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12 Mar 2020, 02:07
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C
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Difficulty:
85%
(hard)
Question Stats:
56% (02:59) correct
44%
(03:12)
wrong
based on 230
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What is the sum of the integers between 1 and 100, inclusive, which are divisible by 3 or 5 or 7?
A. 818
B. 1828
C. 2838
D. 3848
E. 4848
Are You Up For the Challenge: 700 Level Questions
A. 818
B. 1828
C. 2838
D. 3848
E. 4848
Are You Up For the Challenge: 700 Level Questions
12 Mar 2020, 02:41
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Sum of multiples of 3= \((\frac{99}{3})[\frac{(3+99)}{2}]\)= 1683
Sum of multiples of 5 = \((\frac{100}{5})[\frac{(5+100)}{2}]\)= 1050
Sum of multiples of 7 = \((\frac{98}{7})[\frac{(7+98)}{2}]\) = 735
Sum of multiples of 15 = \((\frac{90}{15})[\frac{(15+90)}{2}]\)= 315
Sum of multiples of 21 = \((\frac{84}{21})[\frac{(21+84)}{2}]\)= 210
Sum of multiples of 35 = \((\frac{70}{35})[\frac{(35+70)}{2}]\)= 105
Sum of multiples of 105 = 0
sum of the integers between 1 and 100, inclusive, which are divisible by 3 or 5 or 7= 1683+ 1050+735-315-210-105 + 0= 2838
Sum of multiples of 5 = \((\frac{100}{5})[\frac{(5+100)}{2}]\)= 1050
Sum of multiples of 7 = \((\frac{98}{7})[\frac{(7+98)}{2}]\) = 735
Sum of multiples of 15 = \((\frac{90}{15})[\frac{(15+90)}{2}]\)= 315
Sum of multiples of 21 = \((\frac{84}{21})[\frac{(21+84)}{2}]\)= 210
Sum of multiples of 35 = \((\frac{70}{35})[\frac{(35+70)}{2}]\)= 105
Sum of multiples of 105 = 0
sum of the integers between 1 and 100, inclusive, which are divisible by 3 or 5 or 7= 1683+ 1050+735-315-210-105 + 0= 2838
Bunuel
04 Sep 2020, 08:53
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If you add just the multiples of 3, you get (33)(3 + 99)/2 = 1683. We need to add several large numbers to this (98, 95, 91, etc) so the right answer cannot be A or B.
If we add the multiples of 5, we get (20)(105/2) = 1050, and if we add the multiples of 7 we get (14)(105/2) = 735. If we now just add these three numbers, 1683+1050+735, which gives roughly 3500, we clearly get an answer that is too large, because we are double-counting several things (15, for example, is added twice, because it is a multiple of 3 and of 5). So the correct answer is less than 3400, and since we ruled out A and B, only C could be right.
The answers are so far apart here that estimation is likely going to be a faster approach than exact computation (I would have estimated all the sums above too, but that would make the explanation harder to follow). You definitely don't need to know any of those 3-set Venn formulas for actual GMAT questions.
If we add the multiples of 5, we get (20)(105/2) = 1050, and if we add the multiples of 7 we get (14)(105/2) = 735. If we now just add these three numbers, 1683+1050+735, which gives roughly 3500, we clearly get an answer that is too large, because we are double-counting several things (15, for example, is added twice, because it is a multiple of 3 and of 5). So the correct answer is less than 3400, and since we ruled out A and B, only C could be right.
The answers are so far apart here that estimation is likely going to be a faster approach than exact computation (I would have estimated all the sums above too, but that would make the explanation harder to follow). You definitely don't need to know any of those 3-set Venn formulas for actual GMAT questions.
