If the smallest possible integer such that when the last digit becomes the first digit, the resulting number is twice the original number.
Find .
Input the last three digits of as your answer.
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take the number X = a n … a 1 , then
2 × a n … a 1 = a 1 a n … a 2 ⟹ 1 9 ( ∑ j = 0 j = n − 2 1 0 j a j ) = ( 1 0 n − 2 ) a 1
so, we need to find the lowest n , for which 1 9 ∣ 1 0 n − 2 or 1 0 n ≡ 2 ( m o d 1 9 ) .
1 0 n ≡ 2 ( m o d 1 9 ) ⟹ 5 . 1 0 n − 1 ≡ 1 ( m o d 1 9 ) ⟹ 1 0 n − 1 ≡ 5 − 1 ≡ 4 ( m o d 1 9 )
1 0 n − 1 ≡ 4 ( m o d 1 9 ) ⟹ 2 5 . 1 0 n − 3 ≡ 1 ( m o d 1 9 ) ⟹ 1 0 n − 3 ≡ 2 5 − 1 ≡ 1 6 ( m o d 1 9 )
1 0 n − 3 ≡ 1 6 ⟹ 5 4 . 1 0 n − 7 ≡ 1 ⟹ 1 0 n − 7 ≡ 9 ( m o d 1 9 )
1 0 n − 7 ≡ 9 ≡ − 1 0 ⟹ 1 0 n − 8 ≡ − 1 ( m o d 1 9 ) .
So, we need to find the smallest number of form 1 0 k + 1 that is divisible by 1 9 . Luckily, for k = 9 , 1 0 k + 1 is divisible by 1 9 and we don't have to go through more trouble. So, n − 8 = 9 ⟹ n = 1 7 . We now go back to
1 9 ( ∑ j = 0 j = n − 2 1 0 j a j ) = ( 1 0 n − 2 ) a 1 ⟹ ∑ j = 0 j = n − 2 1 0 j a j = 1 9 ( 1 0 1 7 − 2 ) a 1 ⟹ 1 0 X − a 1 = 1 9 ( 1 0 1 7 − 2 ) a 1
since a 1 = 2 . a 1 6 , a 1 needs to be even and since its multiplication by 1 9 ( 1 0 1 7 − 2 ) determines the number, without its last digit a 1 , we take the minimum a 1 = 2 . then, we only need to know the last two digits of 1 9 ( 1 0 1 7 − 2 ) to know the last three digits of X . for that, we calculate
( 1 0 1 7 − 2 ) . ( 1 9 ) − 1 ≡ ( − 2 ) . 7 9 ≡ 4 2 ( m o d 1 0 0 ) ⟹ a 3 = 2 . 4 , a 2 = 2 . 2