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# What is the sum of all multiples of 3 between 1 and 200?

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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What is the sum of all multiples of 3 between 1 and 200?  [#permalink]

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28 Aug 2018, 02:07
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Question Stats:

76% (01:40) correct 24% (01:36) wrong based on 80 sessions

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[Math Revolution GMAT math practice question]

What is the sum of all multiples of 3 between 1 and 200?

A. 3300
B. 6600
C. 6633
D. 10100
E. 20100

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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" NUS School Moderator Joined: 18 Jul 2018 Posts: 1021 Location: India Concentration: Finance, Marketing WE: Engineering (Energy and Utilities) Re: What is the sum of all multiples of 3 between 1 and 200? [#permalink] ### Show Tags 28 Aug 2018, 02:31 1 1 $$T_n$$ (Last Term) = a+(n-1)d a - First term d - Common Difference n - Number of terms The multiples of 3 from 1 to 200 are 3,6,9,12,....,198 a = 3 d = 3 $$T_n$$ = 198 Now, Using the formula 198 = 3+(n-1)3 Solving Gives, n = 66 Now, Sum of all the terms in AP is given by $$S_n$$ = $$\frac{n}{2}$$[2a+(n-1)d] $$S_{66}$$ = $$\frac{66}{2}$$[2*3+(66-1)3] $$S_{66}$$ = 33[6+195] $$S_{66}$$ = 6633 C is the answer. _________________ Press +1 Kudos If my post helps! VP Joined: 09 Mar 2016 Posts: 1230 What is the sum of all multiples of 3 between 1 and 200? [#permalink] ### Show Tags Updated on: 28 Aug 2018, 03:20 MathRevolution wrote: [Math Revolution GMAT math practice question] What is the sum of all multiples of 3 between 1 and 200? A. 3300 B. 6600 C. 6633 D. 10100 E. 20100 here is my fantastic approach Step One: NUMBER OF MULTIPLES using the formula below NUMBER OF MULTIPLES X IN THE RANGE $$\frac{last..multiple..of..x - first..multiple..of...x}{x}+1$$ $$\frac{198-3}{3}+1 = 66$$ Step Two: FIND THE SUM OF TERMS SUM OF N TERMS = $$\frac{n}{2} (2a+(n-1)d)$$ $$a$$ = 3 the first term $$d$$ = 3 distance $$n$$ = 66 how many terms to add up $$\frac{66}{2} (2*3+(66-1)3)$$= $$6,633$$ For those who are curious and want to learn more about similar formulas, check this post https://gmatclub.com/forum/arithmetic-p ... 61976.html Originally posted by dave13 on 28 Aug 2018, 03:17. Last edited by dave13 on 28 Aug 2018, 03:20, edited 1 time in total. Director Joined: 06 Jan 2015 Posts: 689 Location: India Concentration: Operations, Finance GPA: 3.35 WE: Information Technology (Computer Software) Re: What is the sum of all multiples of 3 between 1 and 200? [#permalink] ### Show Tags 28 Aug 2018, 03:20 MathRevolution wrote: [Math Revolution GMAT math practice question] What is the sum of all multiples of 3 between 1 and 200? A. 3300 B. 6600 C. 6633 D. 10100 E. 20100 n=(198-3)/3 +1 = 66 Sum = (first term + last term)*(n/2) = (3+198)66/2 = 6633 Hence C _________________ आत्मनॊ मोक्षार्थम् जगद्धिताय च Resource: GMATPrep RCs With Solution Manager Joined: 20 Feb 2017 Posts: 162 Location: India Concentration: Operations, Strategy WE: Engineering (Other) Re: What is the sum of all multiples of 3 between 1 and 200? [#permalink] ### Show Tags 28 Aug 2018, 04:05 2 I saw the timer: it shows around 4 minutes, it took me 1 minute Looking at the question you should know it's asking the sum of an AP the series starts from 3 and ends with 198 (both inclusive) Hence number of terms : 198-3/3+1 = 66 the sum = n/2(first term +last term ) =66/2(3+198) = 33*201 (you don't' need to multiply here, check the last digit, it must end with 3, hence answer is C ) _________________ If you feel the post helped you then do send me the kudos (damn theya re more valuable than$)
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What is the sum of all multiples of 3 between 1 and 200?  [#permalink]

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28 Aug 2018, 04:26
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Raksat wrote:
I saw the timer: it shows around 4 minutes

it was meeeeeeeeeee
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What is the sum of all multiples of 3 between 1 and 200?  [#permalink]

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28 Aug 2018, 05:54
1
MathRevolution wrote:
[Math Revolution GMAT math practice question]

What is the sum of all multiples of 3 between 1 and 200?

