Question

How many 3 digit numbers are there in which leaves reminder 2 on division by 5

Answer 1

Answer:

Step-by-step explanation:no .of three digit 5 multiples a nth=a1 +(n-1)*d

995=100+(n-1)*5

895/5=n-1

179+1=n=180

180 numbers divisible by 5

And if 2 added to each multiple __ 180 leavingvremainder 2

Answer 2

Answer:

Since numbers leave a remainder 2, so it should be in form of 5q + 2, q be any integer.

First 3 digit number is 100, which is divisible by 5. So the required number should be 100 + 2 = 102.

Last 3 digit number divisible by 5 = 995. So the required number is 995 + 2 = 997.

This forms an AP with first term 102 and last term 997 and common difference 5.

In APs nth term is given by a + ( n - 1 )d

.

Let the required number of terms be n.

→ 997 = 102 + ( n - 1 )5

→ 895 = ( n - 1 )5

→ 179 = n - 1

→ 180 = n

Number of terms is 180.