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CAT 2022 Slot 1 — QA Question 17

Basics (Functions)Easy

For any real number x, let [x] be the largest integer less than or equal to x. If ∑n=1N[15+n25] = 25, then N is

Answer & solution

Answer: 44

Solution

Easy

The bracket [15+n25]\left[\frac15+\frac{n}{25}\right] is a floor, so it only ever jumps between integer values. Find exactly when it equals 00 and when it equals 11, then count how many 11's you need to make the running total hit 2525.

1

When is the floor 00? The term inside is positive but below 11:

0≤15+n25<1⇒ n25<45⇒ n<20⇒ [15+n25]=0for n=1,2,…,19\begin{aligned} &0\le \frac15+\frac{n}{25}\lt 1\\ &\Rightarrow\ \frac{n}{25}\lt \frac45\\ &\Rightarrow\ n\lt 20\\ &\Rightarrow\ \left[\frac15+\frac{n}{25}\right]=0\quad\text{for } n=1,2,\dots,19 \end{aligned}
2

When is the floor 11? The term sits in [1,2)[1,2):

1≤15+n25<2⇒ 45≤n25<95⇒ 20≤n<45⇒ [15+n25]=1for n=20,21,…,44\begin{aligned} &1\le \frac15+\frac{n}{25}\lt 2\\ &\Rightarrow\ \frac45\le \frac{n}{25}\lt \frac95\\ &\Rightarrow\ 20\le n\lt 45\\ &\Rightarrow\ \left[\frac15+\frac{n}{25}\right]=1\quad\text{for } n=20,21,\dots,44 \end{aligned}
3

Add up the terms until the sum reaches 2525. The first 1919 terms contribute nothing, then each new term adds 11:

∑n=1N[15+n25]=0+⋯+0⏟n=1 to 19+1+1+⋯+1⏟n=20 to N⇒ number of 1’s=N−19=25⇒ N=44\begin{aligned} &\sum_{n=1}^{N}\left[\frac15+\frac{n}{25}\right]=\underbrace{0+\dots+0}_{n=1\text{ to }19}+\underbrace{1+1+\dots+1}_{n=20\text{ to }N}\\ &\Rightarrow\ \text{number of }1\text{'s}=N-19=25\\ &\Rightarrow\ N=44 \end{aligned}
N=44N=\mathbf{44}

The floor is 00 up to n=19n=19 and 11 from n=20n=20 onwards. You need 2525 ones, so stop 2525 terms after n=20n=20: n=20n=20 to n=44n=44. Hence N=44N=44.

CAT 2022 Slot 1 QA Q17: For any real number x, let [x] be the largest integer less than or equal to x. If &sum; n = 1 N 1 5 + n 25 = 2 — Solution | TheCATExam