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Re: Consecutive Integers (m06q28) [#permalink]
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19 Aug 2011, 00:30
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!



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Re: Consecutive Integers (m06q28) [#permalink]
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01 Sep 2011, 19:25
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 + (n1) *d Or, = 1 + (n1) *2 (because common difference is 2) Or, K = 2n1
Now solving for K, we get K = 2 * 21 1 = 41
So , the definite answer is B.



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Re: Consecutive Integers (m06q28) [#permalink]
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04 Sep 2011, 01:37
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This post received KUDOS
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 + (n1) d > K= 1+ (211)*d > K= 41



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Re: Consecutive Integers (m06q28) [#permalink]
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04 Sep 2011, 02:57
petrifiedbutstanding wrote: I didn't get this one.. Which part you didn't get? My post is here: consecutiveintegersm06q287290820.html#p958668There 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: \(S_n=\frac{n}{2}(2*A_1+(n1)d)\) For the above series: \(n=\frac{k1}{2}+1=\frac{k+1}{2}\) {Number of elements in a evenly spaced sequence:(First TermLast 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: sequencesprogressions101891.html
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Re: Consecutive Integers (m06q28) [#permalink]
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15 Mar 2012, 11:42
fluke wrote: Average of elements*Number of Elements=Sum of elements
\(\frac{k+1}{2}*(\frac{k1}{2}+1)=441\)
\(k=41\)
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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Re: Consecutive Integers (m06q28) [#permalink]
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10 Aug 2012, 05:21
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 (first term+last term)/2. \(average=\frac{first \ term+last \ term}{2}=\frac{1+k}{2}\); \(# \ of \ terms=\frac{k1}{2}+1=\frac{k+1}{2}\) (# of terms in an evenly spaced set is \(\frac{first \ termlast \ term}{common \ difference}+1\)) \(sum=\frac{1+k}{2}*\frac{k+1}{2}=441\) > \((k+1)^2=4*441\) > \(k+1=2*21=42\) > \(k=41\). Answer: B.
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Re: Consecutive Integers (m06q28) [#permalink]
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10 Aug 2012, 05:33
Sn = N/2(f+l) lets choose a number i choose 41 it has 21 odd integers 441= N/2(1+L) 882=21(1+L) = 41 ... It took 2 .4 mins if i had chosen other options I would have taken more time
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Re: Consecutive Integers (m06q28) [#permalink]
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12 Aug 2012, 12:42
sum of odd, consecutive, positive integers from 1 to K, that would be: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19.....K As we see it is an arithmetic progression with following details: A1 = 1, d (difference) = 2 and An = k We know that n member of the sequence above can be find as follow: An=A1+d*(n1), but we know that an = K, from question stem so we can find the number of elements in the sequence, i.e. n: K=1+2*(n1) from here => n = (K+1)/2 So the sum of sequence as arithmetic sequence would be : Sn = (1+K)^2 =4* 441, From here we get: K1 = 41 and K2= 43 (negative), thus the answer would be (B) 41 Please, correct me if I am wrong
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Re: Consecutive Integers (m06q28) [#permalink]
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15 Aug 2012, 06:50
Pluggin method is the best here ;
sum consecutive = Number of items x mean
Number of items=(firstlast)/multiple+1, here multiple is 2 mean = (first+last)/2
start with 37 Number items: (371)/2+1=19 mean: (1+37)/2=19 sum = 19 x 19 = 361
thus, you deduct that k should be higher. two choices remain : 41 and 47. It is unecessary to check both. if on satisfies the condition (sum=441), then it is the answer. If not, the answer is the other.
pick 41 Number items: (411)/2+1=21 mean: (41+1)/2=21 sum: 441
thus, the answer is B



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Re: Consecutive Integers (m06q28) [#permalink]
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29 Aug 2012, 08:56
I am not sure if my approach is correct and that if it can be applied to other questions as well. I started of by trying to use the arithmetic progression formula but could not recall it. So I followed the below mentioned alternative approach. The sum required is 441. If we add the first few numbers : 1+3+5+7+9=25. Thus the value of "n" needs to be something like X1 more so with there being a set of 2 such collections of numbers since 5+5=10. Thus came down to seeing which of the answer choices had a "1" in their units digit with the possible options being 21,41,61 etc. 41 was the available option so went ahead with choice B.
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Re: Consecutive Integers (m06q28) [#permalink]
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12 Aug 2013, 11:33
To solve the 'sum' of a sequence of numbers, I use the trapezoid formula. Think each number as little dots: Sum of 2 + 3 + 4: o o o o o o o o o \( Area = \frac{(Top+Bottom)*Height}{2} \) \( Area = \frac{(2+4)*3}{2} \)(you can verify this by counting) In this case: \(Top = 1\) \(Bottom = k\) \( Height = \frac{(k1)}{2}+1 \) [divide by 2 because it's every other number] \( 411 = \frac{(1+k)*Height}{2} \) \( 411 = \frac{(1+k)*(1+k)}{2*2} \) \( 21*21 = \frac{(1+k)}{2}*\frac{(1+k)}{2} \) \( 21 = \frac{(1+k)}{2} \) \( 41 = k \) answer is (b)Another easy way is to substitute numbers from answer choices: Starting with (b) 41 \( Height = \frac{(411)}{2}+1 = 21 \) \( Area (sum) = \frac{(1+41)*21}{2} = 411 \) correct answer is (b)



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Re: Consecutive Integers (m06q28) [#permalink]
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14 Aug 2013, 04:39
If you use the formula n/2*[2a+(n1)d] to find d, you will arrive at n = 21. Hence, there are 21 terms in the A.P.
Now, k is the 21st term.
nth of A.P = a+(n1)d, here a=1 n=21 d=2
21st term of this A.P i.e k = 1+40 = 41
Answer is B



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Re: Consecutive Integers (m06q28) [#permalink]
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28 Jul 2014, 05:53
sum of n odd consecutive integers = n^2 =441. n=21. 2(21)  1 = 41




Re: Consecutive Integers (m06q28)
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