Question

A-4. Find the sum of all those integers between 100 and 800 each of which on division by 16 leaves the remainder 7.

Answer 1

19668

Answer 2

Let the integer be $N$. The condition is that when $N$ is divided by 16, the remainder is 7. This can be written as $N \equiv 7 \pmod{16}$, or $N = 16k + 7$ for some integer $k$.

We are looking for integers $N$ such that $100 < N < 800$. Substituting the form of $N$, we get:

$100 < 16k + 7 < 800$

Subtract 7 from all parts of the inequality:

$100 - 7 < 16k < 800 - 7$

$93 < 16k < 793$

Now, divide by 16:

$\frac{93}{16} < k < \frac{793}{16}$

$5.8125 < k < 49.5625$

Since $k$ must be an integer, the possible values for $k$ are $6, 7, 8, \dots, 49$.

The integers are $16(6)+7, 16(7)+7, \dots, 16(49)+7$.

These integers form an arithmetic progression (AP) with:

First term, $a_1 = 16(6) + 7 = 96 + 7 = 103$.

Last term, $a_n = 16(49) + 7 = 784 + 7 = 791$.

The common difference is $d = 16$.

The number of terms, $n$, is the number of integers from $k=6$ to $k=49$, which is $49 - 6 + 1 = 44$.

The sum of an arithmetic progression is given by the formula $S_n = \frac{n}{2}(a_1 + a_n)$.

Substituting the values, we get:

$S_{44} = \frac{44}{2}(103 + 791)$

$S_{44} = 22(894)$

$S_{44} = 19668$.

The sum of all such integers between 100 and 800 is 19668.