A. 3300
B. 6600
C. 6633
D. 10100
E. 20100

First Multiple = 3

Last Multiple = 198

3 + 6 + 9 + 12 + 15 + 18 + 21 ...................+ 198.

= 3 ( 1 + 2 + 3 + 4 + 5 + 6.........................+ 66)

Now we can find out the sum of consecutive integers up to 66 as follows;

$$\frac{n(n+1)}{2}$$

= $$\frac{66(66+1)}{2}$$

= 33*67

Just multiply the unit digits. 3*7*3 = 3

Only option is C.

A 40 sec approach .
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Re: What is the sum of all multiples of 3 between 1 and 200?  [#permalink]

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28 Aug 2018, 06:39
Top Contributor
MathRevolution wrote:
[Math Revolution GMAT math practice question]

What is the sum of all multiples of 3 between 1 and 200?

A. 3300
B. 6600
C. 6633
D. 10100
E. 20100

Useful formula: 1 + 2 + 3 + 4 + . . . n = (n)(n + 1)/2

We want to find the sum: 3 + 6 + 9 + 12 + . . . 198
Factor out the 3 to get: 3(1 + 2 + 3 + 4 + . . . 66)
Apply formula to get: 3(66)(67)/2
Evaluate to get: 6633

Cheers,
Brent
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Re: What is the sum of all multiples of 3 between 1 and 200?  [#permalink]

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28 Aug 2018, 07:24
MathRevolution wrote:
What is the sum of all multiples of 3 between 1 and 200?

A. 3300
B. 6600
C. 6633
D. 10100
E. 20100

? = 3+6+9+ ... + 195 + 198

In any finite arithmetic progression, the (global) average is equal to the median and to the average of symmetric elements (to the median).

The first and last elements are always symmetrical, hence their average (3+198)/2 is equal to the (global) average.

How many terms are involved?

$$3 \leqslant 3M \leqslant 198\,\,\left( { = 180 + 18} \right)\,\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{:\,\,\,3} \,\,\,\,\,\,1 \leqslant M\,\,\, \leqslant \,\,\,66\,\,\,\,\, \Rightarrow \,\,\,\,66\,$$

Using the homogeneity nature of the average we have:

$$? = 66 \cdot \frac{{\left( {3 + 198} \right)}}{2} = 33 \cdot 201\,\,\,\,\,\mathop \Rightarrow \limits^{choices} \,\,\,\,\,\left\langle ? \right\rangle = \left\langle {33 \cdot 201} \right\rangle = \left\langle 3 \right\rangle$$

$$\left\langle N \right\rangle \,\,\, = \,\,\,\,{\text{units}}\,\,{\text{digit}}\,\,{\text{of}}\,\,N$$

The above follows the notations and rationale taught in the GMATH method.
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Re: What is the sum of all multiples of 3 between 1 and 200?  [#permalink]

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28 Aug 2018, 12:27
MathRevolution wrote:
[Math Revolution GMAT math practice question]

What is the sum of all multiples of 3 between 1 and 200?

A. 3300
B. 6600
C. 6633
D. 10100
E. 20100

greatest multiple of 3=198
198/3=66 total multiples of 3
(198+3)/2=100.5
100.5*66=6633
C
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Re: What is the sum of all multiples of 3 between 1 and 200?  [#permalink]

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30 Aug 2018, 00:51
=>

We need to find the sum of $$3, 6, …, 198$$.
The number of data values is $$\frac{(198 – 3)}{3} + 1 = \frac{195}{3} + 1 = 65 + 1 = 66.$$
Thus the sum of $$3, 6, …, 198$$ is $$3 + 6 + … + 198 = 3+6+…+198=3(1+2+…+66)=\frac{3(66)(66+1)}{2}=3(33)(67)=6,633.$$

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Re: What is the sum of all multiples of 3 between 1 and 200?   [#permalink] 30 Aug 2018, 00:51
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