Question

Find the Number of all positive integers 'n' less than 17 for which n! + (n+1)! + (n+2)! is an integral multiple of 49. Give Reasoning.

Answer 1

n! + (n +1) ! + ( n +2) ! = 49K
where K is integar .

n! + ( n +1)! + ( n +2) ! = 49K
n! +( n +1)n! + ( n+2)(n+1)n! = 49K
n! { 1 + n +1 + n² +3n +2} = 49K
n! { n² +4n +4 } = 49K
n! ( n +2)² = 49K
n! = { 49K/(n +2)²}

here you see LHS is factorial part
and RHS polynomial part .
factorial part never will be fractional or negative .
we also know
1! = 1
2!= 2
3! =6
4! =24
5!=120
hence , factorial never will be 49 or its integral hence, we have require to cancelled it .
so, we let n in such a way that 49 will be divisible by them .

this is possible when ,

take n = 5 and K = 120

then,
LHS = 5! = 120
RHS = { 49 × 5!/(5+2)²}
= { 49×120/49 } = 120
LHS = RHS
so, n = 5 is possible

next n = 12 and K = 4× 12!
then,
LHS = 12!

RHS = { 49×4×12! /( 12 + 2)² }

= { 49×4× 12!/196 }

= 12!

LHS = RHS

so, n = 12 possible .

hence , n < 17 there are two value possible n = 5 and 12

Answer 2

Answer: 5

Step-by-step explanation:

As an add on for @abhi 's answer,

Any n>=14, there are two factors of 7, one from that of 7 and 14 in the expansion of factorial, so any integer above 14 will satisfy this criteria..

So, the final answers for n=5,12,14,15,16,i.e., 5 possible values

Hope you understood my answer