IPMAT Indore 2023 · Quantitative Ability · Question 35

Step 1 — split x into its parts. Write x=\left[ x \right]+\{x\} and substitute:

2\left[ x \right]=\left( \left[ x \right]+\{x\}\right)+n\{x\}

\left[ x \right]=\left( n+1 \right)\{x\}\Rightarrow \{x\}=\frac{\left[ x \right]}{n+1}

Step 2 — impose the range of the fractional part. By definition 0\le \{x\}<1, so

0\le \frac{\left[ x \right]}{n+1}<1 \Rightarrow 0\le \left[ x \right]<n+1

Since \left[ x \right] is an integer, \left[ x \right]\in \{0,1,2,\ldots ,n\} — exactly n+1 solutions.

Step 3 — write each solution. For \left[ x \right]=m,

x=m+\frac{m}{n+1}=m\,\frac{n+2}{n+1}

Step 4 — sum over all m.

\sum_{m=0}^{n}{m}\,\frac{n+2}{n+1}=\frac{n+2}{n+1}\cdot \frac{n\left( n+1 \right)}{2}=\frac{n\left( n+2 \right)}{2}

The \left( n+1 \right) cancels.

Check with n=1: solutions are \left[ x \right]\in \{0,1\}, giving x=0 and x=3/2; sum =3/2. The formula gives 1\cdot 3/2=3/2 ✔

Answer: Option 1 — \frac{n\left( n+2 \right)}{2}

Shortcut. Test n=1 and read the answer off the options. With n=1 the solutions are x=0 and x=1.5, summing to 3/2.

Two survive, so test n=2: solutions are x=0,4/3,8/3, summing to 4. Option (a) gives 2\cdot 4/2=4 ✔; option (b) is stuck at 3/2 ✘.

Answer (a), reached entirely by substitution. Option (b) is placed to catch anyone who tests only n=1 — always try a second value when two options survive the first.