Question

+ Find the smallest positive integer N such that N(N + 1)(N + 2) is divisible by 247.​

Answer 1

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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