The value of a two-digit number is twice as large as the sum of its digits. If the digits were reversed, the resulting number would be 9 less then 5 times the original number. Find the original number

Question

Let t= ten's digit

Let u= units' digit

Answer 1

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

1. The value of tu is twice as large as t + u
2. ut is 9 less than 5 times tu

If we create formulas based around this, we can simplify these statements to be:

1. tu = 2 * (t + u)
2. ut = (5 * tu) - 9

If we also consider that tu = 10t + u, and ut = 10u + t, we can substitute these values and simplify in order to get:

1. 10t + u = 2 * (t + u)
2. 10t + u = 2t + 2u
3. 8t = u
4. 10u + t = (5 * (10t + u)) - 9
5. 10u + t = 50t + 5u - 9
6. 5u = 49t - 9

If we substitute u = 8t into the equation 5u = 49t - 9, we get:

1. 5(8t) = 49t - 9
2. 40t = 49t - 9
3. 9 = 9t
4. t = 1

Using this and our previous equation, we find that tu = 18.

To double-check this, we can substitute it back into our initial equations to ensure that it works:

1. tu = 2 * (t + u)
2. 18 = 2 * (1 + 8)
3. 18 = 2 * 9
4. ut = (5 * tu) - 9
5. 81 = (5 * 18) - 9
6. 81 = 90 - 9

So, with our confirmation out of the way, we can say safely that our answer is 18.