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I tried to write it as $$\overline{abc}=7k+1$$ $$\overline{abc}=8p+4$$ $$\overline{abc}=9t+7$$ but I have no idea what to next since I can't add or substract any value to find the least common multiple.

Another approach I have tried was to rewrite those relations as $$\overline{abc}=7(k+1)-6$$ $$\overline{abc}=8(p+1)-4$$ $$\overline{abc}=9(t+1)-2$$

I can see that $\overline{abc}$ must be an even number divisible by $4$ so $c\in{0,2,4,6,8}$ but here I got stuck.

N. F. Taussig
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Emil Cohen
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    Hint : You need the chinese remainder theorem – Peter Jan 20 '23 at 09:46
  • There are two $3$ digit-numbers with this property. – Peter Jan 20 '23 at 09:49
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    As @Peter mentioned, the Chinese remainder theorem is the best option. Are you familiar with it so we can use it in the answer? Could you provide the source of the problem (a course, homework, exercise, a puzzle by a friend, a random question in your mind) and if you've learnt any theorems or methods in number theory before? – D S Jan 20 '23 at 09:53
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    This was a subject from an olympiad. I didn't know about Chinese remainder theorem, but I don't mind to learn it, quite the contrary, I'll study. – Emil Cohen Jan 20 '23 at 10:14
  • Do you know basic operations in modular arithmetic? Then I may be able to post an answer that will be coherent to you. – D S Jan 20 '23 at 10:18
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    The exercise is formulated as if the number would be unique. If it was actually formulated exactly this way , it is a bad exercise for olympia-level. – Peter Jan 20 '23 at 10:19
  • @DS, yes, I know, I would really appreciate – Emil Cohen Jan 20 '23 at 10:39
  • then here you go – D S Jan 20 '23 at 10:39
  • As here in the linked dupe, it can be solved in $10$ seconds of minute of mental arithmetic as follows: $!!\underbrace{\bmod 7!+!k}_{\textstyle \color{#c00}{k\equiv -7}}!:\ x \equiv 1!+!3\color{#c00}k \equiv 1!+!3(\color{#c00}{-7})\equiv -20,,$ so $,x\equiv -20\equiv 484\pmod{!7!\cdot! 8!\cdot! 9},$ by CCRT or lcm $\ \ $ – Bill Dubuque Jan 20 '23 at 16:42
  • Alternative to the above CRT arithmetic progression optimization: we can instead (mechanically) apply (Easy) CRT to solve it by repeatedly replacing a pair of congruences by a single equivalence congruence, as explained in the prior linked post. $\ \ $ – Bill Dubuque Jan 20 '23 at 16:43

1 Answers1

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Let your number be $x$. We have: $$x \equiv 1 \mod 7 \tag{1}\label{1}$$ $$x\equiv 4 \mod 8\tag{2}\label{2}$$ $$x \equiv 7 \mod 9\tag{3}\label{3}$$ Let's rewrite the last equivalence: $$x = 9p+7\tag{4}\label{4}$$for some integer $p$. Put $\eqref{4}$ in $\eqref{2}$: $$9p+7 \equiv 4 \mod 8$$ $$p \equiv 5 \mod 8$$ Rewriting: $$p = 8q+5\tag{5}\label{5}$$ for some integer q. Put $\eqref{5}$ in $\eqref{4}$: $$x=9(8q+5)+7$$ $$x=72q+45+7$$ $$x = 72q+52\tag{6}\label{6}$$ Put $\eqref{6}$ in $\eqref{1}$: $$72q+52\equiv 1 \mod 7$$ $$2q+3 \equiv 1 \mod 7$$ $$2q \equiv 5 \mod 7$$ $$2q \equiv 12 \mod 7$$ $$q \equiv 6 \mod 7$$ Rewriting: $$q = 7k+6\tag{7}\label{7}$$ for some integer $k$. Put $\eqref{7}$ in $\eqref{6}$: $$x = 72(7k+6) + 52$$ $$\color{green}{x = 504k+484}$$It is trivial from here. There are $2$ such numbers.
As an exercise, try to see why this general form of $x$ would satisfy the conditions.

D S
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