## anonymous 4 years ago 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. check your answer. HELP~

1. Mr.Math

If the number has the form $$a_0a_1$$, then we can make the system: $a_0+a_1=12$ $10a_0+a_1=10a_1+a_0+18$

2. pokemon23

hola turningtest goodnight people

3. UnkleRhaukus

57,75

4. anonymous

I'm lost :c

5. TuringTest

...this exhibits my love/hate relationship with Diophantine equations. Nice formulation Mr. Math :)

6. Mr.Math

Let me explain. You have a number of two digits, call it $$a_0a_1$$, where $$a_0$$ is the 10's digit and $$a_1$$ is the 1's digit. Now we're given that the sum of the two digits is 12. We can write that as $$a_0+a_1=12$$. Following so far?

7. anonymous

ohh yeah, ohkay keep going..

8. Mr.Math

Good. The second information we are given is that when reversing the digits [that's writing it as $$a_1a_0$$], the new number is 18 less than the original number. First you need to notice that if the original number is x, we can write $$x=10a_0+a_1$$. Call the new number y, then $$y=10a_1+a_0$$. Hence the second equation of the system is: $10a_0+a_1=10a_1+a_0+18.$

9. Mr.Math

All you're left to do here is to solve the system for $$a_0$$ and $$a_1$$.

10. anonymous

Can you do the steps and explain to me please? I seem to learn better by looking at the steps done.

11. TuringTest

$a_0+a_1=12\to a_0=12-a_1$$10a_0+a_1=10a_1+a_0+18\to9a_0-9a_1=18$multiply first eqn by 9$9a_0+9a_1=108$$9a_0-9a_1=18$add these together...$18a_0=126\to a_0=7$now we can find the other number with the formula for a_1 on the first line$a_0+a_1=12\to a_1=12-a_0=12-7=5$hence $a_0=7$$a_1=5$so the numbers are$57,75$as UnkleRhaukus said

12. anonymous

the numbers are 57, and 75 but which is the original?

13. UnkleRhaukus

the original is 75