xhimi wrote:
If N is a positive three-digit number that is greater than 200, and each digit of N is a factor of N itself, what is the value of N?
(1) The tens digit of N is 5.
(2) The units digit of N is 5.
Official Solution (Credit: Manhattan Prep)
(1) SUFFICIENT: If 5 is a digit of N, then N is a multiple of 5. All multiples of 5 end with a units digit of either 5 or 0. However, the units digit of N cannot be 0, since 0 is not a factor of any number. Therefore, the units digit of N must also be 5. Therefore N has the form _55, with only the hundreds digit left to consider.
Consider the possible cases for the hundreds digit:
It can’t be 1, since N > 200.
It can’t be an even number, because N is odd (it ends in 5) and thus doesn’t have any even factors.
If it were 3, then N would be 355—but that doesn’t work, since 355 is not a multiple of 3. (Check: 3 + 5 + 5 = 13, which is not divisible by 3.)
If it were 5, then N would be 555. This is a possible value.
If it were 7, then N would be 755—but that doesn’t work, since 755 is not a multiple of 7. (Check: 700 + 55 = 755. 700 is divisible by 7 but 55 is not, so the whole thing is not.)
If it were 9, then N would be 955—but that doesn’t work, since 955 is not a multiple of 9. (Check: 9 + 5 + 5 = 19, which is not divisible by 9.)
The only possible value for N is 555, so statement 1 is sufficient.
(2) INSUFFICIENT: If the units digit of N is 5, then 5 must be a factor of N. All integers ending in 5 are multiples of 5, though, so this fact doesn’t narrow the possibilities any further.
N is a multiple of 5 and any number is a multiple of 1. Using only these digits, try to formulate two numbers that satisfy the statement. Both 515 and 555 satisfy statement (2) so it is not sufficient to answer the question.
The correct answer is (A).