Question

If \([XX]\) denotes the greatest integer less than or equal to \(X\), find the value of \[ \left[\frac{1}{3}\right] + \left[\frac{1}{3}+\frac{1}{99}\right] + \left[\frac{1}{3}+\frac{2}{99}\right] + \cdots + \left[\frac{1}{3}+\frac{98}{99}\right] \]

• A. 33
• B. 34
• C. 66
• D. 67

Hint:

Work out for which values of k, from 0 to 98, the term 1/3 + k/99 is still less than 1, since those terms round down to 0.

Answer 1

The Correct Option is A

A cleaner way to handle this greatest-integer sum is to clear the fractions first. Multiply every term inside the brackets so we are comparing whole numbers instead of fractions, which avoids fraction arithmetic mistakes.

1. Rewrite each term: $\dfrac{1}{3} + \dfrac{k}{99} = \dfrac{33}{99} + \dfrac{k}{99} = \dfrac{33+k}{99}$. So the $k$-th term is $\left[\dfrac{33+k}{99}\right]$, for $k$ running from $0$ to $98$.
2. Find where this floor value switches from 0 to 1: $\left[\dfrac{33+k}{99}\right] = 0$ as long as $33 + k < 99$, that is $k < 66$. So $k = 0$ through $k = 65$ give a floor value of $0$: that is $66$ values of $k$.
3. Find where it switches from 1 to 2: the term equals $1$ once $99 \le 33+k < 198$, that is $66 \le k < 165$. Since $k$ only goes up to $98$, every $k$ from $66$ to $98$ falls in this equals 1 band: that is $98 - 66 + 1 = 33$ values of $k$.
4. Confirm the top end never reaches 2: the highest possible value is at $k = 98$: $\dfrac{33+98}{99} = \dfrac{131}{99}$, about $1.32$, still under $2$, so the floor stays at $1$ and never jumps to $2$ within this range.

Now add the contribution of each band: the $66$ terms with floor value $0$ contribute nothing, and the $33$ terms with floor value $1$ contribute $33 \times 1 = 33$ to the running total.

Let's summarize:

• Rewriting $\dfrac{1}{3}+\dfrac{k}{99}$ as $\dfrac{33+k}{99}$ turns the problem into counting how many values of $33+k$ fall below $99$ (floor $0$) versus between $99$ and $198$ (floor $1$).
• That split is $66$ terms of value $0$ and $33$ terms of value $1$, giving a total of $33$.

So the value of the sum is $33$.