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I'm having some trouble understading how to use congruence to solve this type of exercises:

a) Determinate the remainder of the division of $2173451$ with $17$

b) Determinate the remainder of the division of $1522^{1000}$ with $19$

For example, in b) I know that $1522=80\times19+2$, so $1522 \equiv 2(mod19)$ , so $1522^{1000} \equiv 2^{1000}(mod19)$. What now?

5 Answers 5

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You will want to use Fermat's Little Theorem which states that given a prime $p$ and $a$ such that $gcd(a,p)=1$, $a^{p-1}\equiv1(modp)$.

So in your first question for example, you will want to figure out how to express $a=2173451$ with an exponent of $p-1$ which is $16$. So you will write $2173451$ as some $16q+r$ where q is the biggest multiple of $16$ less than $2173451$ and $r$ is the remainder.

Whatever remainder is given will be the remainder of $2173451$ when divided by $17$.

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    Why the negative rating may I ask? – 2017-02-28
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In exercise $b)$ , reduce the exponent modulo $\phi(19)=18$

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(a) $$17 \times 1 = 17$$ $$17 \times 2 = 34$$ $$17 \times 3 = 51$$ $$17 \times 6 = 102$$ $$17 \times 6 - 2 = 100$$

Hence $$100 \equiv -2 \mod 17$$ \begin{align}2173451&= 2000000+170000+3400+51 \\&\equiv2\times(10^2)^3+0+0+0 \mod 17\\ &\equiv 2\times (-2)^3 \mod 17 \\ &\equiv -16 \mod 17 \\& \equiv 1 \mod 17\end{align}

(b) $$19 \times 5 = 95$$ $$19 \times 5 + 5 = 100 \equiv 5 \mod 19$$ $$1522 = 1500+22 \equiv (-4)\times 10^2+3 \equiv -20+3 \equiv -17 \equiv 2 \mod 19$$

Now, let's examine the exponent, $1000 \equiv 10 \times 100\equiv 10 \times (18 \times 5+10)\equiv 100 \equiv 10 \mod 18$

Hence, we want to compute $2^{10} \mod 19$

$$2^{10} \equiv 1024 \equiv 1000+24 \equiv 10(5)+5 \equiv 55 \equiv 57-2 \equiv -2 \equiv 17 \mod 19 $$

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    How do you go from $1024$ to $10(5)+5$ . I didn't understand. Thanks – 2017-03-01
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    $1024 = 10\times 100 + 24 \equiv 10 \times 5 + 5 \mod 19$ since we know $100 \equiv 5 \mod 19$ and $24 \equiv 5 \mod 19$. – 2017-03-01
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A general technique for problems of type a) is to divide a number into its decimal expansion: $$n=a_0+10a_1+100a_2+\cdots $$ and find remainder of each summand and then sum the remainders.

for $n=2173451 $, we have $$n=1+5\times 10+4\times100+3\times1000+7\times10^4+1\times10^5+2\times10^6 $$ which is congruence modulo $17$ to $$1+(-1)+(-8)+(-9)+ (-6)+(-11)+1\equiv -51\equiv0$$ you may use the following congruences: $$10\equiv 10\pmod{17} \\ 100\equiv -2\pmod{17}\\ 1000\equiv -3\pmod{17}, \cdots$$

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note that $$127850\cdot 17=2173450$$ and $$1522^9 \equiv -1 \mod 19$$