Question

how many two digit numbers leave the remainder 1 when divided by 5 ( in detail A.P)

Answer 1

The two-digit numbers which leave the remainder 1 when divided by 5 are ;

11,16,21,26,.......86,91,96

In this A.P :
a = 11
d = 16-11 = 5
l = 96
n = ?

Tn = a+(n-1)d

96 = 11+(n-1)5
n-1 = 85/5
n = 17+1
n = 18
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Hence their are 18 two digit numbers which when divided by 5 leaves remainder 1.

Answer 2

Given,

The numbers are in AP

Numbers, when divided by 5, leave the remainder 1

The numbers in AP are 2 digit numbers

To Find,

Total numbers  =?

Solution,

First 2 digit number = 10

a = 10 + 1 [It will leave 1 as the remainder when divided by 5]

Last digit of AP <= 99 [100 is a 3 digit number ]

The last number divisible by 5 before 100 is 95

Therefore, the last number of the AP  = 96 [It will leave 1 as the remainder when divided by 5]

The common difference in the AP = 5

AP = 11 , 16 , 21 , ........ 96

From the formula of the nth term in AP, we have

$a_ n = a +(n-1)d$

We have to find n, for the last term that is

$96 = 11 +(n-1)5\\96 - 11 = (n-1)5\\(n-1)5 = 85\\n-1 = 85 / 5\\n-1 = 17\\n = 17 + 1\\n = 18$

Last place in the AP is 18

Hence, there are 18  two-digit numbers leaving the remainder 1 when divided by 5 in AP.