Finding the fractional part of $\frac{5^{200}}{8}$.
I've had this problem given to me (we're learning the Binomial Theorem and all.) So obviously I thought I'd apply the binomial theorem to it, somehow.
I've tried writing $5$ as $8-3$, and expanding the numerator, so that all but one term would be divisible by $8$, (the one term being the constant term.) and hence the fractional part would only (possibly) arise from the constant term being divided by 8. But apparently this is not working because the constant term itself is of the order of magnitude of the original numerator.
Any ideas?
$\endgroup$ 44 Answers
$\begingroup$Note that $$5^2 \equiv 1 \pmod8 \implies 5^{200} = (5^2)^{100} \equiv 1^{100} \pmod8 \equiv 1 \pmod8$$ We hence have $$5^{200} = 8M + 1 \implies \dfrac{5^{200}}8 = M + \dfrac18$$ where $M \in \mathbb{Z}^+$. Hence, the fractional part is $1/8$.
EDIT Another equivalent way is to write $5^{200}$ as $(4+1)^{200}$ and then use binomial theorem, i.e., $$5^{200} = (1+4)^{200} = \sum_{k=0}^{200} \dbinom{200}k 4^k$$ Note that $8$ divides $4^k$ for all $k \geq 2$. Hence, we have $$5^{200} = 8N + \dbinom{200}1 4^1 + \dbinom{200}{0} 4^0 =8N + 800 + 1 = 8M+1$$ We hence have $$5^{200} = 8M + 1 \implies \dfrac{5^{200}}8 = M + \dfrac18$$ where $M \in \mathbb{Z}^+$. Hence, the fractional part is $1/8$.
$\endgroup$ 8 $\begingroup$$5^2\equiv 1\mod 8$, hence $5^{200}\equiv 1\mod 8$. Thus the fractional part is $\dfrac18=0·125$.
$\endgroup$ $\begingroup$Here's a answer without modular arithmetic if you require it,
$5^{200}=25^{100}=(1+24)^{100}$
Now expanding binomially,
$1+^{100}C_1*24+^{100}C_2*(24)^2+(O)$
Now simply divide by 8 and you got the fractional part.
$\endgroup$ 0 $\begingroup$Let's prove by induction that the fractional part of $5^{2n}$ is $\frac18$.
First, show that this is true for $n=1$:
$5^2=8\cdot3+1$
Second, assume that this is true for $n$:
$5^{2n}=8k+1$
Third, prove that this is true for $n+1$:
$5^{2(n+1)}=$
$5^{2n+2}=$
$5^{2n}\cdot5^2=$
$\color{red}{5^{2n}}\cdot25=$
$(\color{red}{8k+1})\cdot25$
$200k+25=$
$200k+24+1=$
$8(25k+3)+1$
Please note that the assumption is used only in the part marked red.
$\endgroup$ 2
