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What are all the two-digit positive integers in which the difference between the integer and the product of its two digits is $12$ ?

What I did so far:

$10a+b-ab=12$,

$10a+b(1-a)=12$,

$-10(1-a)+b(1-a)=2$,

$(b-10)(1-a)=2$

Then I solve and get $b=12$ or $11$, but this can't be right.

Please help.

DeepSea
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jonyoung2002
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  • Did you already consider and choose to reject the possibility that a two digit number could exist which is $12$ less than the product of its digits? Your initial equation implicitly assumes that is impossible; is that actually a justified assumption? – Wildcard Dec 30 '16 at 10:55

3 Answers3

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from the equation $$(a-1)(10-b)=2$$ we obtain $$a-1=2$$ and $$10-b=1$$ from here we get $$a=3,b=9$$ or $$a-1=1$$ and $$10-b=2$$ from here we get $$a=2,b=8$$

Dr. Sonnhard Graubner
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Hint: The equation $10a+b-ab-12=0$ is equivalent to $b(a-1)=2(5a - 6)$. Note that $a-1=0$ is impossible.

Dietrich Burde
  • 130,978
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You have done correctly so far. To proceed further, we can divide it into following cases:

Case 1. $(b-10) = 2$ and $(1-a) = 1$
Case 2. $(b-10)=1$ and $(1-a)=2$
Case 3. $(b-10)=-1$ and $(1-a)=-2$
Case 4. $(b-10)=-2$ and $(1-a)=-1$

We can easily check the values of $a,b$ satisfying the conditions and we thus get the permissible values as: $39$ and $28$. Hope it helps.