2 a b c = c a b 2abc = cab

If A A the smallest possible integer such that when the last digit becomes the first digit, the resulting number is twice the original number.

Find A A .

Input the last three digits of A A as your answer.

For more such interesting questions check out Presh Talwalkar's amazing YouTube Channel MindYourDecisions .


The answer is 842.

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2 solutions

take the number X = a n a 1 X=\overline{a_n \dots a_1} , then

2 × a n a 1 = a 1 a n a 2 19 ( j = 0 j = n 2 1 0 j a j ) = ( 1 0 n 2 ) a 1 2\times \overline{a_n \dots a_1}= \overline{a_1 a_n\dots a_2} \implies 19(\sum_{j=0}^{j=n-2}10^{j}a_j)=(10^n-2)a_1

so, we need to find the lowest n n , for which 19 1 0 n 2 19 | 10^n-2 or 1 0 n 2 ( m o d 19 ) 10^n\equiv 2 \ (mod \ 19) .

1 0 n 2 ( m o d 19 ) 5.1 0 n 1 1 ( m o d 19 ) 1 0 n 1 5 1 4 ( m o d 19 ) 10^n\equiv 2 \ (mod \ 19) \implies 5.10^{n-1}\equiv 1\ (mod \ 19) \implies 10^{n-1}\equiv 5^{-1} \equiv 4\ (mod 19)

1 0 n 1 4 ( m o d 19 ) 25.1 0 n 3 1 ( m o d 19 ) 1 0 n 3 2 5 1 16 ( m o d 19 ) 10^{n-1} \equiv 4 \ (mod \ 19) \implies 25.10^{n-3} \equiv 1 \ (mod \ 19) \implies 10^{n-3} \equiv 25^{-1} \equiv 16 \ (mod \ 19)

1 0 n 3 16 5 4 . 1 0 n 7 1 1 0 n 7 9 ( m o d 19 ) 10^{n-3} \equiv 16 \implies 5^4.10^{n-7} \equiv 1 \implies 10^{n-7} \equiv 9 \ (mod \ 19)

1 0 n 7 9 10 1 0 n 8 1 ( m o d 19 ) 10^{n-7} \equiv 9 \equiv -10 \implies 10^{n-8} \equiv -1 \ (mod 19) .

So, we need to find the smallest number of form 1 0 k + 1 10^k+1 that is divisible by 19 19 . Luckily, for k = 9 k=9 , 1 0 k + 1 10^k+1 is divisible by 19 19 and we don't have to go through more trouble. So, n 8 = 9 n = 17 n-8=9 \implies n=17 . We now go back to

19 ( 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 0 17 2 ) 19 a 1 X a 1 10 = ( 1 0 17 2 ) 19 a 1 19(\sum_{j=0}^{j=n-2}10^{j}a_j)=(10^n-2)a_1 \implies \sum_{j=0}^{j=n-2}10^{j}a_j=\frac{(10^{17}-2)}{19}a_1 \implies \frac{X-a_1}{10}=\frac{(10^{17}-2)}{19}a_1

since a 1 = 2. a 16 a_1=2.a_{16} , a 1 a_1 needs to be even and since its multiplication by ( 1 0 17 2 ) 19 \frac{(10^{17}-2)}{19} determines the number, without its last digit a 1 a_1 , we take the minimum a 1 = 2 a_1=2 . then, we only need to know the last two digits of ( 1 0 17 2 ) 19 \frac{(10^{17}-2)}{19} to know the last three digits of X X . for that, we calculate

( 1 0 17 2 ) . ( 19 ) 1 ( 2 ) . 79 42 ( m o d 100 ) a 3 = 2.4 , a 2 = 2.2 (10^{17}-2).(19)^{-1} \equiv (-2).79 \equiv 42 \ (mod \ 100) \implies a_3=2.4,a_2=2.2

Ajit Athle
Nov 23, 2018

The smallest n-parasitic numbers are also known as Dyson numbers, after a puzzle concerning these numbers posed by Freeman Dyson. In the present case, n=2 and A=105263157894736842, the derivation of which may be read here: https://en.wikipedia.org/wiki/Parasitic_number

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