Suppose x, y, z are three positive integers such that x ≤ y ≤ z and xyz = 72. Which one of the following values of S yields more than one solution to the equation x + y + z = S?
13
14
15
16
Answer is Wrong!
Answer is Right!
This question was previously asked in
UPSC CAPF – 2018
– We need to find which sum S = x+y+z corresponds to more than one such distinct tuple (x, y, z).
– Let’s list all valid tuples (x, y, z) and their corresponding sums S:
1. (1, 1, 72): S = 1+1+72 = 74
2. (1, 2, 36): S = 1+2+36 = 39
3. (1, 3, 24): S = 1+3+24 = 28
4. (1, 4, 18): S = 1+4+18 = 23
5. (1, 6, 12): S = 1+6+12 = 19
6. (1, 8, 9): S = 1+8+9 = 18
7. (2, 2, 18): S = 2+2+18 = 22 (since 2*18=36 and 2*2*18=72, 2<=2<=18) 8. (2, 3, 12): S = 2+3+12 = 17 (since 3*12=36 and 2*3*12=72, 2<=3<=12) 9. (2, 4, 9): S = 2+4+9 = 15 (since 4*9=36 and 2*4*9=72, 2<=4<=9) 10. (2, 6, 6): S = 2+6+6 = 14 (since 6*6=36 and 2*6*6=72, 2<=6<=6) 11. (3, 3, 8): S = 3+3+8 = 14 (since 3*8=24 and 3*3*8=72, 3<=3<=8) 12. (3, 4, 6): S = 3+4+6 = 13 (since 4*6=24 and 3*4*6=72, 3<=4<=6) - Checking for x=4: xyz=72 => yz=18, 4 <= y <= z. Possible (y,z) pairs for yz=18, y<=z are (1,18), (2,9), (3,6). None satisfy y >= 4. No solution starts with x=4.
– Checking for x>=5: Smallest product with x>=5 is 5*5*z. 5*5*z=72 => 25z=72, no integer z. Or 5*y*z=72 with 5<=y<=z. Minimum product 5*5*z means z>=5. 5*5*5=125 > 72. So no solutions for x>=5.
– Now, examine the sums S from our list:
– S = 13 corresponds to (3, 4, 6) – 1 solution.
– S = 14 corresponds to (2, 6, 6) and (3, 3, 8) – 2 solutions.
– S = 15 corresponds to (2, 4, 9) – 1 solution.
– S = 16 does not appear in the list of sums for any valid (x,y,z) tuple.
– The value of S that yields more than one solution is 14.