How many natural numbers n are there such tha | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) We are looking for natural numbers $n$ such that $n! + 10 = k^2$ for some integer $k$.Case $1$: If $n=1$,$1! + 10 = 11$ (not a perfect square).Cas

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) We are looking for natural numbers $n$ such that $n! + 10 = k^2$ for some integer $k$.Case $1$: If $n=1$,$1! + 10 = 11$ (not a perfect square).Case $2$: If $n=2$,$2! + 10 = 12$ (not a perfect square).Case $3$: If $n=3$,$3! + 10 = 6 + 10 = 16 = 4^2$ (a perfect square).Case $4$: If $n=4$,$4! + 10 = 24 + 10 = 34$ (not a perfect square).Case $5$: If $n=5$,$5! + 10 = 120 + 10 = 130$ (not a perfect square).Case $6$: If $n \ge 5$,then $n!$ is a multiple of $10$ because $n!$ contains the factors $2$ and $5$.Thus,$n! + 10 = 10m + 10 = 10(m+1)$ for some integer $m$.For $n \ge 5$,$n!$ is a multiple of $100$ (since $n!$ contains $2 	imes 5 	imes 10 = 100$ for $n \ge 10$,and for $n=5, 6, 7, 8, 9$,we check manually: $5!=120, 6!=720, 7!=5040, 8!=40320, 9!=362880$).Specifically,for $n \ge 5$,$n! + 10$ ends in the digit $0$ (since $n!$ ends in $0$ for $n \ge 5$).For a number ending in $0$ to be a perfect square,it must end in $00$. However,$n! + 10$ for $n \ge 5$ ends in $10, 20, 30, 40, 50, 60, 70, 80, 90$ (specifically $130, 730, 5050, 40330, 362890$).None of these are perfect squares.Therefore,only $n=3$ satisfies the condition.

How many natural numbers $n$ are there such that $n! + 10$ is a perfect square?

Similar Questions

How many words can be formed using $3$ consonants and $2$ vowels from $5$ consonants and $4$ vowels?

${ }^{34} C_5+\sum_{r=0}^4{ }^{(38-r)} C_4=$

The number of ways to distribute $30$ identical candies among four children $C_{1}, C_{2}, C_{3}$ and $C_{4}$ such that $C_{2}$ receives at least $4$ and at most $7$ candies,and $C_{3}$ receives at least $2$ and at most $6$ candies,is equal to

If $^n{P_r} = 840$ and $^n{C_r} = 35$,then $n$ is equal to

The number $(49^{2}-4)(49^{3}-49)$ is divisible by (in $!$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module