+ Find the smallest positive integer N such that N(N + 1)(N + 2) is divisible by 247.
Answers
Given :- Find the smallest positive integer N such that N(N + 1)(N + 2) is divisible by 247. ?
Solution :-
→ N * (N + 1) * (N + 2) ÷ 247 = Remainder 0
→ N * (N + 1) * (N + 2) ÷ (13 * 19) = Remainder 0 .
so,
→ N * (N + 1) * (N + 2) ÷ 13 = Remainder 0 .
→ N * (N + 1) * (N + 2) ÷ 19 = Remainder 0 .
[ if a number is divisible by x , then it will also divisible by prime factors of x .]
since N, (N + 1) and (N + 2) are consecutive positive integers , check smallest multiple of 13 and 19 which are near to each other .
as we know that, 13 * 3 is 39 and 19 * 2 is 38 . Both 38 and 39 are consecutive positive integers .
then,when (n + 2) = 39
→ N = 37
therefore,
→ 37 * 38 * 39 = Divisible by 13 and 19 both
→ 37 * 38 * 39 ÷ 247
→ 54834 ÷ 247 = Remainder 0 .
when (n + 1) = 39,
→ 38 * 39 * 40 = Divisible by 13 and 19 both
→ 59280 ÷ 247 = Remainder 0 .
since we need smallest positive integer . Hence, N is equal to 37 .
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