Question

Find the sum of all two digit number divisible by 4 or 3​

Answer 1

Answer

Answer: The sum of 2 digit numbers which are divisible by 3 and not divisible by 4 is 1233.

Answer 2

Answer:

$\sf{The \ sum \ of \ two \ digit \ numbers \ which}$

$\sf{are \ divisible \ by \ 4 \ or \ 3 \ is \ 432.}$

To find:

$\sf{The \ sum \ of \ all \ two \ digit \ number \ divisible}$

$\sf{by \ 4 \ or \ 3.}$

Solution:

$\sf{Least \ common \ multiple \ of \ 4 \ and \ 3}$

$\sf{is \ 12.}$

$\sf{Multiples \ of \ 12 \ which \ are \ two \ digits}$

$\sf{numbers \ are}$

$\sf{12, \ 24, \ ..., \ 96}$

$\sf{It \ forms \ an \ A.P.}$

$\sf{Here, \ t_{1}=a=12, \ d=24-12=12, \ t_{n}=96}$

$\boxed{\sf{t_{n}=a+(n-1)d}}$

$\sf{\therefore{96=12+(n-1)\times12}}$

$\sf{\therefore{12(n-1)=84}}$

$\sf{\therefore{n-1=\dfrac{84}{12}}}$

$\sf{\therefore{n=7+1}}$

$\sf{\therefore{n=8}}$

$\sf{Their \ are \ 8 \ two \ digits \ numbers}$

$\sf{which \ are \ divisible \ by \ 4 \ or \ 3}$

$\boxed{\sf{S_{n}=\dfrac{n}{2}[t_{1}+t_{n}]}}$

$\sf{\therefore{S_{8}=\dfrac{8}{2}[12+96]}}$

$\sf{\therefore{S_{8}=4\times108}}$

$\sf{\therefore{S_{8}=432}}$

$\sf\purple{\tt{\therefore{The \ sum \ of \ two \ digit \ numbers \ which}}}$

$\sf\purple{\tt{are \ divisible \ by \ 4 \ or \ 3 \ is \ 432.}}$