Please with the explanation of the editorial

Yet another meme problem:

Educational Codeforces Round 80

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conc(A,B) is technically (10^no. of digits in B)*A+(B) (this is the mathematical definition
let k =no of digits in b
we get:
A+B+A*B=(10^k)

*A+B*

A+AB=(10^k)*A

A+A

1+B=10^k

Therefore A can be any number in the given range but B should be of the form 99…9.

Count the no of numbers of the form 99…9 and multiply it with the number of possible A values to get the answer

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Another way to do this is by trying to do it for small numbers like till 1000

like try to generate all numbers at A=1000 and B=1000.

you will see that B always uses 9,99,999 and A is going from to 1-1000 three times.

This is the observation part of it in case you aren’t able to prove it with maths.

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