Needgmat 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.
OAA
Please explain
Statement 1:
Since each digit of N must be a factor of N, 5 must be a factor of N.
Put another way, N must be a MULTIPLE OF 5.
A multiple of 5 has a units digit of 0 or 5.
Since 0 is not a factor of any number and thus cannot be a factor of N, the units digit of N must be 5.
Since N = X55, N is ODD.
Since an odd integer has only ODD factors, and every digit of N must be a factor of N, the hundreds digit of N must be ODD.
Options for N:
355,
555, 755, 955.
Rules:
An integer is a multiple of 3 only if the sum of its digits is a multiple of 3.
An integer is a multiple of 9 only if the sum of its digits is a multiple of 9.
Since the sum of the digits of 355 is not a multiple of 3, 355 is not a multiple of 3.
Thus, the hundreds digit of 355 -- 3 -- is not a factor of 355.
Since the sum of the digits of 955 is not a multiple of 9, 955 is not a multiple of 9.
Thus, the hundreds digit of 955 -- 9 -- is not a factor of 955.
755 is not divisible by its hundreds digit of 7.
Only one option for N remains:
N=555.
SUFFICIENT.
Statement 2:
It's possible that N=555, as in Statement 1.
Since the sum of the digits of 315 is a multiple of 3, 315 is a multiple of 3.
Thus, all 3 digits of 315 -- 3, 1, and 5 -- divide into 315, implying that N=315 is a viable option.
Since N can be different values, INSUFFICIENT.
The correct answer is
A.
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