2023 IOQM/Problem 1

Problem

Let $n$ be a positive integer such that $1 \leq n \leq 1000$ . Let $M_n$ be the number of integers in the set

$X_n = \left\{\sqrt{4n + 1}, \sqrt{4n + 2}, \ldots, \sqrt{4n + 1000}\right\}$ . Let $a = \max\{M_n : 1 \leq n \leq 1000\}$ , and $b = \min\{M_n : 1 \leq n \leq 1000\}$ .

Find $a - b$ .

Solution 1(Spacing of squares)

If for any integer $n$ , if $\sqrt{n}$ is an integer this means $n$ is a perfect square. Now the problem reduces to finding the difference between maximum and minimum no. of perfect squares between $4n+1, 4n+2 .... 4n+1000.$ There are 1000 numbers here.

The idea is that for the same range of numbers, the no. of perfect squares becomes less when the numbers become larger.

For example, there are 3 perfect squares between 1 and 10 but none between 50 and 60.

⇒ The maximum value of $M_n$ occurs when $n$ is minimum and the minimum value of $M_n$ occurs when $n$ is maximum.

Minimum value of $n$ = 1 So, the numbers are 5,6...1004. there are 29 perfect squares here, so $a$ = $max.$ ( $M_n$ )= $29$

Maximum value of $n$ = 1000 So, the numbers are 4001,4002...5000. there are 7 perfect squares here, so $b$ = $min.$ ( $M_n$ )= $7$

⇒ $a-b= 29-7 = \boxed{22}$

~SANSGANKRSNGUPTA

Video Solutions

Video solution by cheetna: [[1]]

Video solution by Unacademy Olympiad Corner: [[2]]

Video solution by Vedantu Olympiad School: [[3]]

Video solution by Olympiad Wallah: [[4]]

Video solution by : Motion Olympiad Foundation Class 5th - 10th: Motion Olympiad Foundation Class 5th - 10th

Please note that above videos solutions are in Hindi, some in English and some in mixed(Hindi + English). ~SANSGANKRSNGUPTA

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2023 IOQM/Problem 1

Problem

Solution 1(Spacing of squares)

Video Solutions