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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