3 years ago

HELP MEE MY FRIKIN POINTSSSS

Mathematics

1 answer:

SSSSS [86.1K]3 years ago

3 0

Answer:

Step-by-step explanation:

7/20+3/5=19/20

7/20+12/20=19/20 YES

3/5+3/10=3/15

6/10+3/10=3/15 NO

1/2+2/3=3/5

3/6+4/6=3/5 NO

1/10+1/2=3/5

1/10+5/10=3/5

6/10=3/5 YES

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WHAT IS THE REMAINDER WHEN <img src="https://tex.z-dn.net/?f=32%5E%7B37%5E%7B32%7D%20%7D" id="TexFormula1" title="32^{37^{32} }"

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Recall Euler's theorem: if $\gcd(a,n) = 1$ , then

$a^{\phi(n)} \equiv 1 \pmod n$

where $\phi$ is Euler's totient function.

We have $\gcd(9,32) = 1$ - in fact, $\gcd(9,32^k)=1$ for any $k\in\Bbb N$ since $9=3^2$ and $32=2^5$ share no common divisors - as well as $\phi(9) = 6$ .

Now,

$37^{32} = (1 + 36)^{32} \\\\ ~~~~~~~~ = 1 + 36c_1 + 36^2c_2 + 36^3c_3+\cdots+36^{32}c_{32} \\\\ ~~~~~~~~ = 1 + 6 \left(6c_1 + 6^3c_2 + 6^5c_3 + \cdots + 6^{63}c_{32}\right) \\\\ \implies 32^{37^{32}} = 32^{1 + 6(\cdots)} = 32\cdot\left(32^{(\cdots)}\right)^6$

where the $c_i$ are positive integer coefficients from the binomial expansion. By Euler's theorem,

$\left(32^{(\cdots)\right)^6 \equiv 1 \pmod9$

so that

$32^{37^{32}} \equiv 32\cdot1 \equiv \boxed{5} \pmod9$

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