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Which is the least number that should be added to 1321 to make it completely divisible by 29 ?
1321 = 29 x 45 + 16
Since 16 is the remainder when 29 divides 1321, 29-16 = 13 should be added to 1321 to make it divisible by 29.
Find the number of divisors of 10800
$$10800 = 2^4 * 5^2 * 3^3$$
Number of divisors = (4+1) (2+1) (3 + 1) = 60
How many numbers up to 800 are divisible by 24?
Quotient when 800 is divided by 24 is 33.
There are 33 numbers up to 800 divisible by 24
What is the unit digit in (476 x 198 x 359 x 242)?
Unit digit in 476 x 198 x 359 x 242
= unit digit in 6 x 8 x 9 x 2 = 4
Find the LCM of the fractions $$\frac{3}{7}, \frac{4}{3}$$ and $$\frac{2}{9}$$.
LCM of the fractions = (LCM of the numerator/ HCF of the denominator)
LCM of 3, 4 and 2 = 3*4 = 12
HCF of 7, 3, 9 = 1
Therefore, LCM of the given fractions = 12/1 = 12.
What is the greatest number which divides the numbers 595, 853 and 486 and leaves remainders 3, 2 and 5 respectively?
Let the required number be x.
When numbers 595, 853 and 486 is divided by x remainders are 3, 2 and 5 respectively.
So 595 - 3 = 592, 853 - 2 = 851 and 486 - 5 = 481 must be divisible by x
Greatest value of ‘x’ = HCF(592, 851, 481) = 37
Hence, option D is the right answer.
P is the smallest number which leaves a remainder of 3 in each case when divided by 4,5,6 and 8. What is the remainder when P is divided by 9 ?
P will be in the form of LCM of n(4,5,6,8)+3
LCM of 4,5,6 and 8 is 120
Therefore required number is 120+3=123
When 123 is divided by 9 remainder is 6
What is the product of all the prime numbers between 0 to 100 ?
Prime numbers from 0 to 100 are 2,3,5,7,11,........
The multiplication will be very complicated. So, the question requires you to find number of zeros at the end.
Product = 2*3*5*7.....
No. of zeros at end = highest power of 10, which is 1 in this case(since 2,5 occur only once each and no higher powers of 2 and 5 occur).
So, the answer is option D as it has only 1 zero at the end.
How many even integers 'n', where, $$ 100 \leq n \leq 200 $$, are divisible neither by 7 nor by 9?
Total even integers between 100 and 200 = $$\frac{200-100}{2}\ \ +\ \ 1\ =\ 51$$
Even integers divisible by 7 = Number of multiple of 14.
There are 7 integers which are of form 14k(112,126,...) between 100-200
Even integers divisible by 9 = Number of multiple of 18.
There are 6 integers which are of form 18k(108,126,...) between 100-200
Even integers divisible by both 7 and 9 = No. of multiple of LCM(14,18) = No. of multiple of 126k = 1
So, Required answer = 51-(7+6)+1 = 39.
Option C is the answer.
The product of :
17 X 17 X 18 X 18 X 19 X 19
shall be fully divisible by which of the following number ?
17*17*18*18*19*19 = $$2^2\times\ 3^4\times\ 17^2\times\ 19^2$$
971 is not divisible by 2 or 3 or 17 or 19
173,23,5813 are also not divisible by 2 or 3 or 17 or 19
646=2*19*17. So, it can divide the given no.
Option C is the answer.
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 the no. be x
LCM(x,990) = 6930
6930=$$2\times3^2\times5\times7\times11$$
990=$$2\times3^2\times5\times11$$
Factor of 7 is missing in 990, so x has to be multiple of 7.
x=7a
GCD(x,550)=110
x=110b
x=7a=110b. So, x=770k
Least value of x is 770.
Sum of digits of 770=7+7+0=14
Option B is the answer.
Find the remainder when :
(1719 X 1721 X 1723 X 1725 X 1727) is divided by 18?
