# How do I approach a question like this?

If $j$, $k$, and $n$ are consecutive integers such that $0<j<k<n$ and the units (ones) digit of the product $jn$ is $9$, what is the units digit of $k$?

(A) $0$
(B) $1$
(C) $2$
(D) $3$
(E) $4$

I'm thinking the answer is (B) because the only factors of $9$ is $1,\, 9$, and $3$ ($\large1\times9\text{ and }3\times3$) so we can take out (A), (C), and (E). All that's left is (B) and (D). But since $k$ is less than $n$, they both can't be $3$ for $jn=9$. And that leaves (B) as the answer?

Am I right in doing it this way?

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But then $j=k$...? And I'm not sure what you mean by "units digit"? To me all solution $\ge (C)$ look good... – draks ... Oct 10 '12 at 20:59

Rather than working backwards from knowing what $jn$ is, you can also work forwards by trial and error. The crucial point of the problem - which amounts to the same thing as clark's discussion of numbers mod $10$ - is that the units digit of a product only depends on the units digit of the factors. This means that you can ignore the other digits of the numbers entirely. From here on out, it's just a matter of cases. Since $j$, $k$, and $n$ are consecutive, then $j$ and $n$ must be two numbers apart, so they're either both even or both odd; obviously they can't both be even (because the product of two even numbers is always even), so they're both odd. This leaves five possibilities for their last digits (for simplicity from here on out, I'll just use $j$ and $n$ to refer to their last digits, since as noted above that's all that we care about):

• $j=1, n=3$. In this case, the product must end in $3$, not $9$.
• $j=3, n=5$. Here the product ends in $5$.
• $j=5, n=7$. Again, the product ends in $5$.
• $j=7, n=9$. This time the product (e.g. $63$) must end in $3$.
• $j=9$. This is the crux of it all: if $j$ ends in $9$, then whatever number comes two numbers after it must end in $1$ (again, we're effectively working 'mod $10$'.) In this case, the product of $j$ and $n$ must end in $9$ (for instance, $9\times 11=99$ or $29\times 31=899$). Since this is the only case that ends in $9$, then this must be the case, and since $k$ is between $j$ and $n$ it must end in $0$.
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I understood yours more. But I don't get how $k$ could be $0\;$ (I understand how $j\bmod{10}=9$ and $n\bmod{10}=1$). But you said that $k$ is between $j$ and $n$ but still how does that make it end in $0$?? – David Oct 10 '12 at 21:53
@david: remember, all three numbers are consecutive. If $j$ is, say, 49 and $n$ is thus 51, $k$ has to be 50. – Steven Stadnicki Oct 10 '12 at 22:06
Okay, got it! Thanks. – David Oct 10 '12 at 22:13

From the problem I understand $jn=9 \mathrm{mod}10$ that means either $$n=3\mathrm{mod}10, j=3 \mathrm{mod}10$$ But if $j=3 \mathrm{mod}10$ then $n=5\mathrm{mod}10$ simirarly we cannot have

$$n=9\mathrm{mod}10, j=1 \mathrm{mod}10$$ so we are with the case $$j=9\mathrm{mod}10, n=1 \mathrm{mod}10$$ so $$k=0 \mathrm{mod}10$$

Instead of using \mathrm{mod} you could use \bmod or \pmod or \mod (different spacing and parenthesis between those). You can use k\bmod{10} to produce this: $$k\bmod{10}$$ – Asaf Karagila Oct 10 '12 at 21:35