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What is the sum of the multiples of 3 from 11–110, inclusive?

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What is the sum of the multiples of 3 from 11–110, inclusive?  [#permalink]

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New post 23 Jan 2019, 01:15
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A
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Difficulty:

  35% (medium)

Question Stats:

71% (02:01) correct 29% (01:37) wrong based on 30 sessions

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Re: What is the sum of the multiples of 3 from 11–110, inclusive?  [#permalink]

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New post 23 Jan 2019, 01:33
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\(12+15+18+ ... +108\)
Last term \(T_n=108=a+(n-1)d\)
\(a=12 , d=3\)
\(108=12+(n-1)3\)
\(n-1 = 96/3 = 32\)
\(n=33\)
Total sum \(S_n=n/2 [first term + last term]\)

\(Sum = 33/2 * [12+108] = 33/2 * 120 = 1980\)
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Re: What is the sum of the multiples of 3 from 11–110, inclusive?  [#permalink]

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New post 23 Jan 2019, 06:53
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Bunuel wrote:
What is the sum of the multiples of 3 from 11–110, inclusive?

A 1,980

B 2,057

C 3.228

D 4,573

E 6,050



multiples of 3 would be 12,15... 108
total multiples ; 108-12 = 96/3 = 32+1 = 33 ; +1 since we are adding inclusive terms

sum formula
sn=n/2 + ( 2a + (n-1) d)

n = 33
a=12
d=3
s33= 33/2 + ( 2*12 + 32* 3)
= 1980
IMO A
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Re: What is the sum of the multiples of 3 from 11–110, inclusive?  [#permalink]

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New post 27 Jan 2019, 18:20
Bunuel wrote:
What is the sum of the multiples of 3 from 11–110, inclusive?

A 1,980

B 2,057

C 3.228

D 4,573

E 6,050


We find the greatest multiple of 3 and the smallest multiple of 3 for the range of values given. We see that the average of the multiples of 3 from 11 to 110, inclusive, is:

(108 + 12)/2 = 60

The number of multiples of 3 from 11 to 110, inclusive, is:

(108 - 12)/3 + 1 = 96/3 + 1 = 33

Since sum = average x quantity, the sum of those multiples is:

60 x 33 = 1,980

Answer: A
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Re: What is the sum of the multiples of 3 from 11–110, inclusive?   [#permalink] 27 Jan 2019, 18:20
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