Find 3 perfect squares completely divisible by 15,17 and 20 and whose HCF is 5.

Question

goten77 · Answer

im interested 2 c how this is done

anonymous · Answer

see i knopw howto find a square divisible by 15,17 and 19. But don't know how to to satisfy 2nd condition (HCF).

shubhamsrg · Answer

any no. divisible by 15,17 and 20 is always of the form 1020n right ? for some natural no.n
1020n = 17 * 5* 3* (2^2) *n
for this to be a perfect square, n= 5*3*17 *(m^2) , for some natural no. m
for m=1, we get our no. (17*5*3*2)^2
for m=2 ,we get our no. (17*5*3*2*2)^2
for m=3, we get our no. (17*5*3*2*3)^2

GCF is ,infact in every case, (17*5*3*2)^2 ..i dont know if any 3 such nos. exist with GCF=5 ..
i may be wrong..

agent0smith · Answer

@shubhamsrg, I wasn't sure if it meant what you said, or something more like that the three numbers are divisible by 15, 17 and 19 (or is it 20? both were mentioned), ie x/15, y/17, and z/19, and x y and z have a GCF 5.