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# Find the sum of all the integers from $1$ and $100$ that are divisible by $2$ or $5$.

$\begin{array}{1 1}3050 \\ 3600 \\ 4150 \\ 3750 \end{array}$

Toolbox:
• $n^{th}$ term of an A.P.$t_n=a+(n-1)d$
• Sum of $n$ terms of the A.P. =$\large\frac{n}{2}$$(t_n+a) Numbers from 1 and 100 that are divisible by 2 and 5 are 2,4,5,6,8,10,12,14,15,........... These numbers are grouped into three groups, 2,4,6,8,10,...........100...........(i) 5,10,15,20,........100........(ii) and 10,20,30.........100.......(iii) Here group (iii) contains numbers that are common in both group (i) and (ii) \therefore\:Sum of the required numbers = um of (i) + sum of (ii) - sum of (iii). (i) is an A.P. with 1^{st} term a=2, common difference d=2 and t_n=100 \Rightarrow\:a+(n-1)d=100 \Rightarrow\:2+(n-1)2=100 or n-1=49 or n=50 \therefore Sum of the numbers in ((i) = \large\frac{n}{2}$$(t_n+a)=25(100+2)=2550$.
Step 2
(ii) is an A.P. with $a=5$, $d=5$ and $t_n=100$
$\Rightarrow\:a+(n-1)d=100$
$\Rightarrow\:5+(n-1)5=100$ or $n=19+1=20$
$\therefore$ Sum of the numbers in (ii) $=\large\frac{n}{2}$$(t_n+a)=10(100+5)=1050. Step 3 Similarly (ii) is also an A.P. with a=10, d=10 and t_n=100 \Rightarrow\:a+(n-1)d=100 \Rightarrow\:10+(n-1)10=100 or n=10 \therefore Sum of the numbers in (iii) =\large\frac{n}{2}$$(t_n+a)=5(100+10)=550$
Step 4
$\therefore\:$ Sum of the required numbers
$=2550+1050-550=3050$.
edited Mar 29, 2014