The sum of the digits of a two digit number is 12. When the digits are reversed the new number is 18 less than the original number. How do you find the original number?

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Answer 1 · 2015-09-24T21:41:07

Express as two equations in the digits and solve to find original number $75$ $75$ .

Suppose the digits are $a$ $a$ and $b$ $b$ .

We are given:

$a + b = 12$ $a + b = 12$

$10 a + b = 18 + 10 b + a$ $10a + b = 18 + 10 b + a$

Since $a + b = 12$ $a+b = 12$ we know $b = 12 - a$ $b = 12 - a$

Substitute that into $10 a + b = 18 + 10 b + a$ $10 a + b = 18 + 10 b + a$ to get:

$10 a + (12 - a) = 18 + 10 (12 - a) + a$ $10 a + (12 - a) = 18 + 10 (12 - a) + a$

That is:

$9 a + 12 = 138 - 9 a$ $9a+12 = 138-9a$

Add $9 a - 12$ $9a - 12$ to both sides to get:

$18 a = 126$ $18a = 126$

Divide both sides by $18$ $18$ to get:

$a = \frac{126}{18} = 7$ $a = 126/18 = 7$

Then:

$b = 12 - a = 12 - 7 = 5$ $b = 12 - a = 12 - 7 = 5$

So the original number is $75$ $75$