R($$\ \frac{\ 1719*1721*1723*1725*1727}{18}$$) = R($$\frac{9\times\ 11\times\ 13\times\ 15\times\ 17\ }{18}$$) = 9
What number shall you get in the unit digit on solving the following problem :
$$\left\{ (6374)^{1793}\times(625)^{317}\times (341^{491}) \right\}$$
Unit digit of $$6374^{1793}\times625^{317}\times341^{491}\ $$ = Unit digit of $$4^{1793}\times5^{317}\times1^{491}\ $$ = unit digit of 4*5*1 = 0
Option A is the answer.
What is the sum of the factors of the number 3240?
We see that the number 3240 can be factorised into: $$3240=2^3\times\ 3^4\times\ 5$$
We use the sum of factors formula:
$$\frac{\left(2^{3+1}-1\right)}{2-1}\times\ \frac{\left(3^{4+1}-1\right)}{3-1}\times\ \frac{\left(5^{1+1}-1\right)}{5-1}$$
$$15\times\ 121\times6$$
10890
What is the maximum power of 24 that can divide 124!?
24 = $$\ 2^3\times3$$
We need to see the maximum power of both.
Maximum power of 2 in 124! = $$\left[\dfrac{124}{2}\right]+\left[\dfrac{124}{4}\right]+\left[\dfrac{124}{8}\right]+\left[\dfrac{124}{16}\right]+\left[\dfrac{124}{32}\right]+\left[\dfrac{124}{64}\right]$$ = 119
So, the maximum power of $$2^3$$ in 124! is $$\left[\dfrac{119}{3}\right]=39$$
Now, let's check for 3: $$\left[\dfrac{124}{3}\right]+\left[\dfrac{124}{9}\right]+\left[\dfrac{124}{27}\right]+\left[\dfrac{124}{81}\right]=41+13+4+1=59$$
Since the maximum power that can be common for both $$2^3$$ and 3 is 39, the maximum power for 24 will also be 39.
What is the maximum possible sum of two numbers whose LCM is 1296?
The maximum sum is possible when two numbers equals to the LCM.
So, the maximum sum = 1296+1296 = 2592.
What is the minimum sum of two natural numbers whose lowest common multiple is 729?
We can see that 729 is 9 times 81, meaning that $$729=9^3=3^6$$
We want to find the minimum sum of the two numbers, in order to get the minimum sum, we would want the numbers to be as close to each other a possible, which would be when both are equal to $$3^3=27$$, giving the sum to be 54
But in that case, 729 would not be the lowest common multiple; it would be 27 itself.
We see that since the 729 is a prime number multiplied repeatedly, the only number that can have LCM as 729 are 1 and 729 itself.
Therefore, the minimum sum would be 730.
Hence, Option E is the correct answer.
What would be the last two digits of the value of $$217^{432}$$?
When finding the last two numbers, we would want to change the number so that it has 1 as its unit digit.
This can be done if we take $$217^{432}=\left(217^4\right)^{108}$$
Since the 7 raised to power 4 ends in 1, following the cycle of 7, 9, 3, 1,7....
To calculate the value of $$217^4$$, we would only have to consider the value of $$17^4$$, since the 200 would not impact the last two digits.
17 squared is 289. to find the last two digits of 17^4, we only have to find the last two digits of 89 times 89, which would be 21
So we essentially have a number of the form $$\left(....21\right)^{108}$$
When the number is of the form $$\left(...a1\right)^n$$, the last two digits of the value are $$\left(a\times\ n\right)1$$
Hence, the last two digits of $$\left(...21\right)^{108}$$ would be 61
Therefore, Option B is the correct answer.
Find the remainder when $$ 47^{121}$$ is divided by 31.
$$47^{121}$$ can be written as $$47^{30}\times\ 47^{30}\times\ 47^{30}\times\ 47^{30}\times\ 47^{1}\times\ 47$$
We know by Fermat's theorem that
when $$a^{p-1}$$ is divide by $$p$$ where $$p$$ is prime and both $$a$$ and $$p$$ are co-prime, the remainder is 1.
Using the theorem in our expression, we get
$$47^{30}mod31=1$$
So the remainder when $$47^{121}$$ is divided by 31 is
$$1\times\ 1\times\ 1\times\ 1\times\ 1\times\ 47mod31=16$$
Hence, the answer is 16.
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