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I recently came across this question:

If C is a digit such that the product of the three-digit numbers $2C8$ and $3C1$ is the five-digit number $90C58$, what is the value of $C$?


I start by multiplying out the two three digit numbers and simplifying: $$2C8\cdot 3C1=60000+5000C+2000+100C^2+600+90C+8=90000+100C+50+8$$ $$\implies6000+500C+200+10C^2+60+9C=9000+10C+5$$ $$\implies10C^2+499C-2745=0$$ $$C=5$$ This quadratic was a pain of solving, due to the extremely large numbers. Is there any way I can avoid getting a quadratic but still arrive at my answer?

Thanks!
Max0815

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  • $\begingroup$ When you say "...such that the three-digit numbers 2C8 and 3C1 is the five-digit number..." Why was that translated to 2C8 X 3C1? $\endgroup$ – NoChance Aug 16 '19 at 22:55
  • $\begingroup$ @NoChance it was a typo. I fixed it $\endgroup$ – Max0815 Aug 16 '19 at 22:56
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We may check each possible value of $C$, brute force, quick answer.

Well, it is simpler maybe to check the relation modulo nine, i.e. find possible values for $C$ so that $$ (2+C+8)(3+C+1) = (9+0+C+5+8)\text{ modulo nine.} $$ We get the simpler equation $(C+1)(C+4)=(C+4)$ modulo nine, so $C(C+4)=0$ modulo nine. If one factor is divisible by $3$, the other is not. So we have only two cases, $C=0,9$ or $C=5$, so that the one or the other factor is divisible by $9$. We check the $5$ first,

sage: 258*351
90558

and of course, $298\cdot 391=300\cdot 391-2\cdot 391> 117300-1000$ has too many digits. Later edit: The $0$ is ruled out because (working modulo $100$, we get $1\times 8=8$) the last two digits of $208\cdot 301$ are $08$, not $58$.

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  • $\begingroup$ I am curious on how you got C(C+4)=0 [mod 9] from (C+1)(C+4)=(C+4) [mod 9]. Could you mind elaborating more? $\endgroup$ – Max0815 Aug 16 '19 at 22:55
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    $\begingroup$ We build the difference, $0=(C+1)(C+4)-(C+4)=(C+1-1)(C+4)=C(C+4)$ modulo nine. $\endgroup$ – dan_fulea Aug 16 '19 at 22:56
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    $\begingroup$ Why did you leave $C=0$ out? $\endgroup$ – N. S. Aug 16 '19 at 22:59
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    $\begingroup$ @N.S. Oh, that was thought and not typed, modulo $100$, $*01$ times $*08$ does not fit in the pattern, thanks a lot, i will add it! $\endgroup$ – dan_fulea Aug 16 '19 at 23:02
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    $\begingroup$ @Max0815: The $5$ "must work" since it is the final solution, let us see all levels, first: $258=2+5+8=15=155=6$ mod nine, $351=3+5+1=9=0$ modulo nine, $90558=9+0+5+5+8=5+5+8=18=0$ modulo nine, so $6\cdot 0=0$ modulo nine is verified. Then $C(C+4)$ mod nine becomes $5(5+4)=5\cdot 9=5\cdot 0=0$ mod nine. $\endgroup$ – dan_fulea Aug 16 '19 at 23:10
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Here is an easy way to finish the problem once you got the quadratic:

$$499 C = 2745- 10C^2$$ Now since the RHS is a multiple of $5$ so is the LHS, and hence $5|C$.

Since $C$ is a digit, $C=0$ or $C=5$, but $0$ is not a solution to your equation.

All you have to do is check if $5$ is a solution.

P.S. The solution by dan_fulea becomes much shorter if you look $\pmod{11}$ instead of $\pmod{9}$:

$$(2-C+8)(3-C+1) = (9-0+C-5+8) \pmod{11} \\ (-1-C)(4-C)=(1+C) \pmod{11}\\(1+C)(5-C) = 0 \pmod{11}$$

This means that $$C =10 \pmod{11} \mbox{ or } C=5 \pmod{11}$$ and since $C$ is a digit....

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  • $\begingroup$ Thanks for the answer. I see how to get my solving speed faster! $\endgroup$ – Max0815 Aug 16 '19 at 23:14
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Once you've built that quadratic, you can avoid factoring it by taking it modulo 10, since you know the answer you want is a single digit integer.

$$10C^2+499C−2745\equiv5-C\equiv0\pmod {10}$$

From which C=5 quickly.

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Just render the tens digit of the product.

Since $1×8<10$, there is no carry from the product of units digits into the tens place. Then the tens digit of the product must satisfy

$8C+1C=9C\equiv 5\bmod 10$

where the $5$ comes from the given tens digit on the right side. The only digit that could possibly work for $C$ is then $5$.

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If you keep it simple, I realized that

2C8 x 3C1= 90C58 where C^2= has a 5 in the ones place meaning that it has to be 5 because it is the only number that squares to make a 5 in the one's place.

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