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What is the sum of integers from 1 to 100 divisible by 2 or 5?

What is the sum of integers from 1 to 100 divisible by 2 or 5?
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ReportAnswer #1

The sum is #3050#.

Explanation:

Ths sum of arithmetric progression is
#S=n/2(a+l)#, where #n# is the number of terms, #a# is the first term and #l# is the last term.

The sum of integres #1# to #100# which is divisible by #2# is
#S_2=2+4+6+…100 = 50/2*(2+100)=2550#
and, the sum of integers divisible by #5# is
#S_5=5+10+15+…100 =20/2*(5+100)=1050#

You may think the answer is #S_2+S_5=2550+1050=3600# but this is wrong.

#2+4+6+…100# and #5+10+15+…100# have common terms.
They are integers divisible by #10#, and their sum is
#S_10=10+20+30+…100=10/2*(10+100)=550#

Therefore, the answer for this question is #S_2+S_5-S_10=2550+1050-550=3050#.

answered by: yumean1119
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