Question

find the sum of all two digit natural numbers divisible by 8​

Answer 1

Answer:

2 digit number divisible by 8 are

16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96

n=11

a=16

d=8

an=96

Sn=n/2(a+an)

S11=11/2(16+96)

=11/2(112)

=11(56)

S11=616

Answer 2

Answer:

The answer to the given question is the sum of all two-digit natural numbers divisible by 8 is 616

Step-by-step explanation:

Given :

Natural numbers divisible by 8

To find :

The sum of all two-digit natural numbers that are divisible by 8.

Solution :

A natural number is a number that starts from 1, 2,3 and goes on.

The two-digit natural number starts from 10 and ends at 99.

let's find the two-digit numbers that are divisible by 8.

It is found by obtaining the multiple of 8.

$16 \\ 24 \\ 32 \\ 40 \\ 48 \\ 56 \\ 64 \\ 72 \\ 80 \\ 88 \\ 96$

The total number is 11.

the sum of the numbers is found by the formula

$sn = \frac{n}{2} ( first \: term+ last \: tem)$

substituting the values in the formula we get the answers as.

$\frac{11}{2} (16 + 96) \\ \frac{11}{2}(112)$

on multiplying the above value we get the answers as

$11(56) = 616$

The value obtained is 616.

Their two-digital answer to the given question is

the sum of all two digit natural numbers divisible by 8 is 616.

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