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Find the Remainder when $792379237923\ldots$upto 400 digits is divided by $101$?

MyApproaach

when ($792379237923\ldots$400 digts)/$101$=

I learned this approach that I have to calculate(let say U)=Is the sum all of all the alternate groups starting with the rightmost

and (let say)Th=Is the the sum all of all the alternate groups starting with the second rightmost

Rem(U-Th)/$101$=?

But I am not following how to calculate U and Th

Can anyone guide me how to approach this problem?

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Let $N$ be the number. Then we really want to find $N \pmod {101}$.

Note that $N=7923\cdot10^{396}+7923\cdot10^{392}+\cdots+7923$.

Next note that $7923\equiv 45 \pmod {101}$

Also, for example, $10^{100}=100^{50}\equiv(-1)^{50}\equiv 1 \pmod {101}$

We get the same result for each term, and there are $100$ of these terms so $N\equiv 45\cdot 100\equiv 56$.

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Since $100\equiv-1$ mod $101$, your number, mod $101$ is:$$-79+23-79+23-\cdots$$ which is $100$ copies of $-56$. And mod $101$, that makes $56$.

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  • $\begingroup$ @alex.jorden can you look at this question too math.stackexchange.com/questions/1510638/… $\endgroup$ – Jack Nov 3 '15 at 5:17
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    $\begingroup$ @alex.jorden $7923$%$10$=-$79$?How $\endgroup$ – Jack Nov 3 '15 at 5:22
  • $\begingroup$ All mod $101$: $100^0\equiv1$. $100^1\equiv-1$. $100^2\equiv1$ and so on. So $23\cdot100^0+79\cdot100^1+23\cdot100^2+\cdots\equiv23-79+23-\cdots$. $\endgroup$ – alex.jordan Nov 3 '15 at 5:26
  • $\begingroup$ Actually I don't understand your question. I didn't say $7923\%10$ is $-79$. It's more like I said $7900\%101$ is $-79$. $\endgroup$ – alex.jordan Nov 3 '15 at 5:36
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    $\begingroup$ @Charc248 Thanks! Others rejected your suggested edit but I implemented it. $\endgroup$ – alex.jordan Jan 12 '17 at 17:58

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