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

Step 1: Understand the notation and the sum.
\([XX]\) is the greatest integer function, also called the floor function: it gives the biggest whole number that is less than or equal to \(X\). For example, \([2.7] = 2\) and \([33] = 3\).
The sum runs over 99 terms, one for each \(k = 0, 1, 2, \ldots, 98\), and each term is \(\left[\dfrac{1}{3} + \dfrac{k}{99}\right]\).

Step 2: Work out when a term equals 0.
A term \(\left[\dfrac{1}{3} + \dfrac{k}{99}\right]\) equals \(0\) as long as the value inside the brackets stays at least \(0\) but strictly less than \(1\).
The largest \(k\) for which \(\dfrac{1}{3} + \dfrac{k}{99} < 1\) is found from \(\dfrac{k}{99} < \dfrac{2}{3}\), which gives \(k < 66\).
So for \(k = 0, 1, 2, \ldots, 65\), the term is \(0\); that is \(66\) terms in total.

Step 3: Work out when a term equals 1.
At \(k = 66\): \(\dfrac{1}{3} + \dfrac{66}{99} = \dfrac{1}{3} + \dfrac{2}{3} = 1\), and \([11] = 1\), so this term is exactly \(1\).
For \(k = 67, 68, \ldots, 98\), the value keeps climbing but stays below \(2\) all the way up to \(k=98\): checking the last term, \(\dfrac{1}{3} + \dfrac{98}{99} = \dfrac{33}{99} + \dfrac{98}{99} = \dfrac{131}{99} \approx 1.32\), which floors to \(1\), not \(2\).
So every term from \(k = 66\) to \(k = 98\) equals \(1\); that is \(98 - 66 + 1 = 33\) terms in total.

Step 4: Add up the two groups.
There are \(66\) terms worth \(0\) each, contributing \(66 \times 0 = 0\).
There are \(33\) terms worth \(1\) each, contributing \(33 \times 1 = 33\).
Adding these two groups gives a total of \(0 + 33 = 33\).

Step 5: Check the wrong options.
34 would come from an off by one slip, for example counting \(k=66\) to \(98\) as 34 terms instead of the correct 33; careful counting (\(98-66+1=33\)) rules this out.
66 mistakenly treats the 66 zero-terms as if each contributed 1, instead of 0, to the total.
67 and 98 both come from similar miscounts of how many terms equal 1 versus how many equal 0, and neither matches the correct grouping of 66 zeros and 33 ones.

Final Answer:
The sum equals \(33\).
\[ \boxed{33} \]