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Question 1
If $$10^{68}$$ is divided by 13, the remainder is
CAT Remainders Questions
Solution
There are multiple ways of solving such questions involving remainders; one easy way is to look for a power of numerator that leaves a remainder of 1 or 01 when divided by the denominator.
In this instance, 1000, when divided by 13, leaves a remainder of -1
We can rewrite the numerator as $$\frac{10^{66}\times\ 100}{13}$$
The remainder would be $$\left[\frac{10^{66}}{13}\right]_R\times\ \left[\frac{100}{13}\right]_R$$
$$\left(-1\right)^{22}\times\ 9$$
9
Therefore, Option C is the correct answer.
Question 2
When $$10^{100}$$is divided by 7, the remainder is
Solution
To find the value of $$10^{100}mod\left(7\right)$$
When 10 is divided by 7, it leaves a remainder 3, so the above equation can be written as,
$$3^{100}mod\left(7\right)$$
Now looking at the cyclicality of powers of 3 when divided by 7,
$$3^1mod 7=3$$
$$3^2mod 7=2$$
$$3^3mod 7=6$$
$$3^4mod 7=4$$
$$3^5mod 7=5$$
$$3^6mod 7=1$$
From this calculation, it is evident that the powers of 3 modulo 7 repeat every 6 steps. This forms a cycle: 3, 2, 6, 4, 5, 1
$$3^{100}=\left(3^6\right)^{16}\times\ \left(3^4\right)$$
Since $$3^6mod 7=1$$
We just need to consider $$3^4mod 7$$ which equals 4
Hence the answer is 4.
Question 3
When $$3^{333}$$ is divided by 11, the remainder is
Solution
There are multiple ways of solving these sorts of questions. One method is to look for powers of the term in the numerator that leave a remainder of 1 or -1 when divided by the denominator.
Noting down the powers of 3, 3, 9, 27, 81, 243
243 is one such number, 242 is multiple of 11 (11 times 22), hence 243 will leave a remainder of 1 when divided by 11.
243 is 3 raised to power 5; we can rewrite the given term as $$\dfrac{3^{330}\times\ 3^3}{11}$$
The overall remainder will be $$\left[\dfrac{3^{330}}{11}\right]_R\times\ \left[\dfrac{3^3}{11}\right]_R$$
$$\left[\dfrac{3^{5\times\ 66}}{11}\right]_R\times\ \left[\dfrac{3^3}{11}\right]_R$$
$$\left[\dfrac{243^{66}}{11}\right]_R\times\ \left[\dfrac{3^3}{11}\right]_R$$
$$1^{66}\times\ \left[\dfrac{27}{11}\right]_R$$
$$1\times\ 5$$
$$5$$
Therefore, Option A is the correct answer.
Question 1
How many of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7?
Solution
The number of multiples of 2 between 1 and 120 = 60
The number of multiples of 5 between 1 and 120 which are not multiples of 2 = 12
The number of multiples of 7 between 1 and 120 which are not multiples of 2 and 5 = 7
Hence, number of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7 = 120 - 60 - 12 - 7 = 41
Question 1
The number of integers x such that $$0.25 \leq 2^x \leq 200$$ and $$2^x + 2$$ is perfectly divisible by either 3 or 4, is
Correct Answer :
5
Solution
At $$x = 0, 2^x = 1$$ which is in the given range [0.25, 200]
$$2^x + 2$$ = 1 + 2 = 3 Which is divisible by 3. Hence, x = 0 is one possible solution.
At $$x = 1, 2^x = 2$$ which is in the given range [0.25, 200]
$$2^x + 2$$ = 2 + 2 = 3 Which is divisible by 4. Hence, x = 1 is one possible solution.
At $$x = 2, 2^x = 4$$ which is in the given range [0.25, 200]
$$2^x + 2$$ = 4 + 2 = 6 Which is divisible by 3. Hence, x = 2 is one possible solution.
At $$x = 3, 2^x = 8$$ which is in the given range [0.25, 200]
$$2^x + 2$$ = 8 + 2 = 3 Which is not divisible by 3 or 4. Hence, x = 3 can't be a solution.
At $$x = 4, 2^x = 16$$ which is in the given range [0.25, 200]
$$2^x + 2$$ = 16 + 2 = 18 Which is divisible by 3. Hence, x = 4 is one possible solution.
At $$x = 5, 2^x = 32$$ which is in the given range [0.25, 200]
$$2^x + 2$$ = 32 + 2 = 34 Which is not divisible by 3 or 4. Hence, x = 5 can't be a solution.
At $$x = 6, 2^x = 64$$ which is in the given range [0.25, 200]
$$2^x + 2$$ = 64 + 2 = 66 Which is divisible by 3. Hence, x = 6 is one possible solution.
