Which of the following is a multiple of 6^2?

For a number to be a multiple of 6^2 it has to divide by the factors of 6^2. Accordingly we can immediately eliminate options A and B as they are odd numbers (and thus clearly don't divide by 2, 6 etc). We can also eliminate option E as it doesn't divide by 6 (the sum of the number is even but it doesn't divide by 3). So we left with options C and D. As option C doesn't divide by 9 up to the some of its digit number, D is the correct answer as the sum of its digit is 36 and thus divide by 9.

Answer: D

sashiim20 - How did you figure out quickly how many 2's and 3's each of those big numbers has?

Just check if number is divisible by 2 and 3 and also divisible by 4 and 9... We just need to find if the number has at least two 2 (ie;4) and two 3 (ie; 9)... this is enough to prove number is divisible by 36... we don't need to do full division...

we can also check the divisibility of 4 (last 2 digits divisible by 4 which is same as 2^2) and 9 (sum of digits divisible by 9 which is same as 3^2)

My question is not whether the number is divisible by two or three but how many 2's and 3's each of the numbers has (up to prime factorization). In big numbers it seems to be time consuming.

You don't need to do prime factorization... Please check the post above i have mentioned, divisibility of 2,3, 4 and 9 is enough (Or in other words, check if 2,3,4 and 9 are factors of number)... divisibility of 4 will automatically prove number is divisible by 2.

Divisibility of 3 - sum of numbers are divisible by 3
Divisibility of 4 - last 2 digits are divisible by 4
Divisibility of 9 - sum of numbers are divisible by 9

From the given options - A, B, E are out.
Only left options are C and D...
Following the divisibility rules we can figure out which is divisible by 36.
C is divisible by 3 but not by 9. Hence Eliminated. Answer D...

Hope its clear now.. Let me know if you have any other doubts...

Thank you

\(6^2 or 36\) when multiplied by any number, cannot end with a 3 or 5.
So straight away A and B are out.

Of Options C,D and E - numbers 52224 and 92004 are not divisible by 9,
which is a necessary condition for a number to be divisible by \(36(2^2 * 3^2)\)

We are left with Option D(89892), which is the correct answer.

Options A, B, E fail the divisibility test of numbers 2 and 3, and hence are out of contention. Out of option C and D, only D is divisible by 9 (because \(6^2\) can also be written as \(9 * 4\)). Answer is Option D.

6^2 = 3^2 * 2^2 = 9 * 4
So to check which of the following is multiple of 6^2 ., we need to check which one is divisible by both 4 and 9.

Divisibility by 4 is easier than 9 , So first eliminate the choices which are not divisible by 4.
Check the last 2 terms of the numbers.
Clearly A and B are not divisible by 4

Now lets check divisibility of C D E by 9
C. 5+2+2+2+4 = 15 not divisible by 9.

D. 8+9+8+9+2 = 36 is divisible by 9.

E. 9+2+0+4+4 = 19 is not divisible by 9.

So, Answer D

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