Question

15) Find the sum of integers from 1 to 100 which are divisible 2 or 3​

Answer 1

Answer:

From 1 to 100, there are 33 multiples of 3 and 20 multiples of 5. Multiples of 3 are in A.P. and sum of the sequence is given by (n/2)(first term + last term), where n= no. of terms. For multiples of 3, n=33, first term= 3 and last term= 99. Similarly we calculate for multiples of 5, where n=20, 1st term= 5 and last term= 100. However there are certain number that are multiples of both 3 & 5, so we need to subtract them and those are multiples of 15. For multiples of 15, n=6, 1st term= 15, last term=90. So using the formula mentioned above for A.P., we can solve. Required Sum = (Sum of multiples of 3) + (Sum of multiples of 5) - (Sum of multiples of 15)

Step-by-step explanation:

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Answer 2

Sum of the integers that are divisible by 2:

2+4+6+?+100

=2(1+2+3+?+50)

=2(50∗51)/2

=2550

Sum of the integers that are divisible by 3:

3+6+9+?+99

=3(1+2+3+?+33)

=3(33∗34)/2

=1683

Now, the sum of integers that are divisible by both 2 and 3:

6+12+18+?+96

=6(1+2+3+?+16)

=6∗(16∗17)/2

=816

Now from the set theory:

n(A∩B) = n(A)+n(B) - n(A∪B)

= 2550 + 1683 - 816

= 3417

so answer is 3417

hope it helps

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