At $$x = 7, 2^x = 128$$ which is in the given range [0.25, 200]
$$2^x + 2$$ = 128 + 2 = 130 Which is not divisible by 3 or 4. Hence, x = 7 can't be a solution.
At $$x = 8, 2^x = 256$$ which is not in the given range [0.25, 200]. Hence, x can't take any value greater than 7.
Therefore, all possible values of x = {0,1,2,4,6}. Hence, we can say that 'x' can take 5 different integer values.
Question 1
Let $$n!=1*2*3* ...*n$$ for integer $$n \geq 1$$.
If $$p = 1!+(2*2!)+(3*3!)+... +(10*10!)$$, then $$p+2$$ when divided by 11! leaves a remainder of
Solution
According to given condiiton we have p = (1 × 1!) + (2 × 2!) + (3 × 3!) + (4 × 4!) + … + (10 × 10!) . So n × n! = [(n + 1) - 1] × n! = (n + 1)! - n!. So equation becomes p = 2! - 1! + 3! - 2! + 4! - 3! + 5! - 4! +… + 11! - 10!. So p = 11! - 1! = 11! - 1. p + 2 = 11! + 1 .So when it is divided by 11! gives a remainder of 1. Hence, option 4.
Question 2
If x = $$(16^3 + 17^3+ 18^3+ 19^3 )$$, then x divided by 70 leaves a remainder of
Solution
We know that x = $$16^3 + 17^3 + 18^3 + 19^3 = (16^3 + 19^3) + (17^3 + 18^3)$$
= $$(16 + 19)(16^2 - 16 * 19 + 19^2) + (17 + 18)(17^2 - 17 * 18 + 18^2)$$ = 35 × odd + 35 × odd = 35 × even = 35 × (2k)
=> x = 70k
=> Remainder when divided by 70 is 0.
Question 1
The remainder, when $$(15^{23} + 23^{23})$$ is divided by 19, is
Solution
The remainder when $$15^{23}$$ is divided by 19 equals $$(-4)^{23}$$
The remainder when $$23^{23}$$ is divided by 19 equals $$4^{23}$$
So, the sum of the two equals$$(-4)^{23}+(4)^{23}=0$$
Question 1
How many even integers n, where $$100 \leq n \leq 200$$ , are divisible neither by seven nor by nine?
Solution
Between 100 and 200 both included there are 51 even nos. There are 7 even nos which are divisible by 7 and 6 nos which are divisible by 9 and 1 no divisible by both. hence in total 51 - (7+6-1) = 39
There is one more method through which we can find the answer. Since we have to find even numbers, consider the numbers which are divisible by 14, 18 and 126 between 100 and 200. These are 7, 6 and 1 respectively.
Question 2
The number of positive integers n in the range $$12 \leq n \leq 40$$ such that the product (n -1)*(n - 2)*…*3*2*1 is not divisible by n is
Solution
positive integers n in the range $$12 \leq n \leq 40$$ such that the product (n -1)*(n - 2)*…*3*2*1 is not divisible by n, implies that n should be a prime no. So there are 7 prime nos. in given range. Hence option B.
Question 1
When $$2^{256}$$ is divided by 17, the remainder would be
Solution
$$2^4 = 16 = -1$$ (mod $$17$$)
So, $$2^{256} = (-1)^{64} $$(mod $$17$$)
$$= 1$$ (mod $$17$$)
Hence, the answer is 1. Option a).
Question 2
After the division of a number successively by 3, 4 and 7, the remainders obtained are 2, 1 and 4 respectively. What will be the remainder if 84 divides the same number?
Solution
Since after division of a number successively by 3, 4 and 7, the remainders obtained are 2, 1 and 4 respectively, the number is of form ((((4*4)+1)*3)+2)k = 53K
Let k = 1; the number becomes 53
If it is divided by 84, the remainder is 53.
Option d) is the correct answer.
Alternative Solution.
Consider only for 3 and 4 and the remainders are 2 and 1 respectively.
So 5 is the first number to satisfy both the conditions. The number will be of the form 12k+5. Put different integral values of k to find whether it will leave remainder 5 when divided by 7. So the first number to satisfy such condition is 48x4+5= 53
Question 3
$$7^{6n} - 6^{6n}$$, where n is an integer > 0, is divisible by
Solution
Consider n=1 we have $$7^{6} - 6^{6}$$ which is = $$(7^{3} + 6^{3})(7^{3} - 6^{3})$$ = 13 * 127 * 43 which is divisible by all the 3 options.
Option d) is the correct answer.
Question 1
Let $$b$$ be a positive integer and $$a = b^2 - b$$. If $$b \geq 4$$ , then $$a^2 - 2a$$ is divisible by
Solution
We know that a=$$b^2-b$$.
