One term of (2π+5)^n = 288000π^8, what's n?

Question

Without using calculator what's the value of n?

Using binomial expansion I get:

ⁿC_p * 2^n-p * π^n-p * 5^p = 288000π⁸

Easily I know that n-p=8, by the π's power

Then the power of 2 is also 8, so I can divide both sides

ⁿC_p * 5^p = 1125

Well I know that

• n-p=8
• n-8=p
• n=p+8

And with "pascal triangle's" rule(ⁿC_p = ⁿC_n-p)

ⁿC₈ * 5^p = 1125

ⁿC₈ * 5^n-8 = 1125

Well there are definitively many ways to equate this, but how can I get the value of n without calculator or guessing(Trial and error)?

Or is there any other approach?

The answer is n=10 and p=2...

Sorry for no LaTeX(I don't know how to work with it

Answer 1

The term of degree $8$ in the expansion of $(x+a)^n$ is $$ \binom{n}{8}a^{n-8}x^8 $$ so you want to solve $$ \binom{n}{8}5^{n-8}2^8=288000=2^83^25^3 $$ hence $$ \binom{n}{8}5^{n-11}=9 $$ We can exclude $n>11$, because the right hand side is not divisible by $5$.

Therefore you just have to check $n=8,9,10,11$: \begin{align} \binom{8}{8}&=1 \\ \binom{9}{8}&=\binom{9}{1}=9 \\ \binom{10}{8}&=\binom{10}{2}=\cdots\\ \binom{11}{8}&=\binom{11}{3}=\cdots \end{align}

Answer 2

${n\choose 8} * 5^{n-8} = 1125$

But $1125 = 9*5^3$ so $5^{n-8}|5^3$ and $8\le n \le 11$

And $5^{n-11}{n\choose 8} = 9$.

$9\not \mid 5$ so either $n-11=0$ or $5^{11-n}|{n\choose 8}$.

Now as $8\le n\le 11$ and ${n\choose 8}=\frac {9*..n}{(n-8)!}$ then $11 - n$ is at most $1$.

So $n = 10$ or $11$. Now as no prime but $3,5$ divide $\frac {9*..*n}{(n-8)!}$ we can not have $n = 11$ so

so $n =10$.