Find the two-digit numeral whose units' digit is 5 more than its tens' digit and whose value equals 3 times the sum of these digits.

Question

Let t= ten's digit

Let u= units' digit

Answer 1

A two digit positive integer can be written as 10x + y, where x is the tens digit and y is the ones digit. For example, 37 = 10(3) + 7.

So, y = x+5 and 10x + y = 3(x + y)

x - y = -5

7x - 2y = 0

Multiply the first equation by -2:

-2x + 2y = 10

7x - 2y = 0

Add the equations to get 5x = 10

So, x = 2 and y = x + 5 = 7

The number is 27

Answer 2

Starting off, the question asks us to find a two-digit number, tu, satisfying two conditions.

1. u is 5 more than t
2. tu is 3 times t + u

If we list out all possibilities for #1, we get the list {16, 27, 38, 49}. No other two-digit numbers satisfy this condition (since 05 isn't technically a two-digit number). Using this list, we can then try to see which numbers satisfy the second condition:

1. 16 = 3 * (1 + 6)
2. 16 = 3 * 7
3. 16 = 21
4. 27 = 3 * (2 + 7)
5. 27 = 3 * 9
6. 27 = 27
7. 38 = 3 * (3 + 8)
8. 38 = 3 * 11
9. 38 = 33
10. 49 = 3 * (4 + 9)
11. 49 = 3 * 13
12. 49 = 39

So from this, it becomes apparent that our original answer is 27.