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What is the sum of the integers between 1 and 100, inclusive, which ar 12 Mar 2020, 02:07
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C
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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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C
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What is the sum of the integers between 1 and 100, inclusive, which ar 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
Bunuel
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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What is the sum of the integers between 1 and 100, inclusive, which ar 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.
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What is the sum of the integers between 1 and 100, inclusive, which ar 15 Mar 2020, 04:56
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Bunuel
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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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

Answer: C
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What is the sum of the integers between 1 and 100, inclusive, which ar 04 Sep 2020, 06:38
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Can anyone explain this method?
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What is the sum of the integers between 1 and 100, inclusive, which ar 04 Sep 2020, 07:02
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Can anyone explain this method?
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    • An approach is already discussed above so I am trying to give an alternate one.

Given

    • All the integers between 1 and 100, inclusive.


To Find

    • The sum of the integers that are divisible by 3 or 5 or 7.



Approach and Working Out

    • We need to take all the numbers that are multiples of any of these three numbers.

    • It means we need to remove all the numbers which are not multiples of 3, 5, and 7.
      o So these are the numbers which are co-primes to 3, 5, or 7.

    • Take LCM of 3,5, and 7.
      o It is 105.

    • It is easier to calculate till 105 so we will do that.

    • Number of co-primes to 3,5 and 7 = 105 (1 – \(\frac{1}{3}\)) (1 – \(\frac{1}{5}\)) (1 – \(\frac{1}{7}\))
      o = 2 × 4 × 6
      o = 48

    • Now observe this,
      o The least such number is 1 and the highest is 104.
      o The second least is 2 and the second highest is 103.
      o So, we can a pair which adds up to 105.
      o The sum will be \(\frac{48}{2 }\)× 105 = 2520

    • Sum of all the numbers from 1 to 105 = \(\frac{(105 × 106)}{2} \)= 5565
      o So, our answer would have been 5565 – 2520 = 3045 had it been till 105.

    • Actual answer = 3045 – 105 – 102
      o = 2838


Correct Answer: Option C
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What is the sum of the integers between 1 and 100, inclusive, which ar 04 Sep 2020, 07:45
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Divisible by 3 or 5 or 7: It means divisible by 3 or 5 or 7 or 15 or 21 or 35 or LCM of (3,5,7) 105

Number of terms divisible by 3 from 1 to 100:

=> 99 = 3 + ( n-1) 3
=> \(\frac{99 }{ 3}\) = n
=> n = 33

Number of terms divisible by 5 from 1 to 100:

=> 100 = 5 + ( n-1) 5
=> \(\frac{100 }{ 5}\) = n
=> n = 20

Number of terms divisible by 7 from 1 to 100:

=> 98 = 7 + ( n-1) 7
=> \(\frac{98 }{ 7}\) = n
=> n = 14

Similarly,

=> for 15 => n = \(\frac{90}{15}\) = 6

=> for 21 => n = \(\frac{84}{21}\) = 4

=> for 35 => n = \(\frac{70}{35}\) = 2

=> for 105 => n = 0


Sum = \(\frac{n}{2}\)[first term + last term]

=> for 3 = \(\frac{33}{2}\) [3+99] = 1,683

=> for 5 = \(\frac{20}{2}\) [5+100] = 1,050

=> for 7 = \(\frac{14}{2}\) [7+98] = 735

=> for 15 = \(\frac{6}{2}\) [15+90] = 315

=> for 21 = \(\frac{4}{2}\) [21+84] = 210

=> for 35 = \(\frac{2}{2}\) [35+70] = 105

=> for 105 = 0

Total: 1683 + 1050 + 735 - 315 - 210 - 105 - 0 = 2838

Answer C
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What is the sum of the integers between 1 and 100, inclusive, which ar 25 Nov 2024, 19:05
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