$, and $$3x + y^{2} - 30$$, are four consecutive terms of an increasing arithmetic sequence. The sum of the four number is divisible by:
The terms $$x, 17, 3x - y^{2} - 2$$ and $$3x + y^{2} - 30$$ are in A.P.
=> Common difference = $$d = 17 - x$$ ----------Eqn(I)
$$d = 3x - y^2 - 19$$ ----------Eqn(II)
$$d = 2y^2 - 28$$ ----------Eqn(III)
From eqn(I) & (II), => $$17 - x = 3x - y^2 - 19$$
=> $$4x - y^2 = 36$$ -------Eqn(IV)
From eqn(II) & (III), => $$3x - y^2 - 19 = 2y^2 - 28$$
=> $$x - y^2 = -3$$ ---------Eqn(V)
Solving eqn(IV) & (V), we get :
$$x = 13 , y^2 = 16$$
=> Terms are = $$13,17,21,25$$
$$\therefore$$ Sum = $$13+17+21+25 = 76$$, which is divided by 2. (among the given options)
Consider four natural numbers: x, y, x + y, and x - y. Two statements are provided below:
I. All four numbers are prime numbers.
II. The arithmetic mean of the numbers is greater than 4.
Which of the following statements would be sufficient to determine the sum of the four numbers?
Natural numbers = $$x , y , (x+y) , (x-y)$$
Statement I : As all the numbers are prime, therefore, either x or y has to be 2 because otherwise (x+y) cannot be prime.
Case 1 : If x = 2, then (x-y) cannot be prime
Case 2 : If y = 2, numbers = $$(x-2) , x , (x+2)$$
These numbers are prime, hence all possibility = 3,5,7
$$\therefore$$ Sum = 2+3+5+7 = 17
Using statement II, we cannot find the required sum, as no specific value of mean is given.
Thus, statement I alone is sufficient.
Two numbers, $$297_{B}$$ and $$792_{B}$$ , belong to base B number system. If the first number is a factor of the second number then the value of B is:
In Base B, $$297_B = 2B^2 + 9B + 7$$
and $$792_B = 7B^2 + 9B + 2$$
It is given that $$297_{B}$$ is a factor of $$792_{B}$$
=> $$\frac{7B^2 + 9B + 2}{2B^2 + 9B + 7}$$ must be an integer
=> $$\frac{(2B^2 + 9B + 7) + (5B^2 - 5)}{2B^2 + 9B + 7}$$
=> $$\frac{5B^2 - 5}{2B^2 + 9B + 7} + 1 = k$$
=> $$5B^2 - 5 = (2B^2 + 9B + 7) k$$ (where $$k$$ is factor)
Put $$k = 1$$
=> $$5B^2 - 5 = 2B^2 + 9B + 7$$
=> $$B^2 - 3B - 4 = 0$$
=> $$(B - 4) (B + 1) = 0$$
=> $$B = 4 , -1$$
Since, B is a base,so B must be greater than 9. Hence, it is not possible
Put $$k = 2$$
=> $$5B^2 - 5 = 4B^2 + 18B + 14$$
=> $$B^2 - 18B - 19 = 0$$
=> $$(B - 19) (B + 1) = 0$$
=> $$B = 19 , -1$$
$$\therefore B = 19$$
Read the following instruction carefully and answer the question that follows:
Expression $$\sum_{n=1}^{13}\frac{1}{n}$$ can also be written as $$\frac{x}{13!}$$ What would be the remainder if x is divided by 11?
