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

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

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New post 28 Aug 2018, 01:07
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Question Stats:

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

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New post 28 Aug 2018, 01:31
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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.
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What is the sum of all multiples of 3 between 1 and 200?  [#permalink]

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New post Updated on: 28 Aug 2018, 02:20
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, 02:17.
Last edited by dave13 on 28 Aug 2018, 02:20, edited 1 time in total.
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Re: What is the sum of all multiples of 3 between 1 and 200?  [#permalink]

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New post 28 Aug 2018, 02:20
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
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Re: What is the sum of all multiples of 3 between 1 and 200?  [#permalink]

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New post 28 Aug 2018, 03:05
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 )
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What is the sum of all multiples of 3 between 1 and 200?  [#permalink]

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New post 28 Aug 2018, 03:26
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Raksat wrote:
I saw the timer: it shows around 4 minutes



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

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New post 28 Aug 2018, 04:54
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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

So, 3(67*33) is our answer.

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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New post 28 Aug 2018, 05:39
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

Answer: C

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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New post 28 Aug 2018, 06:24
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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New post 28 Aug 2018, 11:27
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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New post 29 Aug 2018, 23:51
=>

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 \(3 + 6 + … + 198 = 3+6+…+198=3(1+2+…+66)=\frac{3(66)(66+1)}{2}=3(33)(67)=6,633.\)

Therefore, the answer is C.
Answer: C
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Re: What is the sum of all multiples of 3 between 1 and 200? &nbs [#permalink] 29 Aug 2018, 23:51
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