Asked by cherryl
“ y and z are integers greater than 1
5y is the square of a number
75yz is the cube of a number.
The smallest value for y + z is = ?”
5y is the square of a number
75yz is the cube of a number.
The smallest value for y + z is = ?”
Answers
Answered by
drwls
The smallest 5y (with y an integer) that is a square is 25 (y = 5). The next smallest is 100 (y = 20). Let's try y=5 and see what z's will work.
Try various integer z's with y = 5
z = 1, 75yz = 375 (not a cube)
z = 2, 75yz = 750 (not a cube)
z = 3, 75yz = 1125 (not a cube)
z = 4, 75yz = 1500 (not a cube)
z = 5, 75yz = 1875 (not a cube)
z = 6, 75yz = 2250 (not a cube)
z = 7 and 8 don't work either, but
For z=9 75yz = 3375 = 15^3
Therefore y = 5, z = 9 and y + z = 14
No lower integers will work. If I had tried y = 20, the sum y+z would have had to exceed 14.
I used a "trial and error" method. I don't see any easier method, but there might be one.
Try various integer z's with y = 5
z = 1, 75yz = 375 (not a cube)
z = 2, 75yz = 750 (not a cube)
z = 3, 75yz = 1125 (not a cube)
z = 4, 75yz = 1500 (not a cube)
z = 5, 75yz = 1875 (not a cube)
z = 6, 75yz = 2250 (not a cube)
z = 7 and 8 don't work either, but
For z=9 75yz = 3375 = 15^3
Therefore y = 5, z = 9 and y + z = 14
No lower integers will work. If I had tried y = 20, the sum y+z would have had to exceed 14.
I used a "trial and error" method. I don't see any easier method, but there might be one.
Answered by
Reiny
All we can do is perhaps shorten some of the trial-and-error steps that drwls had to make
Start with his determination that if
5y is a perfect square, then y = 5
then if 75yz is a cube
then 75(5)z is a cube
If 375z is a cube
375 = 5 x 5 x 5 x 3 x z
for a perfect cube, each factor should show up 3 times
I see three 5's and one 3, so it looks like we need another pair of 3's, or a 9
so z=9
Start with his determination that if
5y is a perfect square, then y = 5
then if 75yz is a cube
then 75(5)z is a cube
If 375z is a cube
375 = 5 x 5 x 5 x 3 x z
for a perfect cube, each factor should show up 3 times
I see three 5's and one 3, so it looks like we need another pair of 3's, or a 9
so z=9
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