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Updated on: 09 Oct 2025, 08:07
Originally posted by BrentGMATPrepNow on 10 Jul 2019, 10:21.
Last edited by Bunuel on 09 Oct 2025, 08:07, edited 1 time in total.
Last edited by Bunuel on 09 Oct 2025, 08:07, edited 1 time in total.
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
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Question Stats:
50% (02:28) correct
50%
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N is a positive 3-digit integer in which all three digits have the same value. The product 53N is a 5-digit integer comprised of five of the following six digits: 1, 2, 4, 6, 7, 9. Which of the following digits is NOT in the product 53N?
A. 1
B. 2
C. 4
D. 6
E. 7
A. 1
B. 2
C. 4
D. 6
E. 7
11 Jul 2019, 06:21
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GMATPrepNow
KEY PROPERTY #1: There's a nice divisor property that says: If n is divisible by d, then kn is divisible by d (assuming n, d and k are INTEGERS)
For example, since 33 is divisible by 11, we also know that (33)(7) is divisible by 11
Since 111 is divisible by 3, we know that (2)(111) is divisible by 3, and (3)(111) is divisible by 3, and (4)(111) is divisible by 3, and (5)(111) is divisible by 3, etc.
In other words, we know that N is divisible by 3.
So, from property #1, we also know that 53N is divisible by 3.
KEY PROPERTY #2: If a number is divisible by 3, the sum of its numbers is also be divisible by 3
So, we must choose 5 numbers (from 1, 2, 4, 6, 7, 9) so that the sum of the digits is divisible by 3.
1 + 2 + 4 + 6 + 7 + 9 = 29
So, if we DON'T select the digit 2, the remaining digits add to 27, which IS divisible by 3.
Answer: B
Cheers,
Brent
10 Jul 2019, 12:40
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N is a multiple of 111, so N is divisible by 3, and 53N is divisible by 3. So the digits of 53N must sum to a multiple of 3. The six possible digits sum to 29, and it's only by subtracting the '2' that we get a multiple of 3. So '2' is not one of the digits.
