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In two digit number, the product og digits is $20$ and if $9$ is added to the number, the digits will be reversed. What is the number?

Can anyone help me with this? Thanks.

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    $\begingroup$ How many different ways can $20$ be written as the product of two single-digit numbers? $\endgroup$ – John Wayland Bales Jan 8 '17 at 2:11
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    $\begingroup$ @John Wayland Bales, $4*5$. $\endgroup$ – user292114 Jan 8 '17 at 2:12
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    $\begingroup$ So do you see what the answer must be now? $\endgroup$ – John Wayland Bales Jan 8 '17 at 2:14
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    $\begingroup$ @John Wayland Bales, yea. $\endgroup$ – user292114 Jan 8 '17 at 2:15
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There is only one possible combination of two single digits that give $20$ when multiplied together. They can be put after one another in two different ways. One of those ways makes the resulting number fulfill the second criterion, the other does not.

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Let the two digit number be $xy=10x+y$. Then your conditions mean that:

  • $xy=20$ and
  • $10x+y+9=10y+x \Leftrightarrow x+1=y$

Can you solve the above for $x,y$ non-zero digits?

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$ab=20$.

(Product of digits)

$10a+b+9=10b+a$.

(First digit * 10 (place value) + Second digit = Number)

$9a+9= 9b$

$a+1=b$.

(Simplification)

We can see that |ab| is 45.

* I neglected the case where b = 0. That would mean adding 9 makes b=9. However, it is negligible because if b were 0, that would mean the product is 0, which goes against the product being 20 *

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