Around the World in 80 Questions (Day 3): If a and b are integers and

The number 729 is 3^6.
Therefore, the possible pairs of (a, b) are 7 pairs (1, 729), (3, 243), (9, 81), (27, 27), (81, 9), (243, 3), and (729, 1).
We also have to consider the negative pairs so 7 more pairs. (-1, -729), (-3, -243), (-9, -81), (-27, -27), (-81, -9), (-243, -3), and (-729, -1).
Therefore 7+7=14
Hence A: 14.

729 = 3^6

Hence, we can have 7 factors.

In an order pair, the value of a, and b matter. Hence (1,729) and (729,1) is considered different.

Number of order pairs = 7*2 = 14

IMO A

given that a, b are integers and a*b = 3^6
positive pairs can be
3^6 *1
3^5 * 3
3^4*3^2
3^3*3^3
3^2*3^4
3^1 * 3^5
1*3^6

7 pairs


-ve pairs
(-3)^6*1
(-3)^5 * -3
(-3)^4* (-3)^2
(-3)^3 * ( -3)^3
(-3)^2 * (-3)^4
(-3)^1 * (-3)^5
(1* (-3)^6)

7 pairs

total 14 pairs possible

OPTION A

bb Bunuel

Checking if my response was evaluated. The answer is correct, however kudos wasn't awarded.

Thanks.

https://gmatclub.com/forum/around-the-w ... l#p3236467

Logic is incorrect. Hence no kudos.

[quote="Bunuel"

Logic is incorrect. Hence no kudos.[/quote]

Bunuel

Thank you for the feedback. I looked into some of the responses which were indicated correct. I think you're referring to the negative factors. If so, I agree that I didn' t consider that. But if that's the case, isn't none of the option correct.

The questions asks us to find the ordered pair. In an ordered pair (a,b) \(\neq\) (b,a), that's why the term ordered pair is used. Otherwise the pair is un-ordered. Therefore the correct number of ordered pair should be 28 and not 14.

For example

(1,729), (729,1), (-1, -729) and (-729,-1) are all different ordered pairs. Can you please clarify this.

Sure, here are possible pairs:

1. a = -729 and b = -1
2. a = -243 and b = -3
3. a = -81 and b = -9
4. a = -27 and b = -27
5. a = -9 and b = -81
6. a = -3 and b = -243
7. a = -1 and b = -729
8. a = 1 and b = 729
9. a = 3 and b = 243
10. a = 9 and b = 81
11. a = 27 and b = 27
12. a = 81 and b = 9
13. a = 243 and b = 3
14. a = 729 and b = 1

Hope it helps.

Thank you so much. I realized my mistake.

Since 729 = 3^6, and we need to multiply a and b to get 3^6, we must ensure that a and b are of the form 3^p and 3^q where p and q are integers (note that -3^p and -3^q is also possible, but more of that later)

Thus, we have: 3^p * 3^q = 3^6
=> p + q = 6

Thus, there are 7 non-negative integer solutions: 0+6, 1+5, 2+4, 3+3, 4+2, 5+1, 6+0
Thus, the solutions here are 3^0 * 3^6, 3^1 * 3^5, etc.
Note: Number of non-negative integer solutions for the equation p + q = N is (N + 1)

Considering the negative solutions, there are 7 more solutions; for example: (-3^0) * (-3^6)

Thus, there are 7 + 7 = 14 solutions
Answer A