The number of ordered pairs (m, n), m, n ∈ 1,2, ldots, 50 such that 6m+9n is a multiple of 5 is

Question

The number of ordered pairs (m, n), m, n ∈ 1,2, ldots, 50 such that 6m+9n is a multiple of 5 is (A) 1250 (B) 2500 (C) 500 (D) 625

Tardigrade Admin · Accepted Answer

As the last digit of 6m, m ∈ N is 6,6m+9n will be divisible by 5 if the unit's digit of 9n is 4 or 9 . This is possible when n is odd. ∴ required number of ordered pairs =50 × 25= 1250.

Q. The number of ordered pairs $(m, n), m, n \in {1, 2$ , $\dots, 50}$ such that $6^{m} + 9^{n}$ is a multiple of 5 is

1250

2500

500

625

As the last digit of $6^{m}, m \in N$ is $6, 6^{m} + 9^{n}$ will be divisible by 5 if the unit's digit of $9^{n}$ is 4 or 9 . This is possible when $n$ is odd.
$∴$ required number of ordered pairs $= 50 \times 25 =$ 1250.

Q. The number of ordered pairs (m,n),m,n∈{1,2, …,50} such that 6m+9n is a multiple of 5 is

1250

2500

500

625

As the last digit of 6m,m∈N is 6,6m+9n will be divisible by 5 if the unit's digit of 9n is 4 or 9 . This is possible when n is odd. ∴ required number of ordered pairs =50×25= 1250.

Q. The number of ordered pairs $(m, n), m, n \in {1, 2$ , $\dots, 50}$ such that $6^{m} + 9^{n}$ is a multiple of 5 is

As the last digit of $6^{m}, m \in N$ is $6, 6^{m} + 9^{n}$ will be divisible by 5 if the unit's digit of $9^{n}$ is 4 or 9 . This is possible when $n$ is odd.
$∴$ required number of ordered pairs $= 50 \times 25 =$ 1250.