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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\'manpreet singh
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Now 96 is the product of 3 single digit numbers.'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.
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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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.
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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