Question

A two-digit number is four times the sum of its digits. If the digits are reversed, the new number becomes $k$ times the sum of the digits. What is the value of $k$?

• A. 8
• B. 10
• C. 5
• D. 7

Answer 1

Answer: (D)

7

Answer 2

Let the two-digit number be represented by $10t + u$, where $t$ is the tens digit and $u$ is the units digit. The sum of the digits is $t + u$.

Given that the two-digit number is four times the sum of its digits: $10t + u = 4(t + u)$ $10t + u = 4t + 4u$ $6t = 3u$ $2t = u$

When the digits are reversed, the new number becomes $10u + t$. This new number is $k$ times the sum of the digits: $10u + t = k(t + u)$

Substitute $u = 2t$ into the second equation: $10(2t) + t = k(t + 2t)$ $21t = 3kt$

Since $t$ is a non-zero digit, divide both sides by $3t$: $7 = k$

Therefore, the value of $k$ is 7.