Problem 9 · Place-value equations and divisibility · 5 points
Made by AI and not reviewed by a real person.
A six-digit lock code has distinct, nonzero digits. Multiplying it by 5 moves its last digit to the front while keeping the other digits in order; what is the sum of the original code’s first and last digits?
- 7
- 8
- 12
- 6
- 9
Show a hint
- Write the code as a five-digit prefix followed by its final digit, then compare the place values before and after multiplication.
- Eliminate the prefix and use divisibility to narrow down the final digit.
Show the answer and solution
Answer: B — 8
Let N be the code and d its final digit. Moving d to the front gives 5N = 100000d + (N−d)/10, so 49N = 999999d. Now 999999 = 7×142857, and 142857 leaves remainder 1 when divided by 7. Thus d must be a multiple of 7; as a nonzero digit, it is 7. The equation gives N = 142857, and indeed 5×142857 = 714285. The requested sum is 1+7 = 8. The tempting answer 7 is only the final digit, not the sum.