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Bunuel
The number N is 5H72, the hundred's digit being represented by H. What is the value of H?

(1) N is divisible by 4.
(2) N is divisible by 9.

Is the Statement I correct? How can N be divisible by 4 if the sum of last two digits is not divisible by 4?
I know even if the sum of last two digits was divisible by 4 the answer will still be insufficient. But I am not sure if the statement I is correct.

An integer to be divisible by 4, the last two digits must be divisible by 4, not the sum of the last two digits. For example, 123356 is divisible by 4 because 56 is divisible by 4.
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Bunuel
The number N is 5H72, the hundred's digit being represented by H. What is the value of H?

(1) N is divisible by 4.
(2) N is divisible by 9.

(1) For a number to be divisible by 4, its last two digits must be divisible by 4. So in case of 5H72, since last two digits are divisible by 4 (72 is a multiple of 4), 5H72 WILL be divisible by 4, irrespective of the value of H (meaning H can take any value from 0 to 9 doesnt matter). So Insufficient.

(2) For a number to be divisible by 9, its sum of digits must be divisible by 9. So 5+H+7+2 = 14+H must be divisible by 9. We can see that H can thus take only one value of '4', which will make sum of digits as 18, and thus divisible by 9. Sufficient.

Hence B answer
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Here's an interesting variant of this problem that I thought of while solving this one!

Quote:
The number N is 6H12, the hundred's digit being represented by H. What is the value of H?

(1) N is divisible by 4.
(2) N is divisible by 9.

In this case, the answer is E, even though all I did was change the number slightly... :)
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ccooley
Here's an interesting variant of this problem that I thought of while solving this one!

Quote:
The number N is 6H12, the hundred's digit being represented by H. What is the value of H?

(1) N is divisible by 4.
(2) N is divisible by 9.

In this case, the answer is E, even though all I did was change the number slightly... :)

Thank you. Posted it HERE.
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A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

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