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# What is the sum of the integers between 1 and 100, inclusive, which ar

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Joined: 02 Sep 2009
Posts: 64068
What is the sum of the integers between 1 and 100, inclusive, which ar  [#permalink]

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12 Mar 2020, 01:07
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55% (hard)

Question Stats:

67% (02:28) correct 33% (03:08) wrong based on 27 sessions

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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

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Re: What is the sum of the integers between 1 and 100, inclusive, which ar  [#permalink]

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12 Mar 2020, 01:41
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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

Bunuel wrote:
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

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Re: What is the sum of the integers between 1 and 100, inclusive, which ar  [#permalink]

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15 Mar 2020, 03:56
Bunuel wrote:
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

We can use the formula (note the notation s(m(3)), for example, means the sum of multiples of 3 between 1 and 100, inclusive):

Sum = s(m(3)) + s(m(5)) + s(m(7)) - s(m(3 & 5)) - s(m(3 & 7)) - s(m(5 & 7)) + s(m(3 & 5 & 7))

Now, let’s calculate each term in the above formula, using the general formula: sum = average x number:

s(m(3)) = (99 + 3)/2 x [(99 - 3)/3 + 1] = 51 x 33 = 1683

s(m(5)) = (100 + 5)/2 x [(100 - 5)/5 + 1] = 105/2 x 20 = 1050

s(m(7)) = (98 + 7)/2 x [(98 - 7)/7 + 1] = 105/2 x 14 = 735

s(m(3 & 5)) = (90 + 15)/2 x [(90 - 15)/15 + 1] = 105/2 x 6 = 315

s(m(3 & 7)) = (84 + 21)/2 x [(84 - 21)/21 + 1] = 105/2 x 4 = 210

s(m(5 & 7)) = (70 + 35)/2 x [(70 - 35)/35 + 1] = 105/2 x 2 = 105

s(m(3 & 5 & 7)) = 0 (since LCM(3, 5, 7) = 105, which is already greater than 100)
Therefore, the desired sum is:

1683 + 1050 + 735 - 315 - 210 - 105 + 0 = 3468 - 630 = 2838

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Re: What is the sum of the integers between 1 and 100, inclusive, which ar   [#permalink] 15 Mar 2020, 03:56