Question

{40 points}

Find the sum of 2 digit number which are divisible by 3 and not divisible by 4​

Answer 1

Question: Find the sum of 2 digit numbers which are divisible by 3 and not divisible by 4​.

Solution:

ATQ, we have to find the sum of 2-digit numbers that are divisible by 3, but not by 4.

It can be found out by finding the sum of two digit numbers divisible by 3, and subtracting the sum of two digit numbers divisible by 4 & 3 from it.

Finding the sum of 2-digit numbers divisble by 3.

• A.P $\longmapsto$ 12, 15, 18, 21 . . . . . 99.

• a  $\longmapsto$ 12

• d  $\longmapsto$  3

• a$\sf_{n}$  $\longmapsto$ 99

We know that,

⇒ a$\sf_{n}$ = a + (n - 1)d

⇒ 99 = 12 + (n - 1)3

⇒ 99 - 12 = (n - 1)3

⇒ 87 = 3n - 3

⇒ 87 + 3 = 3n

⇒ 90 = 3n

⇒ n = 30

Sum of 2-digit numbers divisble by 3 is,

$\Longrightarrow \sf S_{30} = \dfrac{n}{2} \ \bigg(2a + (n-1)d\bigg)\\ \\ \\ \\\Longrightarrow \sf S_{30} = \dfrac{30}{2} \ \bigg(2(12) + (30-1)3\bigg)\\ \\ \\ \\\Longrightarrow \sf S_{30} = 15 \ \bigg(24 + (29)3\bigg)\\ \\ \\ \\\Longrightarrow \sf S_{30} = 15 \ \bigg(24 + 87\bigg)\\ \\ \\ \\\Longrightarrow \sf S_{30} = 1665$

Finding the sum of 2-digit numbers divisble by 4, that are divisble by 3 as well.

• A.P $\longmapsto$ 12, 24, 36 . . . . . . 96.

• a  $\longmapsto$ 12

• d  $\longmapsto$  12

• a$\sf_{n}$  $\longmapsto$ 96

We know that,

⇒ a$\sf_{n}$ = a + (n - 1)d

⇒ 96 = 12 + (n - 1)12

⇒ 96 - 12 = (n - 1)12

⇒ 84 = 12n - 12

⇒ 84 + 12 = 12n

⇒ 96 = 12n

⇒ n = 8

Sum of 2-digit numbers divisble by 4 that are divisble by 3 as well is,

$\Longrightarrow \sf S_8 = \dfrac{n}{2} \ \bigg(2a + (n-1)d\bigg)\\ \\ \\ \\\Longrightarrow \sf S_8 = \dfrac{8}{2} \ \bigg(2(12) + (8-1)12\bigg)\\ \\ \\ \\\Longrightarrow \sf S_8 = 4 \ \bigg(24 + (7)12\bigg)\\ \\ \\ \\\Longrightarrow \sf S_8 = 4 \ \bigg(24 + 84\bigg)\\ \\ \\ \\\Longrightarrow \sf S_8 = 432$

Therefore,

⇒ Sum of 2-digit no's. that are divisble by 3 but not 4 = S$\sf _{30}$ - S$\sf _{8}$

⇒ Sum of 2-digit no's. that are divisble by 3 but not 4 = 1665 - 432

⇒ Sum of 2-digit no's. that are divisble by 3 but not 4 = 1233

$\boxed{\boxed{\sf ANSWER: 1233}}$

Answer 2

Answer:

$\huge\pink{\mathfrak{Hello!!!}}$

Let's make an A.P

Sum of all 2 -digit number divisible by 3 be SnSn

Sn=a+a+d+.....a+(n−1)dSn=a+a+d+.....a+(n−1)d

a=12;d=3a=12;d=3

First number is 12 and last is 99

So 99=12+(n−1)d==>n=3099=12+(n−1)d==>n=30

HENCE Sn=30×12+29×30×32=360+29×15×3=1665.Sn=30×12+29×30×32=360+29×15×3=1665.

The numbers which are divisible by 4 and 3 are 4×3,4×2×3,....,4×8×34×3,4×2×3,....,4×8×3

Let Sn1Sn1 denote the sum of The numbers which are divisible by 4 and 3

Sn1=4×3+4×2×3......+4×8×3=12(1+2+....+8)=12((9)(8)2)=12(36)=12×36Sn1=4×3+4×2×3......+4×8×3=12(1+2+....+8)=12((9)(8)2)=12(36)=12×36

The sum of numbers divisible by 3 but not by 4 is Sn−Sn1=1665−432=1233.Sn−Sn1=1665−432=1233.

Answer 1233