BTGmoderatorDC wrote:The difference 942 − 249 is a positive multiple of 7. If a, b, and c are nonzero digits, how many 3-digit numbers abc are possible such that the difference abc − cba is a positive multiple of 7 ?
A. 142
B. 71
C. 99
D. 20
E. 18
OA E
Source: Official Guide
Notice that abc = 100a + 10b + c = 98a + 2a + 7b + 3b + c. Since 98a and 7b are divisible by 7, so when abc is divided by 7, it's equivalent to when 2a + 3b + c is divided by 7. Similarly, since cba = 100c + 10b + a = 98c + 2c + 7b + 3b + c, when cba is divided by 7, it's equivalent to when 2c + 3b + a is divided by 7. Now, if we subtract 2c + 3b + a from 2a + 3b + c , we have a - c and this quantity must be divisible by 7. Therefore, we could have (a, c) as (8, 1) or (9, 2). However, b, the tens digit, can be any digit from 1 to 9. Therefore, we can express abc as 811, 821, ..., 891 or as 912, 922, ..., 992. So there are 18 such numbers (9 numbers in the 800s and another 9 in the 900s).
(Note: For example, if a is 8, c is 1 and b is 5, we have abc = 851 and cba = 158 and 851 - 158 = 693, which is 99 x 7.)
Alternate Solution:
We see that abc = 100a + 10b + c and cba = 100c + 10b + a. If we subtract the second from the first, we have:
100a + 10b + c - (100c + 10b + a)
100a + 10b + c - 100 c - 10b - a
99a -99c
99(a - c)
Now, we know that this difference abc - cba = 99(a - c) must be divisible by 7. Since 99 is not divisible by 7, then (a - c) must be. Also, since a and c must be single-digit numbers, their only possible values are [a = 9 and c = 2] OR [a = 8 and c = 1]. But b can be any nonzero single digit, so b can be any of [1, 2, 3, ..., 9]. Therefore, we can express abc as 912, 922, ..., 992, or as 811, 821, ..., 891. So there are 18 such numbers (9 numbers in the 900s and another 9 in the 800s).
Answer: E
