Can you find those numbers?
Let a be a single-digit positive integer. Let n be another positive integer that adjusts itself so that the unit's place digits of n and a n + 1 are the same. Find the sum of all possible values of a .
The answer is 14.
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1 solution
That's it. :)
How did you reach ( a − 1 ) n ≡ 9 ( m o d 1 0 ) ?
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Just by rearranging the first congruence. In full (and all modulo 1 0 ),
a n + 1 ≡ n - this comes directly from the question
a n − n ≡ − 1 - just moving the terms
( a − 1 ) n ≡ − 1 - factorising
( a − 1 ) n ≡ 9 - since − 1 ≡ 9 ( m o d 1 0 ) .
I hope that helps! The key step is the factorisation. The idea is that if a − 1 is even, you can never multiply it by a whole number and get a result that ends in 9 . If a − 1 is 5 , all of its multiples end in 0 or 5 . If a − 1 is any other odd digit, we can always find a number to multiply it by to get a result that does end in 9 . The last trick in the question is that a − 1 can't be 9 (or − 1 ) as we're told a is a positive single-digit number.
We need a n + 1 ≡ n ( m o d 1 0 ) . Rearranging, this is ( a − 1 ) n ≡ 9 ( m o d 1 0 ) , so the valid a are those such that a − 1 has a multiplicative inverse modulo 1 0 . So a − 1 ∈ { 1 , 3 , 7 , 9 } . a can't be 0 or 1 0 , so the only possible a values are 2 , 4 , 8 with sum 1 4 .