So$$a^2-a$$ = b($$b^3-2b^2-b+2$$) . = (b - 2)(b - 1)( b)(b + 1)
The above given is a product of 4 consecutive numbers with the lowest number of the product being 2(given b >= 4)
In any set of four consecutive numbers, one of the numbers would be divisible by 3 and there would be two even numbers with the minimum value of the pair being (2,4).
Thus, for any value of b >=4, $$a^2-4$$ would be divisible by 3 x 2 x 4 = 24.
Thus, option C is the right choice. Options A and B are definitely wrong as a set of four consecutive numbers need not always include a multiple of 5 eg:(6,7,8,9)
Question 2
Let n be the number of different five-digit numbers, divisible by 4 with the digits 1, 2, 3, 4, 5 and 6, no digit being repeated in the numbers. What is the value of n?
Solution
To be divisible by 4 , last 2 digits of the 5 digit no. should be divisible by 4 . So possibilities are 12,16,32,64,24,36,52,56 which are 8 in number. Remaining 3 digits out of 4 can be selected in $$^4C_3 $$ ways and further can be arranged in 3! ways . So in total = 8*4*6 = 192
Question 1
Let N = 1421 * 1423 * 1425. What is the remainder when N is divided by 12?
Solution
The numbers 1421, 1423 and 1425 when divided by 12 give remainder 5, 7 and 9 respectively.
5*7*9 mod 12 = 11 * 9 mod 12 = 99 mod 12 = 3
Question 2
The integers 34041 and 32506 when divided by a three-digit integer n leave the same remainder. What is n?
Solution
The difference of the numbers = 34041 - 32506 = 1535
The number that divides both these numbers must be a factor of 1535.
307 is the only 3 digit integer that divides 1535.
Question 3
Let N = $$55^3 + 17^3 - 72^3$$. N is divisible by:
Solution
$$55^3 + 17^3 - 72^3$$ = $$(55-72)k + 17^3$$. This is divisible by 17
Remainder when $$55^3$$ is divided by 3 = 1
Remainder when $$17^3$$ is divided by 3 = -1
Remainder when $$72^3$$ is divided by 3 = 0
So, $$55^3 + 17^3 - 72^3$$ is divisible by 3
So, the answer is d) 3 and 17
Question 1
The remainder when $$7^{84}$$ is divided by $$342$$ is :
Solution
$$7^3$$ = 343
$$7^{84}$$ = $$(7^3)^{28}$$ = $$343^{28}$$
$$343^{28}$$ mod 342 = $$1^{28}$$ mod 342 = 1
Question 1
A certain number, when divided by 899, leaves a remainder 63. Find the remainder when the same number is divided by 29.
Solution
Let's say N is our number
N = (899K + 63) or N = ($$29 \times 31$$K) + 63
So when it is divided by 29, remainder will be $$\frac{63}{29}$$ = 5
Question 2
A number is formed by writing first 54 natural numbers next to each other as 12345678910111213 ... Find the remainder when this number is divided by 8.
Solution
For a number to be divisible by 8, last 3 digits must be divisible by 8.
Last 3 digits of this number are 354.
354 mod 8 = 2
Hence, 2 is the remainder.
Question 1
If m and n are integers divisible by 5, which of the following is not necessarily true?
Solution
Let's say m=5k and n=5t
So m-n = 5(k-t) will be divisible by 5.
$$m^2 - n^2 = 25(k^2 - t^2)$$ will be divisible by 5.
$$m+n = 5(k+t)$$ will be divisible by 5 but not necessarily with 10.
Question 1
Find the minimum integral value of n such that the division $$\frac{55n}{124}$$ leaves no remainder.
Solution
As 55 and 124 don't have any common factor, and n has to be a minimum integer, Hence, it should be 124 only. So that given equation won't have a remainder.
Question 2
Let k be a positive integer such that k+4 is divisible by 7. Then the smallest positive integer n, greater than 2, such that k+2n is divisible by 7 equals
Solution
let's say k+4 = 7m
k = 7m-4
Now for k+2n or 7m+(2n-4) is also multiple of 7.
or 2n-4 should be a multiple of 7
So 2n-4 = 7p
or 2n = 7p+4
For p=2; n=9 (p cannot be 1 as n is an integer )
Question 3
A positive integer is said to be a prime number if it is not divisible by any positive integer other than itself and 1. Let $$p$$ be a prime number greater than 5. Then $$(p^2-1)$$ is
Solution
Let the Prime number be 6n+1.
So $$(p^2 - 1)$$ = 6n(6n+2) = 12n(3n+1)
For any value of n , n(3n+1) will have a factor of 2
Hence given equation will be always be divisible by 24
Question 1
The remainder when $$2^{60}$$ is divided by 5 equals
Solution
$$2^{60}$$ or $$4^{30}$$ when divided by 5
So according to remainder theorem
remainder will be $$(-1)^{30}$$ = 1.
