Q. 4 The product of three integers 𝐗, 𝐘 and Z is 𝟏𝟗𝟐. 𝐙 is equal to 4 and P is equal to the average of X and
Y. What is the minimum possible value of P?
(A) 6
(B) 7
(C) 8
(D) 9.5

The product of three integers X, Y, and Z equals 192, with Z fixed at 4, so XY=48. P, the average of X and Y, reaches its minimum value of 7 when X=6 and Y=8 (or vice versa).​

Problem Breakdown

XYZ=192 and Z=4 imply XY=192/4=48, where X and Y are integers. P=(X+Y)/2, so minimizing P requires minimizing the sum X+Y for integer factor pairs of 48. For a fixed product, the sum minimizes when factors are closest, typically near √48≈6.928.​

Factor Pairs and Sums

Positive integer pairs (X,Y) where XY=48, with sums and P values:

Negative pairs like (-6,-8) yield the same P=7, as averages match. Closest pair (6,8) gives minimal sum 14, so P=7.​

Options Analysis

• (A) 6: Impossible, as no pair sums to 12 (e.g., hypothetical 5.196^2=48, but not integers).​

• (B) 7: Correct, from (6,8) or (8,6), sum=14.​

• (C) 8: From (4,12), sum=16, larger than minimal.​

• (D) 9.5: From (3,16), sum=19, not minimal.​

Key Insight

Integer constraint and closeness to square root ensure P=7 minimum among options. Verify: 684=192.​