Question

Consider the set of all positive integers n for which $f\left(n\right)=n!+\left(n+1\right)!+\left(n+2\right)!$ is divisible by $49$.

• A. The number of integers n in (1,15) is 3
• B. The number of integers n in (5,17) is 4
• C. The number of integers n in (1,20) is 8
• D. The number of integer n in (1,20) is 9

Answer 1

Answer: (A), (B), (C)

The correct options are
A The number of integers n in (1,15) is 3
B The number of integers n in (5,17) is 4
C The number of integers n in (1,20) is 8

Solution-

$f\left(n\right)=n!+(n+1)!+(n+2)!$

$=n!(1+(n+1)+(n+1\left)\right(n+2\left)\right)$

$=n!(1+n+1+n^2+3n+2)$

$=n!(n^2+4n+4)$

$=n!(n+2{)}^2$

for divisibility by $49$.

there must be two factors of $7$ in it.

$n=5$ is divisible.

$n=12$ is divisible.

$n=14$ ... are all divisible.

because $14\times 7$ will be there in $n!$

No. of integers in $(1,15)$ are$5,12,14\to 3$

No. of integers in $(5,17)$ are $12,14,15,16\to 4.$

No. of integers in $(1,20)$ are $5,12,19,15,16,17,18,19\to 8.$

A,B,C is correct.