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# M06-28

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Math Expert
Joined: 02 Sep 2009
Posts: 55629

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16 Sep 2014, 00:28
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Difficulty:

55% (hard)

Question Stats:

62% (02:24) correct 38% (02:23) wrong based on 105 sessions

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What is the value of $$k$$ if the sum of consecutive odd integers from 1 to $$k$$ equals 441?

A. 47
B. 41
C. 37
D. 33
E. 29

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Math Expert
Joined: 02 Sep 2009
Posts: 55629

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16 Sep 2014, 00:28
2
7
Official Solution:

What is the value of $$k$$ if the sum of consecutive odd integers from 1 to $$k$$ equals 441?

A. 47
B. 41
C. 37
D. 33
E. 29

Consecutive odd integers represent an evenly spaced set (aka arithmetic progression). Now, the sum of the terms in any evenly spaced set is the mean (average) multiplied by the number of terms, where the mean of the set is $$\frac{\text{first term} + \text{last term}}{2}$$.

$$\text{average}=\frac{\text{first term} +\text{last term}}{2}=\frac{1+k}{2}$$;

$$\text{# of terms}=\frac{k-1}{2}+1=\frac{k+1}{2}$$ (# of terms in an evenly spaced set is $$\frac{\text{last term} - \text{first term}}{\text{common difference}}+1$$)

$$\text{sum} = \frac{1+k}{2}*\frac{k+1}{2}=441$$. Simplify: $$(k+1)^2=4*441$$, so $$k+1=2*21=42$$ giving $$k=41$$.

Alternative Solution:

Use the formula for the sum of the first $$n$$ positive odd integers: $$(\frac{x+1}{2})^2$$ where $$x$$ is the last of these integers. Then $$\frac{k+1}{2} = \sqrt{441} = 21$$, and $$k = 41$$.

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Joined: 20 Jan 2014
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Location: India
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08 Oct 2014, 06:43
Alternate approach:

Sum of n evenly spaced numbers = n/2 [2a + (n-1)d]

Where n is the number of odd no, a = 1st number and d= difference
No of integers - (last - first)/2 + 1

Number of odd no = k-1/2, a=1, d=2

441= k+1/4 [2 + (k+1/2-1)2]
441= k+1/4 [2 + K +1 -2]
441*4 = (K+1)^2 = (441 * 4)
k+1 = 21 * 2 = 42
k=41

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Joined: 14 Jul 2014
Posts: 91

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22 Jan 2015, 04:57
5
5
60 seconds appraoch

Sum of First "n" +ve ODD Numbers = n^2

Quickly check...
20^2 = 400
21^2 = 441 ... Bingo!

Hence (21*2) - 1 = 41 .... or you can even quickly check this manually ... will take only 30 seconds more
Hence k = 41

Just FYI....
Sum of First "n" natural nos = n(n+1)/2
Sum of First "n" ODD natural nos = n^2
Sum of First "n" EVEN natural nos = n (n+1)
Intern
Joined: 28 Dec 2014
Posts: 3

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27 Jan 2015, 14:04
1+3 = 4 (# of terms)^2 = (2)^2
1+3+5 = 9 (# of terms)^2 = (3)^2
1+3+5+7 = 16 (# of terms)^2 = (4)^2
1+3+5+7+....+k=441

Therefore # of terms = square root of 441 = 21
21st consecutive odd term = 41.
Intern
Joined: 13 Feb 2015
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07 Aug 2015, 01:31
1
I've follwed this logic while solving this question on CAT:
1. The odd numbers end with 1, 3, 5, 7, 9, and the sum of them is 25 for the 1st ten (e.g. 1, 3, 5, 7, 9), 75 for the 2nd ( e.g. 11, 13, 15, 17, 19), 125 for the 3rd etc. So the sum ends with number 5.
2. Following this logic, 47 will ends with 4 *5 + 1 + 3+ 5 + 7, {number of tenth in the 47 = 4} * {sum of each tenth ends with 5} + {sum of all other odds till end of 47}
3. Obviously with the info from (1) and (2) only number which ends with 1 e.g. 41 will give a sum which ends with 1 e.g. 441

It took me ~10 seconds to evaluate the correct answer.
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Joined: 21 Jan 2015
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15 Aug 2015, 03:17
1
I think this is a high-quality question and I don't agree with the explanation. You write :
(# of terms in an evenly spaced set is first term−last termcommon difference+1)
but it is the opposite :

(# of terms in an evenly spaced set is Last term−first term/common difference+1)
Math Expert
Joined: 02 Sep 2009
Posts: 55629

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20 Aug 2015, 09:32
Barnal wrote:
I think this is a high-quality question and I don't agree with the explanation. You write :
(# of terms in an evenly spaced set is first term−last termcommon difference+1)
but it is the opposite :

(# of terms in an evenly spaced set is Last term−first term/common difference+1)

Typo edited. Thank you.
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20 Feb 2016, 14:52
I used the second method...
the sum of first x odd integers is x^2.
441 = 21^2.
this means that we are looking for the 21st odd integer.
39 is the 20th odd integer, and 41 is the 21st.
Manager
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24 Nov 2016, 20:29
1
Got this question wrong at first because i couldn't make out that the K is not the number of terms in this series, its just the last term in it.

We already know the formula of the sum of first N odd integers = $$N^2$$
So, $$N^2$$ = 441
N = 21

Hence, the number of terms in this series is 21.

At this point i realised that N and K are different here, for eg:
1,3,5,7 => N=4; K=7
1,3,5,7,9,11,13 => N=7; K=13

We can see that, $$K = 2N-1$$
since we have N=21 for this series, we can find out K
$$K=2*21 - 1$$
$$K=41$$

Side note (not quite reqd here but could be helpful in similar questions): "Sum of odd number of odd numbers is odd and Sum of even number of odd numbers is even"
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Senior Manager
Joined: 16 Nov 2016
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16 Aug 2018, 23:12
hi,

please explain what went wrong with my approach

sum= avg * total number of data points

$$441 = [1+k][/2] * [k+1][/2] * [k-1][/2]$$

$$441 = [k[/2]^2 - [1][/2]^2$$

1765 = k^2

so I guessed that closest is 41
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Joined: 22 May 2018
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08 Jan 2019, 12:26
Hello.

I don't get it why the common difference is 2.

k−12+1=k+12# of terms=k−12+1=k+12 (# of terms in an evenly spaced set is last term−first term / common difference +1

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10 Feb 2019, 14:17
Simple Plug-in Method : ( Plug in 41 )
N/2 (A +L)
21/2 (1 + 41)
21/2 (42)
21 x 21
= 441

Please Hit Kudos if u liked the method, so that i can access gmatclub tests
M06-28   [#permalink] 10 Feb 2019, 14:17
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# M06-28

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