Question

A certain two-digit number has a value that is seven more than six times the sum of its digits. The tens digit is 3 more than the units digit. Find the number.

Answer 1

Explanation:

Let the two-digit number be represented as $10a+b$, where $a$ is the tens digit and $b$ is the units digit. From the problem statement, we have:

$10a+b = 6(a+b)+7$ ....(1)

Also, we know that the tens digit is 3 more than the units digit, so we have:

$a = b+3$ ....(2)

Substituting (2) into (1), we get:

$10(b+3)+b = 6((b+3)+b)+7$

Simplifying the above equation, we get:

$11b+33 = 13b+25$

Subtracting $11b$ from both sides, we get:

$2b+33 = 25$

Subtracting 33 from both sides, we get:

$2b = -8$

Dividing both sides by 2, we get:

$b = -4$

Since $b$ represents a digit, we see that $b$ must be 6 (and not -4), since it is the only digit that satisfies the condition that the tens digit is 3 more than the units digit. Substituting $b=6$ into (2), we get:

$a = b+3 = 6+3 = 9$

Therefore, the two-digit number is $10a+b = 10(9)+6 = \boxed{96}$.