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What elementary number theory methods can I use for solving this type of questions?

One of these kind of problems is to find $x$ where $985242x6565 = 172195\cdot 572167$ without multiplying the numbers again.

I tried doing, let's call $985242x6565$ $N$, $N\text{mod}10^5$ to somehow get to the $5th$ digit but this is doesn't get me anywhere and in hindsight it doesn't make any sense. How should I proceed?

Bill Dubuque
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Taken Spark
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  • Try thinking modulo $11$. If that doesn't work, then additionally try looking at another modulo as well on top of this. Do you know the easy way to find a number's value modulo $11$? – JMoravitz May 06 '21 at 12:09
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    Modulo 9 is also often useful. – Jaap Scherphuis May 06 '21 at 12:10
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    @Jaap modulo $9$ sometimes has a hard time distinguishing between the missing digit being a $9$ versus a $0$ (as is the case here). As $11>10$ we can skip that frustration by using $11$ instead which is a similarly low calculation required aproach. – JMoravitz May 06 '21 at 12:16
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    @JMoravitz I'd still take those odds and would generally try 9 first, just because the mental arithmetic is slightly easier and hence less error-prone. If it gives an ambiguous result or if I was feeling unsure, only then would I go for 11 as well. – Jaap Scherphuis May 06 '21 at 12:26

2 Answers2

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The alternating sum and difference of digits from right to left of a number is equivalent modulo 11 to the original number.

$172195\mapsto 5-9+1-2+7-1=1$

$572167\mapsto 7-6+1-2+7-5=2$

So the first number is equivalent to $1$ modulo $11$ and the second is equivalent to $2$ modulo $11$. Their product should then be $1\cdot 2=2$ modulo $11$.

We have $5-6+5-6+x-2+4-2+5-8+9=x+4$ and we want this to be $2$ more than a multiple of $11$ with $x$ being a digit. This happens precisely when $x=9$.

JMoravitz
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The remainder of the division by 11 is equal to the alternating sum of the digits of the number. Because
$1 \equiv 1$ (mod 11)
$10 \equiv -1$ (mod 11)
$100 \equiv 1$ (mod 11)
$1000 \equiv -1$ (mod 11)
$10000 \equiv 1$ (mod 11)
and so on...

Thus,
$172195 \equiv (-1+7-2+1-9+5) \equiv 1$ (mod 11)
$572167 \equiv (-5+7-2+1-6+7) \equiv 2$ (mod 11)
$172195⋅572167 \equiv 1\cdot2 \equiv 2$ (mod 11)

$985242x6565 \equiv (9-8+5-2+4-2+x-6+5-6+5) \equiv 4+x$ (mod 11)
and $x\equiv9$ (mod 11).

Answer: $x=9$

Anastasija Serova
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