If the 6-digit number N68M83 is divisible by 11, which of the options given…

Concept: A number is divisible by 11 exactly when the alternating digit sum — the sum of digits in odd positions minus the sum of digits in even positions, counted from either end — equals 0 or a multiple of 11.

Application:

1. Number the digits of N68M83 from the right: position 1 = 3, position 2 = 8, position 3 = M, position 4 = 8, position 5 = 6, position 6 = N.

2. Sum the odd positions (1, 3, 5): 3 + M + 6 = M + 9.

3. Sum the even positions (2, 4, 6): 8 + 8 + N = N + 16.

4. Form the alternating difference: (M + 9) − (N + 16) = M − N − 7.

5. This difference must be 0 or a multiple of 11. Since M and N are single digits (N is also the non-zero leading digit, so 1–9), M − N − 7 = 0 gives M − N = 7 (valid digit pairs: N = 1, M = 8 and N = 2, M = 9), and M − N − 7 = −11 gives M − N = −4 (valid digit pairs: N = 4, M = 0; N = 5, M = 1; N = 6, M = 2; N = 7, M = 3; N = 8, M = 4; N = 9, M = 5). These 8 pairs are the only digit assignments that make N68M83 divisible by 11.

6. Checking the other three relations against all 8 of these pairs: M − N = 1 is never true for any of them (the pairs give M − N = 7 or −4, never 1); M = N is never true for any of them either; M + N = 5 is never true for any of them (their sums are 9, 11, 4, 6, 8, 10, 12, 14). Only M − N = 7 is satisfied — by two of the eight pairs — so it is the unique relation, among the four given, that can hold whenever N68M83 is divisible by 11.

Cross-check: Testing M − N = 7 directly confirms it — with N = 1, M = 8 the number is 168883 = 11 × 15353, and with N = 2, M = 9 the number is 268983 = 11 × 24453; both divide exactly by 11.

So M − N = 7 is the relation between M and N that is consistent with N68M83 being divisible by 11.