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Odd even-

by \'manpreet singh » Sun Nov 18, 2012 11:08 pm
The product of the units digit, the tens digit, and the
hundreds digit of the positive integer m is 96. What is
the units digit of m ?
(1) m is odd.
(2) The hundreds digit of m is 8.
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by Anurag@Gurome » Sun Nov 18, 2012 11:12 pm
'manpreet singh wrote:The product of the units digit, the tens digit, and the
hundreds digit of the positive integer m is 96. What is
the units digit of m ?
(1) m is odd.
(2) The hundreds digit of m is 8.
Now 96 is the product of 3 single digit numbers.
96 = 12 * 8 = 3*4*8 = 2*6*8 =4*4*6.
So the units digit, tens digit and hundreds digit can be either 3, 4 and 8 or 2, 6 or 8 or 4, 4, 6 in any order.
Consider first statement (1) alone.
m is odd.
So the units digit of m is odd.
Only possibility is 3 from the combination of 3, 4 and 8; SUFFICIENT.

Next consider (2) alone.
The hundreds digit is 8.
So any of the 2 combinations 3, 4, 8 or 2, 6, 8 are possible.
So if hundreds digit is 8, units and tens digit can be any of 2, 3, 4 or 6.
Since we do not get a unique answer, (2) alone is NOT sufficient.

The correct answer is A.
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by Bill@VeritasPrep » Sun Nov 18, 2012 11:15 pm
96 can be broken down to its primes, 2^5*3, which we can then combine into various single-digit integers: 3*4*8 and 2*6*8, for example.

Starting with Statement 2, we can see that it won't be sufficient. In either example, we can come up with multiple units digits: 843 or 834, and 862 or 826. Insufficient.

Statement 1 is enough. By knowing that m is odd, and knowing that the only odd factor of 96 is 3, we know m must end in a 3. It could be 483 or 843, but it will always end in 3. Sufficient.
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by \'manpreet singh » Mon Nov 19, 2012 12:31 am
Thanks for the quick response!
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