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RenB
Joined: 13 Jul 2022
Last visit: 30 Apr 2026
Posts: 388
Given Kudos: 302
Location: India
Concentration: Finance, Nonprofit
Schools: Stanford '26 Wharton '26 (D) Booth '26 (D) Yale '26 (D) CBS '26 (A$$$) Darden '26 (WL) HBS '26 (D) Ross '26 (A$$$$) LBS '26 (D) Tuck '26 (A$$) Kellogg '26 (A)
GMAT Focus 1: 715 Q90 V84 DI82 
GPA: 3.74
WE:Corporate Finance (Consulting)
Schools: Stanford '26 Wharton '26 (D) Booth '26 (D) Yale '26 (D) CBS '26 (A$$$) Darden '26 (WL) HBS '26 (D) Ross '26 (A$$$$) LBS '26 (D) Tuck '26 (A$$) Kellogg '26 (A)
GMAT Focus 1: 715 Q90 V84 DI82
Posts: 388
24 Oct 2023, 17:48
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A 4-digit number 6a3b, where a and b are nonzero single-digit numbers, is divisible by 12. If b is a multiple of 2, how many possible values can 'a' have?
A. 4
B. 5
C. 6
D. 7
E. 9
A. 4
B. 5
C. 6
D. 7
E. 9
24 Oct 2023, 20:17
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rules to divide by 12: divide by 3 and divide by 4
Divisibility rule:
3 sum of digits divisible by 3. 6+3+a+b = a+b+9 ----> a+b divisible by 3
4 2^2, exponent means the last two digits must be divisible by 4 ---> 3b is divisible by 4
for 3b to be divisible by 4, then 3b must be either 32 or 36 ----> b can be 2 or 6
when b=2, a can be 1,4, or 7
when b=6, a cane be 3,6, or 9
therefore, there can be 6 values for a
Divisibility rule:
3 sum of digits divisible by 3. 6+3+a+b = a+b+9 ----> a+b divisible by 3
4 2^2, exponent means the last two digits must be divisible by 4 ---> 3b is divisible by 4
for 3b to be divisible by 4, then 3b must be either 32 or 36 ----> b can be 2 or 6
when b=2, a can be 1,4, or 7
when b=6, a cane be 3,6, or 9
therefore, there can be 6 values for a
