If N is a positive integer, what is the tens digit of N?

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Source: — Data Sufficiency |

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Tens digit will be the penultimate digit (2nd-to-last digit)

Statement (1) tells us N is a multiple of 25.
If N = 25, then tens digit is "2".
If N = 50, then tens digit it "5".
We're getting different answers to the original question while still satisfying this statement. So Statement (1) is INSUFFICIENT.

Statement (2) tells us N is divisible by 16.
If N = 16, then tens digit is "1".
If N = 32, then tens digit is "3".
Again we're getting different answers to the original question while still satisfying this statement. So statement (2) is INSUFFICIENT.

Now we have to combine statements (things get more interesting here), so N must be a multiple of both 25 and 16. If we start listing out multiples of 25 until we happen to come upon some that are multiples of 16, this would take quite some time (especially since no calculators are allowed).

So this is where PRIME FACTORIZATION comes in handy. 25 = 5x5, and 16 = 2x2x2x2. They don't share any prime factors in common. So the LCM of 25 and 16 would have to be 25x16 itself. Might be easier to do this in stages: 25x4 = 100, and 100x4 = 400. So "N" must be a multiple of 400.

And what is the tens digit of any multiple of 400? "0". We have a definite answer to the original question now, so together the statements are SUFFICIENT.

Answer is (C).
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Statement 1: N is divisible by 5.
This means that N is a multiple of 25. e.g 25, 50, 75, etc. and the tens digit vary. hence, statement 1 is NOT SUFFICIENT.

Statement 2: N is divisible by 16.
This means that N is a multiple of 16. e.g 16, 32, 48, etc and the tens digit vary. Hence, statement 2 is NOT SUFFICIENT

Combining both statement together:
To find a number that is divisible by both 25 and 16, find the LCM or product of 25 and 16, 25*16=400
Therefore, N is a multiple of 400. e.g 400, 800, 1200, etc in all cases, the tens digits of N=0.
Then, both statements combined together are SUFFICIENT. Answer = option C