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For positive integer m, the m-th heptagonal number is given [#permalink]

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24 Apr 2012, 07:20

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For positive integer m, the m-th heptagonal number is given by the formula (5m^2 – 3m)/2. For positive integer n, the n-th 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

For positive integer m, the m-th heptagonal number is given by the formula (5m^2 – 3m)/2. For positive integer n, the n-th 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 n-th triangular number is the sum of the first n positive integers, then n-th 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; ...

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.

Re: For positive integer m, the m-th 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;

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.

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

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20 Apr 2013, 01:57

emmak wrote:

For positive integer m, the m-th heptagonal number is given by the formula (5m2 – 3m)/2. For positive integer n, the n-th 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

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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 m-th heptagonal number is given [#permalink]

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06 Jul 2013, 08:00

Smita04 wrote:

For positive integer m, the m-th heptagonal number is given by the formula (5m^2 – 3m)/2. For positive integer n, the n-th 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

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

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08 Feb 2015, 15:44

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For positive integer m, the m-th heptagonal number is given [#permalink]

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21 Mar 2015, 12:45

Smita04 wrote:

For positive integer m, the m-th heptagonal number is given by the formula (5m^2 – 3m)/2. For positive integer n, the n-th 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 m-th heptagonal number = n-th 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 ≤ 80so 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 ≤ 128nothing lies in between.... 65 ≤ k ≤ 72 So 130≤ 2k ≤ 144so 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 m-th heptagonal number is given [#permalink]

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03 Apr 2016, 09:15

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

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11 Sep 2017, 04:19

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