CAT 2024 Slot 2 · Quant · Question 9

Rewrite each logarithm so that its base matches the base in the denominator below it.

Since 8={2}^{3}, {\log}_{8}\left( a+b \right)=\frac{1}{3}{\log}_{2}\left( a+b \right).

Since 27={3}^{3}, {\log}_{27}\left( a-b \right)=\frac{1}{3}{\log}_{3}\left( a-b \right).

The first fraction becomes \frac{1}{3}\cdot \frac{{\log}_{2}\left( a+b \right)}{{\log}_{2}c}=\frac{1}{3}{\log}_{c}\left( a+b \right), by the change-of-base rule.

The second becomes \frac{1}{3}{\log}_{c}\left( a-b \right) in exactly the same way.

Adding, \frac{1}{3}\left({\log}_{c}\left( a+b \right)+{\log}_{c}\left( a-b \right) \right)=\frac{2}{3}.

So {\log}_{c}\left( \left( a+b \right)\left( a-b \right) \right)=2, which means {a}^{2}-{b}^{2}={c}^{2}.

Rearranging, {a}^{2}={b}^{2}+{c}^{2}, and both b and c are at most 10.

Therefore {a}^{2}\le{10}^{2}+{10}^{2}=200, so a\le \sqrt{200}\approx 14.14.

The greatest integer is 14, and it is attainable: with a=14 and b=10, c=\sqrt{196-100}=\sqrt{96}\approx 9.80, which satisfies 10\ge b\ge c.

Shortcut. The two bases, 8 and 27, are cubes of 2 and 3, and the right-hand side is \frac{2}{3} — the \frac{1}{3} cancels and the whole condition collapses to {a}^{2}-{b}^{2}={c}^{2} in one move. After that it is just the largest hypotenuse with both legs capped at 10.