Let $$a_0, a_1, \ldots, a_{23}$$ be real numbers such that

$$\left(1 + \frac{2}{5}x\right)^{23} = \sum_{i=0}^{23} a_i x^i$$

for every real number $$x$$. Let $$a_r$$ be the largest among the numbers $$a_j$$ for $$0 \leq j \leq 23$$. Then the value of $$r$$ is ______.

The expansion of $$\left(1+\frac{2}{5}x\right)^{23}$$ by the binomial theorem is

$$\left(1+\frac{2}{5}x\right)^{23}=\sum_{i=0}^{23}\binom{23}{i}\left(\frac{2}{5}x\right)^i =\sum_{i=0}^{23}\binom{23}{i}\left(\frac{2}{5}\right)^i x^{\,i}.$$

Comparing with $$\displaystyle\sum_{i=0}^{23} a_i x^i,$$ we identify

$$a_i=\binom{23}{i}\left(\frac{2}{5}\right)^{i},\qquad 0\le i\le 23.$$

To find the largest coefficient, examine the ratio of successive terms:

$$\frac{a_{i+1}}{a_i}= \frac{\binom{23}{i+1}(2/5)^{\,i+1}}{\binom{23}{i}(2/5)^{\,i}} =\frac{23-i}{i+1}\cdot\frac{2}{5}.\quad -(1)$$

The sequence $$a_0,a_1,\dots,a_{23}$$ increases while the ratio in $$(1)$$ exceeds $$1$$ and begins to decrease once the ratio falls below $$1$$. Hence we need to solve

$$\frac{23-i}{i+1}\cdot\frac{2}{5}\gt 1.$$

Multiply both sides by $$5(i+1):$$

$$2\,(23-i)\gt 5(i+1).$$

Simplify:

$$46-2i\gt 5i+5 \;\;\Longrightarrow\;\; 46-5 > 7i \;\;\Longrightarrow\;\; 41>7i.$$

Therefore

$$i<\frac{41}{7}=5.857\ldots$$

Because $$i$$ is an integer, inequality $$(1)$$ holds for $$i=0,1,2,3,4,5.$$ At $$i=5,$$ compute one more ratio:

$$\frac{a_6}{a_5}=\frac{23-5}{6}\cdot\frac{2}{5}=\frac{18}{6}\cdot\frac{2}{5}=3\cdot0.4=1.2\gt1,$$

so the coefficients are still rising up to $$i=6.$$ Next, for $$i=6$$

$$\frac{a_7}{a_6}=\frac{23-6}{7}\cdot\frac{2}{5} =\frac{17}{7}\cdot0.4\approx0.97\lt1,$$

meaning the sequence starts decreasing after $$i=6$$. Hence the maximum coefficient occurs at

$$r=6.$$

Final answer: $$r=6.$$