Question

Fie numarul A=$3^{n+1}$X$7^{n}$X$2^{m}$+$3^{n}$X$7^{n+1}$+5X$21^{n}$,unde m,n∈N*.Sa se determine forma generala a numerelor n si m,astfel incat A div. cu 630.

Answer 1

A = 3ⁿ⁺¹ₓ7ⁿₓ2ᵃ + 3ⁿₓ7ⁿ⁺¹ +5ₓ3ⁿₓ7ⁿ =3ⁿₓ7ⁿ(3ₓ2ᵃ + 7 + 5) = 3ⁿₓ7ⁿ(3ₓ2ᵃ +12) =
= 3ⁿ⁺¹ₓ7ⁿₓ2(2ᵃ⁻¹ + 6)
ptr. a-1 = 2     a = 3     A = 3²ₓ7ₓ10(3ⁿ⁻¹ₓ7ⁿ⁻¹ₓ2) = 630ₓ21ⁿ⁻¹ₓ2 = divizibil cu 630
n ≥ 1