Expression : $$\sum_{n=1}^{13}\frac{1}{n} = \frac{x}{13!}$$
=> $$\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + ...... + \frac{1}{13} = \frac{x}{13!}$$
=> $$x = \frac{13!}{1} + \frac{13!}{2} + \frac{13!}{3} + ......... + \frac{13!}{13}$$
Now, if $$x$$ is divided by 11
=> $$\frac{13! + \frac{13!}{2} + \frac{13!}{3} + ......... + \frac{13!}{13}}{11}$$
All terms are divisible by 11 except $$\frac{13!}{11}$$
$$\therefore$$ Remainder if x is divided by 11 = Remainder of $$\frac{13!}{11 \times 11}$$
= $$(10! \times 12 \times 13) \% 11$$
= $$(10 \times 1 \times 2) \% 11 = 20 \% 11 = 9$$
How many whole numbers between 100 and 800 contain the digit 2?
Let us analyse all numbers from 100 to 200 (both excluded).
Total numbers when unit digit is 2 = 10 {102,112,122, ..., 192}
Total numbers when tens digit is 2 = 10 {120,121,121, ..., 129}
We can see that the there is exactly one number{122} which have both tens and unit digit as 1. So total unique numbers from 100 to 200 (both excluded) which contain at least one 2 digit = 10 + 10 - 1 = 19
From 200 to 299 (both included), all numbers have at least one digit as 2. Total such number = 100.
From 300 to 400, 400 to 500, 500 to 600, 600 to 700 and 700 to 800 we will have 19 number for each pair with at least one 2 digit.
Therefore, total whole numbers between 100 and 800 contain the digit 2 = 100 + 19*6 = 214.
A number is interesting if on adding the sum of the digits of the number and the product of the digits of the number, the result is equal to the number. What fraction of numbers between 10 and 100 (both 10 and 100 included) is interesting?
As the number is between 10 and 100 and 100 cannot be the number we are looking for, we can assume the number to be of two-digits.
Let the number be xy.
According to the question, for the number to be interesting
x + y + xy = 10x + y
On solving, we get
xy = 9x
or, x (9 - y) = 0
x cannot be 0, because we need a number greater than or equal to 10.
So, 9 - y = 0
=> y = 9
For all the numbers whose unit digit is 9 will be an interesting number.
So, the numbers are 19, 29, 39, 49, .....99
There are 9 such numbers out of 91 total numbers between 10 and 100 including both.
Required fraction = $$\dfrac{9}{91}$$ = 0.0989
As this is not given in any of the options, the answer will be "none of the above".
Hence, option E is the correct answer.
p and q are positive numbers such that $$p^q = q^p$$, and $$q = 9p$$. The value of p is
$$p^q = q^p$$.
It has been given that $$q = 9p$$.
Substituting, we get,
$$p^{9p}=(9p)^p$$
$$(p^p)^9 = 9^p*p^p$$
=> $$(p^p)^8 = 9^p$$
$$p^{8p}=9^p$$
Raising the power to $$\frac{1}{p}$$ on both sides, we get,
$$p^8=9$$
$$p=\sqrt[8]{9}$$.
Therefore, option D is the right answer.
Consider the expression $$(xxx)_{b}=x^3$$, where b is the base, and x is any digit of base b. Find the value of b:
$$(xxx)_{b}=x^3$$
=> $$xb^2+xb+x = x^3$$
=> $$b^2+b+1=x^2$$
On substituting b=1,and b=2, we get $$x^2$$ as $$3$$, and $$7$$. Since $$3$$ and $$7$$ are not perfect squares, we can infer that no number satisfies the given condition. Therefore, option E is the right answer.
Please read the following sentences carefully:
I — 103 and 7 are the only prime factors of 1000027
II — $$\sqrt[6]{6!}>\sqrt[7]{7!}$$
III — If I travel one half of my journey at an average speed of x km/h, it will be impossible for me to attain an average speed of 2x km/h for the entire journey.
Let us evaluate the statements one by one:
I: 103 and 7 are the only prime factors of 1000027
On successively dividing 1000027 by 103 and 7, we get 1387 as the answer.
1387 is divisible by 19.
Therefore, statement I is false.
II: $$\sqrt[6]{6!}>\sqrt[7]{7!}$$
Raising the power to 42 on both sides, we get,
$$[6!]^7>[7!]^6$$
$$6!*[6!]^6 > 7^6*[6!]^6$$
$$7^6$$ is greater than $$6!$$.
