Question

Find the sum of all 2-digit numbers which give 3 as remainder when divided by
8.
0 663
O 660
O 559
O 700​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

The correct answer is option (2)660

Step-by-step explanation:

To find,

The sum of all 2-digit numbers which gives 3 as the remainder when divided by 8

Recall the formula

The number of terms of an AP = n = $\frac{t_n - t_1}{d} +1$,

Sum to n- terms of an AP = $S_n = \frac{n}{2} [2t_1+(n-1)d]$

where $t_n$ is the last term $t_1$ is the first term, d is the common difference of the AP

Solution:

The numbers which give the reminder 3 when divisible by 3 is of the form 8n+3

The first two digit number of the form 8n+3 = 8×1+3 = 11

Then numbers which give the remainder 3 when divided by 8 are

11, 19, 27,...................,

To get the last two-digit number in this set

The largest two-digit number = 99

Divide 99 by 8 we get

99 = 8×12 +3, which is of the form 8n+3

The last member in this set is 1, 19, 27,..................., is 99

Hence, the two-digit numbers which give the remainder 3 when divided by 8 are

A = {11, 19, 27,...................,99}

This forms an AP, with first term  $t_1$  = 11 last term  $t_n$ = 99 and common difference = 8

To find the number of terms in this set  A

We have,

Number of terms, n =  $\frac{t_n - t_1}{d} +1$

Substituting the values we get

n = $\frac{99 - 11}{8} +1$

= $\frac{88}{8} +1$

= 12

The number of terms in the set A = 12

To find the sum of elements in the set A

We have  $S_n = \frac{n}{2} [2t_1+(n-1)d]$

Substituting the values, we get

$S_{12}= \frac{12}{2} [2X11+(12-1)8]$

= 6(22 + 11×8)

= 6(22+88)

= 6×110

= 660

∴ The sum of all 2-digit numbers which gives 3 the as the remainder when divided by 8 = 660

The correct answer is option (2)660

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