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For positive integer m, the mth heptagonal number is given [#permalink]
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24 Apr 2012, 07:20
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For positive integer m, the mth heptagonal number is given by the formula (5m^2 – 3m)/2. For positive integer n, the nth triangular number is the sum of the first n positive integers. Which of the following is true for k, the smallest triangular number that is also heptagonal? (A) 33 ≤ k ≤ 40 (B) 41 ≤ k ≤ 48 (C) 49 ≤ k ≤ 56 (D) 57 ≤ k ≤ 64 (E) 65 ≤ k ≤ 72 Bunuel, can you please help with this one? Edit: This question was moved to an archive because this concept is NOT tested on the GMAT Original tags were MGMAT, Algebra, Too hard, and 700+ == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.
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Last edited by bb on 04 Nov 2017, 09:00, edited 2 times in total.
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Re: For positive integer m, the mth heptagonal number is given [#permalink]
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24 Apr 2012, 08:09
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Smita04 wrote: For positive integer m, the mth heptagonal number is given by the formula (5m^2 – 3m)/2. For positive integer n, the nth triangular number is the sum of the first n positive integers. Which of the following is true for k, the smallest triangular number that is also heptagonal?
(A) 33 ≤ k ≤ 40 (B) 41 ≤ k ≤ 48 (C) 49 ≤ k ≤ 56 (D) 57 ≤ k ≤ 64 (E) 65 ≤ k ≤ 72
Bunuel, can you please help with this one? It's been a long time since I've last heard about heptagonal and triangular numbers. Anyway, probably the best way would be to write down the numbers. Since the nth triangular number is the sum of the first n positive integers, then nth triangular number is given by the formulas n(n+1)/2. For example: The 1st triangular number is 1(1+1)/2=1; The 2nd triangular number is 2(2+1)/2=3=1+ 2; The 3rd triangular number is 3(3+1)/2=6=3+ 3; The 4th triangular number is 6+ 4=10; The 5th triangular number is 10+ 5=15; ... So, triangular numbers are: 1, 1+ 2=3, 3+ 3=6, 6+ 4=10, 10+ 5=15, 15+ 6=21, 21+ 7=28, 28+ 8=36, 36+ 9=45, 45+ 10= 55, 55+ 11=66, ... On the other hand, heptagonal numbers are: 1, 7, 18, 34, 55, 81, ... using (5m^2 – 3m)/2. So, as you can see the smallest triangular number that is also heptagonal is 1. Since it's not among answer choices I guess they don't consider 1, so the next one is 55. Answer: C.
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Re: For positive integer m, the mth heptagonal number is given [#permalink]
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29 Apr 2012, 05:44
Hi Bunuel
Do you think this particular concept is tested on GMAT?



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Re: For positive integer m, the mth heptagonal number is given [#permalink]
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29 Apr 2012, 05:46



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Re: For positive integer m, the mth heptagonal number is given [#permalink]
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20 Apr 2013, 01:27
The triangular number = n(n+1)/2 [Sum of positive numbers].
The 1st triangular number is 1(1+1)/2=1; The 2nd triangular number is 2(2+1)/2=3=1+2; The 3rd triangular number is 3(3+1)/2=6=3+3; The 4th triangular number is 6+4=10; The 5th triangular number is 10+5=15;
So, triangular numbers are: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66......
On the other hand, heptagonal numbers are: 1, 7, 18, 34, 55, 81, ... using (5m^2 – 3m)/2.
So, the smallest triangular number that is also heptagonal is 1 which is not present in any choice. Hence, look for the next one. So the next one is 55. Only C satisfied the equation.
Answer: C.



