The 7-digit number 77A444B is divisible by 24. What is the maximum value of (A + B)?
16
24 = 8 × 3, so 77A444B must be divisible by both 8 and 3.
Divisibility by 8 depends on the last three digits, 44B. Values of B making 44B divisible by 8 are B = 0 (440) and B = 8 (448).
Divisibility by 3: sum of digits = 7 + 7 + A + 4 + 4 + 4 + B = 26 + A + B; this must be a multiple of 3, i.e. A + B ≡ 1 (mod 3).
To maximise A + B, take B = 8. Then A + 8 ≡ 1 (mod 3) means A ≡ 2 (mod 3); the largest such single digit A is 8.
Then A + B = 8 + 8 = 16, and 26 + 16 = 42 is divisible by 3, so the number 7784448 is divisible by 24.
Hence, the maximum value of A + B is 16.
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Important Questions from Divisibility Rules
Find the least value of x for which 57x716 is divisible by 9.
Which of the following numbers is NOT divisible by 11?
If 321y72 is a multiple of 6, where y is a digit, what is the least value of y?
From the given numbers A, B, C and D, which number is NOT divisible by 11?
A = 712712
B = 177210
C = 64614
D = 756148
Which of the following numbers is divisible by 7 ?
