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A three digit number 4a3 is added to another three digit number 984 to give the four digit number 13b7 which is divisible by 11. Then $$(a + b)^2 = $$
Now 13b7 is divisible by 11. Hence (b+1)-10 =11m (divisibility rule for 11)
As we can see, this is only possible when m=0 ( do remember that "b" has to be a single-digit)
Hence b+1-10 or b=9 Hence the number is 1397. Subtracting 984 from it we get 413. Hence a=1
So (a+b)^2=100
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