Question

how many numbers of two digits are divisible by 3

Answer 1

Lets write the numbers of two digits which are divisible by 3.

12,15,18,21,24,27......................99.!

This is in the form of AP.

First term a = 12
Common difference d = a2-a1 = 15-12 = 3
Last term an = 99

Use the formula, an = a +( n-1 )d

99 = 12 + (n-1) 3

99-12 = (n-1)3

87 = (n-1)3

n-1 = 87/3

n-1 = 29

n = 29+1

n = 30

Therefore there are 30 two digit number which are divisible by 3.

:)

Answer 2

Answer:

$\huge{\pink{\boxed{\green{\sf{n=30}}}}}$

Step-by-step explanation:

$\huge{\pink{\boxed{\green{\underline{\red{\sf{SOLUTION-}}}}}}}$

☆ A.P =12,15,18,21.........,96,99

${\orange{given}} \\ { \pink{ \boxed{ \green{a = 12}}}} \\ { \pink{ \boxed{ \green{d = 3}}}} \\ { \pink{ \boxed{ \green{ a_{n} = 99}}}} \\ \\ { \blue{to \: find}} \\ { \purple{ \boxed{ \red{n =? }}}}$

☆ According to given question:

$\to a_{n} = a +( n - 1)d \\ \to 99 = 12 + (n - 1) \times 3 \\ \to 99 - 12 = 3n - 3 \\ \to87 = 3n - 3 \\ \to87 + 3 = 3n \\ \to 90 = 3n \\ \ \to n = \frac{90}{3} \\ { \pink{ \boxed{ \green{\therefore n = 30 }}}}$

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