Calculating value of a variable in combination?

Question

Using the formula (bionomial theorem):

$${}_nC_r \cdot a^{n-r} \cdot b^{r}$$

i was trying to find out the term with the coefficient $400$ from $(2x + 5y)^5$? Then i came up with this equation which i have no idea to solve beside trying numbers between 1 and 5. Is there any other way?

$$2^{5-r} \cdot 5^{r} \cdot {}_5C_r = 400$$

can we find the value of r without brute forcing?

Answer 1

$\frac{2^5}{2^r} \cdot 5^r \cdot \binom 5r = 400$

$\frac{32}{2^r} \cdot 5^r \cdot \binom 5r = 400$

$\frac{1}{2^r} \cdot 5^r \cdot \binom 5r = \frac{25}{2}$

$\frac{1}{2^r} \cdot 5^r \cdot \binom 5r = \frac{5 \cdot 5}{2}$

$\frac{1}{2^r} \cdot 5^r \cdot \binom 5r = \frac{5}{2} \cdot \binom 51$

On comparing both sides,

$2^r = 2^1$

$r = 1$

Answer 2

Let's assume there is an exact (positive integer) solution to:

$$2^{5-r} \cdot 5^{r} \cdot {5 \choose r} = 5^2(2^4)$$

Clearly then $0 \leq r \leq 5$.

Then

$$2^{5-r} \cdot 5^{r} \cdot (5)....(5-r+1)=r! (5^2(2^4))$$

If $r=5$ then exponent of $5$ on both sides do not match. So by the fundamental theorem of arithmetic $r \neq 5$. Continuing from there, the exponent of $5$ must be $2$ on the left hand side, hence we must have:

$$5^{r+1}=5^2$$

By inspection $r=1$ and we check to see that this works.