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# Find the number of natural numbers between 10 and 300 each of which gives remainder 3 when divided by 4.

Given:

Numbers between 10 and 300.

To do:

We have to find the number of numbers that lie between 10 and 300, which when divided by 4 leave a remainder 3.

Solution:

We know that,

nth term $a_n= a + (n - 1)d$

Numbers which when divided by 4 that leave a remainder 3 are $(4+3), (8+3), (12+3), (16+3), ......, (296+3), (300+3), ......$

Numbers between 10 and 300 which when divided by 4 that leave a remainder 3 are $11, 15, 19, ......, 299$

Here, first number in the series is 7 and the last number is 299.

Clearly, it is an arithmetic progression with first term $a=7$ and the common difference $d=4$.

Let there are $n$ numbers in the series.

So, $299$ will be the nth term

This implies,

$299=11 + (n-1)\times4$

$4n-4= 299–11$

$4n=288+4$

$4n=292$

$n=\frac{292}{4}$

$n=73$.

Therefore, 73 numbers lie between 10 and 300, which when divided by 4 leave a remainder 3.