Find the sum of all natural numbers between 1 and 100 which are not exactly divisible by 3 or 7
Answers
The question can be solved using the concept of set theory,
Numbers that are divisible by 3, 5 and 7 can be represented as n(3), n(5) and n(7) respectively.
Between 0 to 500,
n(3) = 500/6= 166
n(5) = 500/5 = 100
n(7)= 500/7 = 71
Total numbers = n(3) + n(5) +n(7)= 166+100+71 = 337
Here, we have calculated numbers that are divisible by 3,5 and 7 individually.
There are some numbers that have been repetitively counted, in order to eliminate them we have to find out the common numbers.
Numbers that are divisible by 3 and 5 = n(3&5) = 500/ (LCM of 3 & 5) = 500/15 = 33
Numbers that are divisible by 5 and 7 = n(5&7) = 500/ (LCM of 5 & 7) = 500/35 = 14
Numbers that are divisible by 3 and 7 = n(3&7) = 500/ (LCM of 3 & 7) = 500/21 = 23
Numbers that are divisible by 3,5 and 7 = n(3&5&7) = 500/(LCM of 3&5&7) = 500/ 105 = 4
Formula of set theory can be used,
n(A or B or C) = n(A)+n(B)+n(C)-n(A&B)-n(B&C)-n(A&C)+n(A&B&C)
Substituting the values in the given formula, we obtain
The actual numbers that are divisible by 3 or 5 or 7 = 337–33–14–23+4 = 271
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Ashutosh Sabnekar
Ashutosh Sabnekar
Answered Aug 13, 2017 · Author has 246 answers and 554.2k answer views
A = numbers between 0 - 500 to be exactly divisible by 3 = 166.
B = numbers between 0 - 500 to be exactly divisible by 5 = 100.
C = numbers between 0 - 500 to be exactly divisible by 7 = 71.
A & B = number between 0–500 to be exactly divisible by 3 & 5 = 33
B & C = number between 0–500 to be exactly divisible by 7 & 5 = 14
C & A = number between 0–500 to be exactly divisible by 3 & 7 = 23
A & B & C = number between 0–500 to be exactly divisible by 3 & 5 & 7= 4
Using set theory, we get :
number between 0–500 to be exactly divisible by 3 or 5 or 7 is :
A + B + C - (A & B) - (B & C) - (C & A) + - (A & B & C)
= 166 + 100 + 71 - 33 - 14 - 23 + 4
= 341 - 70 = 271.
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