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Re: For positive integer m, the mth heptagonal number is given [#permalink]
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20 Apr 2013, 01:57
emmak wrote: For positive integer m, the mth heptagonal number is given by the formula (5m2 – 3m)/2. For positive integer n, the nth triangular number is the sum of the first n positive integers. Which of the following is true for k, the smallest triangular number that is also heptagonal? (A) 33 ≤ k ≤ 40 (B) 41 ≤ k ≤ 48 (C) 49 ≤ k ≤ 56 (D) 57 ≤ k ≤ 64 (E) 65 ≤ k ≤ 72
Express appreciation by pressing KUDOS. If m = 1, then the heptagonal number is (5*1^2 – 3×1)/2 = (5 – 3)/2 = 1. If m = 2, then the heptagonal number is (5*2^2 – 3×2)/2 = (20 – 6)/2 = 14/2 = 7. If m = 3, then the heptagonal number is (5*3^2 – 3×3)/2 = (45 – 9)/2 = 36/2 = 18. If m = 4, then the heptagonal number is (5*4^2 – 3×4)/2 = (80 – 12)/2 = 68/2 = 34. If m = 5, then the heptagonal number is (5*5^2 – 3×5)/2 = (125 – 15)/2 = 110/2 = 55. If m = 6, then the heptagonal number is (5*6^2 – 3×6)/2 = (180 – 18)/2 = 162/2 = 81. Using n*(n+1)/2 if n =1, triangle number is 1 if n =2, triangle number is 3 if n =3 triangle number is 6 if n =4, triangle number is 10 if n =5, triangle number is 15 if n =6 triangle number is 21 if n=7, triangle number is 28 if n = 8, triangle number is 36 if n =9, triangle number is 45 if n = 10 triangle is 55.....stop... We have 55 as the answer. the target number must be 34 or 55, A or C now. Now we need to find the smallest value which will exist for both triangle and heptagonal
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Re: For positive integer m, the mth heptagonal number is given [#permalink]
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06 Jul 2013, 08:00
Smita04 wrote: For positive integer m, the mth heptagonal number is given by the formula (5m^2 – 3m)/2. For positive integer n, the nth triangular number is the sum of the first n positive integers. Which of the following is true for k, the smallest triangular number that is also heptagonal? m (A) 33 ≤ k ≤ 40 (B) 41 ≤ k ≤ 48 (C) 49 ≤ k ≤ 56 (D) 57 ≤ k ≤ 64 (E) 65 ≤ k ≤ 72
Bunuel, can you please help with this one? hey thanks for the beautiful question i dont know about questions of this kind till date.. thanks again a here's my approach : heptagonal numbers : 5m^2 3m from choices we can see the total range is only till 72 so first wirte down these by substituting values of k = 1,2, 3 we get 1,7, 18, 34, 5x 22/2 = 55, 27x3 we have reached till 72 out limit so we can't exceed from here now get to 2nd one (nx n+1)/2 so u can easly get 55 by keeping n = 10 and our answer is C hope this helps, please PM for any queries



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Re: For positive integer m, the mth heptagonal number is given [#permalink]
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21 Mar 2015, 12:45
Smita04 wrote: For positive integer m, the mth heptagonal number is given by the formula (5m^2 – 3m)/2. For positive integer n, the nth triangular number is the sum of the first n positive integers. Which of the following is true for k, the smallest triangular number that is also heptagonal?
(A) 33 ≤ k ≤ 40 (B) 41 ≤ k ≤ 48 (C) 49 ≤ k ≤ 56 (D) 57 ≤ k ≤ 64 (E) 65 ≤ k ≤ 72
Bunuel, can you please help with this one? Another approach....though I think that solution by calculating individual series is a faster method. the mth heptagonal number = nth triangular number= k (5m^2 – 3m)/2 = n ( n+1) /2 = k (5m^2 – 3m) = n ( n+1) = 2 k n ( n+1) = 2K = Product of two consecutive integers.33 ≤ k ≤ 40 So 66 ≤2 k ≤ 80 so 8 x 9 = 72 41 ≤ k ≤ 48 So 82 ≤2 k ≤ 96 so 9 x10 = 90 49 ≤ k ≤ 56 So 98 ≤2 k ≤ 112 so 10 x 11 =110 57 ≤ k ≤ 64 So 114≤ 2k ≤ 128 nothing lies in between.... 65 ≤ k ≤ 72 So 130≤ 2k ≤ 144 so 11 x 12 =132 so now we have to check ( cumbersome ... but easy.) 5m^2 – 3m 2 k = 0 The only value of 2k that satisfies the above equation is 110. So K= 55 Hence, C.
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Re: For positive integer m, the mth heptagonal number is given [#permalink]
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14 May 2016, 14:31
What do you mean exactly by the terms ' heptagonal number ' and ' triangular number ' ?



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Re: For positive integer m, the mth heptagonal number is given [#permalink]
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