Therefore, statement II is false.
III:It has been given that the person travels one-half of the journey at x kmph.
Let us assume the distance to be '2d'.
Let us assume the average speed to be 2d and check for the feasibility.
Let the speed at which the person travels the other half of the journey be y.
d/x + d/y = 2d/2x
d/x + d/y = d/x
=> d/y = 0 or y tends to infinity.
Therefore, such a scenario is not possible and hence, statement III is true.
Only statement III is true. Therefore, option C is the right answer.
Little Pika who is five and half years old has just learnt addition. However, he does not know how to carry. For example, he can add 14 and 5, but he does not know how to add 14 and 7. How many pairs of consecutive integers between 1000 and 2000 (both 1000 and 2000 included) can Little Pika add?
Little Pika can add (1000, 1001), (1001, 1002), (1002, 1003), (1003, 1004), (1004, 1005) and (1009, 1010).
Similarly, he can add (1010, 1011), (1011, 1012), (1012, 1013), (1013, 1014), (1014, 1015) and (1019, 1020).
Similarly, he can add (1020, 1021), (1021, 1022), (1022, 1023), (1023, 1024), (1024, 1025) and (1029, 1030).
Similarly, he can add (1030, 1031), (1031, 1032), (1032, 1033), (1033, 1034), (1034, 1035) and (1039, 1040).
Similarly, he can add (1040, 1041), (1041, 1042), (1042, 1043), (1043, 1044) , (1044, 1045) and (1049, 1050).
We can see that there are 30 cases when we have changed unit and tens digit. Now the hundreds digit can be anything from {0, 1, 2, 3, 4}.
Hence, total number of such pairs which Pika can add = 5*30 = 150.
He can also number of form, (1099, 1100), (1199, 1200), (1299, 1300), (1399, 1400) (1499, 1500) and (1999, 2000)
Therefore, we can say that Pika can add 150+6 = 156 numbers. Hence, option C is the correct answer.
The micromanometer in a certain factory can measure the pressure inside the gas chamber from 1 unit to 999999 units. Lately this instrument has not been working properly. The problem with the instrument is that it always skips the digit 5 and moves directly from 4 to 6. What is the actual pressure inside the gas chamber if the micromanometer displays 003016?
The meter skips all the numbers in which there is a 5
From, (1 to 99) -> 5 occurs 10 times in tens place and 10 times in units place (which also includes 55)
=> Total occurrence = $$10 + 10 - 1 = 19$$
Similarly from (100 to 199), from (200 to 299)..., from (400 to 499) , from (600 to 699),...from (900 to 999)
It occurs = $$8 \times 19 = 152$$
Now, from (500 to 599), there are 100 numbers, => micromanometer reading can change from 499 to 600.
Thus, total numbers skipped from (1 to 999) = $$19 + 152 + 100 = 271$$
Similarly, from (1000 to 1999) = 271
and from (2000 to 2999) = 271
Also, 3005 and 3015 are also skipped
=> Total number of skips = $$271 + 271 + 271 + 2 = 815$$
$$\therefore$$ Actual pressure = $$3016 - 815 = 2201$$
Ln a bank the account numbers are all 8 digit numbers, and they all start with the digit 2. So, an account number can be represented as $$2x_1x_2x_3x_4x_5x_6x_7$$. An account number is considered to be a ‘magic’ number if $$x_{1}x_{2}x_{3}$$ is exactly the same as $$x_{4}x_{5}x_{6}$$ or $$x_{5}x_{6}x_{7}$$ or both. $$X_{i}$$ can take values from 0 to 9, but 2 followed by seven $$0_{s}$$ is not a valid account number. What is the maximum possible number of customers having a ‘magic’ account number?
