Question

Find the sum of all two digit numbers which are divisible by 3 but not divisible by 4​

Answer 1

Answer:

1233

Step-by-step explanation:

First let's find the sum of all 2 digit numbers which are divisible by 3.

The least 2 digit multiple of 3 is 12 and the greatest is 99.

Let the multiples are in an AP.

Then the AP will be 12, 15, 18,..., 96, 99.

First term = $T_1$ = 12

$n^{th}$ term = Last term = $T_n$ = 99

Common difference   d = 3

No. of terms = n =

$\frac{99 - 12}{3} + 1 \\ \\ \frac{87}{3} = 1 \\ \\ 29 + 1 \\ \\ 30$

n = 30

Sum of terms = $S_n$ =

$\frac{30}{2}[12 + 99] \\ \\ 15 \times 111 \\ \\ 1665$

$S_n$ = 1665

Now let's find the sum of all 2 digit numbers which are divisible by 12. Because these are the 2 digit numbers which are divisible by not only 3 but also 4. We're going to deduct this sum from 1665 to get the answer. So this question is very simple!!!

The least 2 digit multiple of 12 is 12 and the greatest is 96.

Let these are in an AP.

Then it will be 12, 24, 36,..., 84, 96.

d = 12

$T_1$ = 12

$T_n$ = 96

n =

$\frac{96 - 12}{12} + 1 \\\\\frac{84}{12} + 1 \\\\7 + 1 \\\\8$

n = 8

$S_n$ =

$\frac{8}{2}[12 + 96] \\ \\ 4 \times 108 \\ \\ 432$

$S_n$ = 432

Now subtract this from 1665 to get the answer.

1665 - 432 = 1233

∴ 1233 is the answer.

Thank you. Have a nice day. :-)

#adithyasajeevan

Answer 2

1233
also tell me this answer