When you add 1 to the end of abcde, you get abcde1.

Since 1abcde needs to be one third of abcde1, "e" when multiplied by 3 must end in a 1. The only number that ends in a 1 when multiplied by 3 is 7, there e is 7.

abcd7

7 * 3 = 21, so 2 must be carried over when multiplying abcd7 by 3. This means that (d * 3) + 2, must end in 7. This means that d = 5.

abc57

5 * 3 = 15. So 1 must be carried over when multiplying abc57 by 3. This means the (c * 3) + 1 must end in 5. Therefore, c = 8.

ab857

8 * 3 = 24. So 2 must be carried over when multiplying ab857 by 3. This means that (b * 3) + 2 must end in 8. Therefore, b = 2.

a2857

Since 2 * 3 = 8, there is nothing to carry over. So I just started substituting numbers for a until I got the answer 42857.

## 3 Comments:

42857

142857 * 3 = 428571

Perhaps I should have posted my working out too.

The 5 digit number can be represented by abcde.

When you add 1 to the end of abcde, you get abcde1.

Since 1abcde needs to be one third of abcde1, "e" when multiplied by 3 must end in a 1. The only number that ends in a 1 when multiplied by 3 is 7, there e is 7.

abcd7

7 * 3 = 21, so 2 must be carried over when multiplying abcd7 by 3. This means that (d * 3) + 2, must end in 7. This means that d = 5.

abc57

5 * 3 = 15. So 1 must be carried over when multiplying abc57 by 3. This means the (c * 3) + 1 must end in 5. Therefore, c = 8.

ab857

8 * 3 = 24. So 2 must be carried over when multiplying ab857 by 3. This means that (b * 3) + 2 must end in 8. Therefore, b = 2.

a2857

Since 2 * 3 = 8, there is nothing to carry over. So I just started substituting numbers for a until I got the answer 42857.

Yes, 42857.

10A + 1 = 3 (100000 + A)

10A + 1 = 300000 + 3A

7A = 299999

A = 42857

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