Account number = $$2 x_1x_2x_3x_4x_5x_6x_7$$
Case 1 : $$x_1x_2x_3$$ is exactly same as $$x_4x_5x_6$$
=> $$x_1x_2x_3 = x_4x_5x_6 = 000$$ and $$x_7$$ = 1 to 9 (as '20000000' is not valid)
=> 9 possibilities
Now, $$x_1x_2x_3 = x_4x_5x_6$$ = 001 to 999 and $$x_7$$ = 0 to 9
=> $$999 \times 10 = 9990$$ possibilities.
Case 2 : $$x_1x_2x_3$$ is exactly same as $$x_5x_6x_7$$
=> $$x_1x_2x_3 = x_5x_6x_7 = 000$$ and $$x_4$$ = 1 to 9 (as '20000000' is not valid)
=> 9 possibilities
Now, $$x_1x_2x_3 = x_5x_6x_7$$ = 001 to 999 and $$x_4$$ = 0 to 9
=> $$999 \times 10 = 9990$$ possibilities.
Subtracting common possibilities in above cases.
=> $$x_4x_5x_6 = x_5x_6x_7$$
=> $$x_4 = x_5 = x_6 = x_7$$
Except '0', the possibilities are 1111,2222,....,9999 => 9 possibilities.
$$\therefore$$ Maximum possible number of customers having a ‘magic’ account number
= $$(9 + 9990) + (9 + 9990) - (9)$$
= $$19989$$
If x and y are real numbers, then the minimum value of $$x^{2}+ 4xy+ 6y^{2}-4y+ 4$$ is
Expression : $$x^{2}+ 4xy+ 6y^{2}-4y+ 4$$
= $$(x^2 + 4xy + 4y^2) + (2y^2 - 4y + 4)$$
= $$(x + 2y)^2 + \frac{1}{2} (4y^2 - 8y + 8)$$
= $$(x + 2y)^2 + \frac{1}{2} [(2y)^2 - 2 (2y) (2) + (2)^2 + 4]$$
= $$(x + 2y)^2 + \frac{1}{2} [(2y - 2)^2 + 4]$$
= $$(x + 2y)^2 + \frac{1}{2} (2y - 2)^2 + 2$$
Since, $$x$$ and $$y$$ are real, => Min value of $$(x + 2y)^2 = 0$$
Minimum value of $$(2y - 2)^2 = 0$$
$$\therefore$$ Minimum value of expression = $$0 + 0 + 2 = 2$$
Let X be a four - digit positive integer such that the unit digit of X is prime and
the product of all digits of X is also prime. How many such integers are possible?
Given that unit digit of $$X$$ and product of all 4 digits of $$X$$ are prime.
The product of two numbers to be prime is possible only when one of the numbers is prime and the other is ‘1’.
The possibilities for the prime unit digits are - 2,3,5,7 (Note that 1 is not a prime number)
Hence the possibility of remaining 3 digits, considering the product of all 4 digits to be prime is ‘111’ only.
Hence all the possible numbers are 1112,1113,1115,1117
$$\therefore$$ Total 4 integers are possible.
Yes, Number Systems is one of the most important topics in XAT Quantitative Ability. It forms the foundation for many concepts and frequently appears in the exam through direct and application-based questions.
XAT may include questions on divisibility rules, factors and multiples, HCF and LCM, remainders, cyclicity, units digits, and base system.
Candidates should focus on understanding fundamental concepts such as divisibility, prime numbers, factors, remainders, and number properties. Regular practice of topic-wise questions and mock tests is essential for improving accuracy.
The difficulty level can range from easy to medium. While some questions test basic concepts, others require creative application of concepts and properties.
Cracku's XAT Number Systems Questions are designed according to the latest XAT exam pattern and difficulty level. They provide topic-wise practice, detailed solutions, shortcut techniques, and performance analysis to help aspirants strengthen their fundamentals and solve Number Systems questions efficiently.
You can practice XAT Number Systems questions through topic-wise question banks, previous year papers, sectional tests, and mock tests that provide detailed explanations and multiple solving approaches.
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