Home
Class 12
MATHS
The number of ordered pairs (m,n) where ...

The number of ordered pairs `(m,n)` where `m`, `n in {1,2,3,…,50}`, such that `6^(m)+9^(n)` is a multiple of `5` is

A

`1250`

B

`2500`

C

`625`

D

`500`

Text Solution

Verified by Experts

The correct Answer is:
A

`(a)` `6^(m)+9^(n)`
Unit digit of `6^(m)` is `=6`
Unit digit of `9^(n)` will be `=9` or `1`
For multiple of `5` unit digit of `9^(n)` must be `=9`
It occur when `n=` odd
Total number or ordered pair `=50xx25=1250`
Promotional Banner

Topper's Solved these Questions

  • PERMUTATION AND COMBINATION

    CENGAGE|Exercise Multiple Correct Answer|2 Videos
  • PERMUTATION AND COMBINATION

    CENGAGE|Exercise Comprehension|8 Videos
  • PARABOLA

    CENGAGE|Exercise Question Bank|9 Videos
  • PRINCIPLE OF MATHEMATICAL INDUCTION

    CENGAGE|Exercise Exercise|9 Videos

Similar Questions

Explore conceptually related problems

The number of ordered pairs of positive integers (m,n) satisfying m le 2n le 60 , n le 2m le 60 is

The HCF of numbers of the form 2^(m) and 3^(n) is ____.

How many ordered pairs of (m,n) integers satisfy (m)/(12)=(12)/(n) ?

If two distinct numbers m and n are chosen at random form the set {1, 2, 3, …, 100}, then find the probability that 2^(m) + 2^(n) + 1 is divisible by 3.

If f and g are two functions defined on N , such that f(n)= {{:(2n-1if n is even), (2n+2 if n is odd):} and g(n)=f(n)+f(n+1) Then range of g is (A) {m in N : m= multiple of 4 } (B) { set of even natural numbers } (C) {m in N : m=4k+3,k is a natural number (D) {m in N : m= multiple of 3 or multiple of 4 }

The number of ways of arranging m positive and n(lt m+1) negative signs in a row so that no two are together is a)^m+1p_n b ^n+1p_m c)^m+1C_n d)^n+1c_m

The common ratio of the G.P. a^(m-n) , a^m , a^(m+n) is ………… .

A square matrix M of order 3 satisfies M^(2)=I-M , where I is an identity matrix of order 3. If M^(n)=5I-8M , then n is equal to _______.

The common ratio of the G.P. a^(m-n), a^(m), a^(m+n) is ……………

The number of ways in which we can distribute m n students equally among m sections is given by a. ((m n !))/(n !) b. ((m n)!)/((n !)^m) c. ((m n)!)/(m ! n !) d. (m n)^m