Simple Problem #03
If I add to some number, it becomes a multiple of If I subtract of that number, it becomes a multiple of What is the remainder when evaluating
This is part of my simple problems set .
The answer is 1.
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1 solution
Let the number be n :
As n + 1 is a multiple of 1 0 , n ends in a 9 .
Note that 9 , 2 9 , 4 9 , 6 9 . . . are possible values for n (as well as − 1 1 , − 3 1 , − 5 1 . . . , but 1 9 , 3 9 , 5 9 , 7 9 . . . aren't due to the 4 n − 1 condition.
Therefore all n that satisfy can be written as n = 2 0 k + 9 .
n 2 = ( 2 0 k + 9 ) 2 = 4 0 0 k 2 + 3 6 0 k + 8 1 . When dividing this by 4 0 , all that is left is remainder 1 .