For positive integer m, the m-th heptagonal number is given

No, this concept is not tested on the GMAT.

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.

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

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

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 ≤ 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.

What do you mean exactly by the terms ' heptagonal number ' and ' triangular number ' ?

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