The Division Star
Let A B be a 2-digit integer such that the 8-digit number 4 8 3 2 A B 1 8 is divisible by 11.
Find the maximum value of A B .
The answer is 89.
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3 solutions
A number is divisible by 11 if a1 - a2 + a3 - a4....... is divisible by 11. thus 4 - 8 + 3 - 2 + A - B + 1 - 8 . cong. 0 (mod 11), or A - B -10 .cong. 0(mod(11), or A - B. cong.-1, so A=8, B =9. Ed Gray
Let w = * in our number 4 8 3 2 w 1 8 .
Then:
4 8 3 2 w 1 8 = 4 8 3 2 0 0 1 8 + 1 0 0 w = 4 3 9 2 7 2 8 × 1 1 + 1 0 + 9 9 w + w = 1 1 × ( 4 3 9 2 7 2 8 + 9 w ) + ( w + 1 0 )
Therefore:
4 8 3 2 w 1 8 ≡ w + 1 0 (mod 11)
Now, we have to solve the:
w + 1 0 ≡ 0 (mod 11) ⟺
w ≡ 1 (mod 11) congruence.
The biggest solution for 10 ≤ w ≤ 99 (w is a two digit number) is 8 9 .
Another approach would be to simply use divisibility rule of 11: ( 1 8 + B ) − ( 8 + A ) is divisible by 11. Therefore 1 0 + B − A is divisible by 11. Hence B = 9 and A = 8 .
Taking the algorithm of divisibility by 11 One can derive, A-B=10(mod11) Or,A-B=-1(mod11) 0<A<B<9
So A=8 and B=9. At maximum