Two digit number divided by sum of digits etc.

**Probability1** · Mar 2nd 2011, 06:57 AM

If a 2 digit number is divided by the sum of its digits the result is 6 with a remainder of 6.If the same number has its digits swapped ( if it was 23, it becomes 32) and gets divided by the difference of its digits the result is 23.What's the number?

Could anyone please help me solve this problem with an equation system with 2 unknowns. I would really appreciate it.

**TKHunny** · Mar 2nd 2011, 07:22 AM

Digits are "a" and "b". You must remember how the base 10 positional system works.

"If a 2 digit number is divided by the sum of its digits the result is 6 with a remainder of 6"

$\frac{10\cdot a + b}{a+b} = 6 + \frac{6}{a+b}$

More simply, since a+b is NOT zero

$10\cdot a + b = 6(a+b) + 6$

Let's see what you get for the other.

**Probability1** · Mar 2nd 2011, 08:51 PM

I think I tried solving it this way, but for some reason it did not work.
Guess I'll try again. Thanks for the reply.

**Probability1** · Mar 2nd 2011, 08:54 PM

Will the second one be
(10b + a) : ( b - a ) = 23

**HallsofIvy** · Mar 3rd 2011, 01:47 AM

Originally Posted by Probability1

Will the second one be
(10b + a) : ( b - a ) = 23

I presume you mean $\frac{10b+ a}{b- a}= 23$ which would be the same as $10b+ a= 23(b- a)$ . That would be correct if b> a. If a> b, it would be $\frac{10b+ a}{a- b}= 23$ or $10b+ a= 23(a- b)$ . Try solving each of those (with 10a+ b= 6(a+ b)+ 6 as the second equation) and see which is correct.Of course, a and b must be integers between 0 and 9.

**Probability1** · Mar 3rd 2011, 03:06 AM

Solved it.
The correct one is 10b + a = 23 ( a- b )
The answer is 96.
Thanks for the help.

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