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120secs · **Posted:** Mon Aug 08, 2011 6:18 am

Here is a non-algebraic solution

Put given solutions in ascending order:

29
33
37
41
47

TEST 37 (in this way we can eliminate 2 choices)

S=1+3+5+....+37
#of terms: (37-1)/2+1=19
mean: (37+1)/2=19
Sum of terms: 19*18= 342 < 441 (so we eliminate 29,33,37)

We can either try 41 and 47, or using common sense:

361+ 39 + 41 = 441

Answer is B. Any comments welcome

-120secs

saeedt · **Posted:** Mon Aug 08, 2011 10:09 am

I got the answer but with a longer solution.

(k-1)/2 +1 =N

a) => N= (k+1)/2

b) Sum/N=Average

c) Average in this evenly-spaced sets is (K+1)/2

a,b,c) 441/(k+1)/2 = (K+1)/2

Solving for K, K=41

saeedt · **Posted:** Mon Aug 08, 2011 10:23 am

mustdoit wrote:

bigfernhead wrote:

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

(A) 47
(B) 41
(C) 37
(D) 33
(E) 29

[Reveal] Spoiler: OA

B

Source: GMAT Club Tests - hardest GMAT questions

Sum of 1+3+5+7+9 = 25

Now every next 5 odd number will be greater by 50

for eg....11+13+15+17+19 = 75
21+23+25+27+29 = 125 and so on...

SO reaching 441 is like this...

25+75+125+175 = 400 ( from 1 to 40)

now adding 41 gives 441 so the K is 41....

No need to use formula......heh heh...

What happens that every 5 next numbers add 50? intuition says it must be 25

rongali · **Posted:** Mon Aug 08, 2011 11:02 am

answer is B....n^2/4 or n+1 ^2/4 depend on whether n is odd or even

Jammy1976 · **Posted:** Tue Aug 09, 2011 7:07 am

Simple formula leads 41

Posted from my mobile device

dimri10 · **Posted:** Wed Aug 10, 2011 9:33 am

+1 to the n^2 approach. briliant

ppasuthip · **Posted:** Fri Aug 19, 2011 12:30 am

Not sure if this is the correct way to do so ... and might be quite time consuming but what i did was
try to listing out the odd number out
1 3 5 7 9 11
Then try to add the number
1 + 3 = 4
1 + 3 + 5 = 9
1 + 3 + 5 + 7 = 16
1 + 3 + 5 + 7 + 9 = 25

So now I know the patter!

it shall be the square of concecutive number

then I check the ans for which is the square and could result in end of 1 (the unit digit)

A. 47 + 1 = 48; /2 =24 --> no way this could be the ans
B. 41 + 1 = 42; /2 = 21 --> could be the ans so try to squre it, resulting in 441

Then I applied this to the rest of the choices as well ... only one more possiblity which is choice C : 37 +1 =38'; /2 = 19 but the quare of 19 result in 361 so definately not the ans.

hope this help!

bholakc · **Posted:** Thu Sep 01, 2011 7:25 pm

Explanation:
Sum of positive, consecutive odd integer starting from 1 = n^2 ( i.e. n square)
Therefore, n^2 = 441
Clearly, n = 21 (Avoiding Negative sign because question is limited to positive, consecutive odd integer)

Again,
Starting from 1 to K refers to K is the last term of the series
So, Last term = first term + [Number of terms -1] * Common Difference
Or, K = a + (n-1) *d
Or, = 1 + (n-1) *2 (because common difference is 2)
Or, K = 2n-1

Now solving for K, we get
K = 2 * 21 -1
= 41

So , the definite answer is B.

petrifiedbutstanding · **Posted:** Sun Sep 04, 2011 1:11 am

I didn't get this one..

jagdeepsingh1983 · **Posted:** Sun Sep 04, 2011 1:37 am

sum of first odd numbers which start from 1 to k = n^2
441= n^2 --> n =21

first number =1 .. we know that last number = first number + (n-1) d --> K= 1+ (21-1)*d --> K= 41

fluke · **Posted:** Sun Sep 04, 2011 2:57 am

petrifiedbutstanding wrote:

I didn't get this one..

Which part you didn't get?

My post is here:
consecutive-integers-m06q28-72908-20.html#p958668

There is an arithmetic series like this:

Odd consecutive integers:
{1,3,5,7,9,11,...,k}

Stem says:
1+3+5+7+9+11+...+k=441

In an arithmetic progression; sum of n elements is:

S_n=\frac{n}{2}(2*A_1+(n-1)d)

For the above series:
n=\frac{k-1}{2}+1=\frac{k+1}{2} {Number of elements in a evenly spaced sequence:(First Term-Last Term)/Common Difference+1}

A_1=1

d=2

S_n=441

441=\frac{k+1}{4}(2*1+(\frac{(k+1)}{2}-1)2)

Solve for k.

Read this for more:
sequences-progressions-101891.html

Jdam · **Posted:** Mon Dec 19, 2011 5:27 pm

Any other simpler way to figure this out besides the formula?

fcomorales · **Posted:** Thu Mar 15, 2012 11:42 am

fluke wrote:

Average of elements*Number of Elements=Sum of elements

\frac{k+1}{2}*(\frac{k-1}{2}+1)=441

k=41

Ans: "B"

fluke's explanation is really nice. A variant of this rationale would be the following:
From the question stem we know that the set is an evenly spaced one, so the average of the elements of the set is the average of both extremes. In this case, the smallest term is 1 and the biggest term is k. Therefore, the average is:

avg=\frac{1+k}{2}

On the other hand, we know the sum of all the terms of the set. We can determine the average dividing the sum by the number of elements in the set. If n is the number of elements in the set, then the average is:

avg=\frac{441}{n}

Then,

avg=\frac{1+k}{2}=\frac{441}{n}

n=\frac{441*2}{k+1}

At this point, you can easily realize that 441 is divisible by 3 and by 9 (sum of the digits is divisible by 9).

n=\frac{7*7*9*2}{k+1}

Now, using the answer choices, you have to find the value of k that makes n an integer.

(A) 47 -> 47 + 1 -> 48 = 3*2*8 -> No 8 in the numerator

(B) 41 -> 41 + 1 -> 42 = 3*2*7 -> 42 is a factor of the numerator

(C) 37 -> 37 + 1 -> 38 = 2*19 -> No 19 in the numerator

(D) 33 -> 33 + 1 -> 34 = 2*17 -> No 17 in the numerator

(E) 29 -> 29 + 1 -> 30 = 3*2*5 -> No 5 in the numerator

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