## How many two-digit numbers are divisible by 4 and which leaves the remainder 1.​

Question

How many two-digit numbers are divisible by 4 and which leaves the remainder 1.​

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1 month 2021-08-18T12:14:25+00:00 1 Answer 0 views 0

Step-by-step explanation:

This question can be solved by using AP.

First two digit number which leaves a remainder of 1 when divided by 4 is 13. Let this term be denoted by a

Similarly the next two digit number leaves a when divided by 4 is 17. So the common difference is 17–13 = 4. Let this common difference be denoted by d

In the same pattern last two digit number which leaves a remainder of 1 when divided by 4 is 97. Let this term be denoted by l

And we know that last term of an AP is given by l=a+((n-1)*d) where n is the number of terms in the sequence which is to be found out.

Substituting the values we get,

97=13+((n-1 )*4)

So. 97–13=84=(n-1)*4

So n-1 =84/4 = 21

So n= 22

So there are 22 different two digit numbers which leaves a remainder of 1 when divided by 4

Hope it helps.

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