Question

suppose x y z are three positive integers such that x \leq≤ y\leq≤ z and x y z=72. Which one of the following values of s yields more than one solution to the equation x+y+z=s?
(a) 13
(b) 14
(c) 15
(d) 16​

Answer 1

Answer:

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Answer 2

Answer:

Step-by-step explanation:

We need more than one solution , it means we should have more than one set of (x,y,z) values

Now looking at options

since 13 is prime number , 13=13x1x1 which gives only one solution and this even wont fit in xyz=72

So this can be ruled out

We are given with xyz=72=2^3x3^2

So for making 15 out of the factors we only have choice (9,4,2)

similary it can be checked for 16

for 14=2x7

we have (8,3,3) : 8+3+3=15

and (6,4,3) ;  6+4+3=15

so for S=14 we have more than one solution