Question

If 529200 =2x

x 3y x 5 p x7q then find HCF of (x+y+p+q) and xypq​

Answer 1

2x×3y×5p×7q=529200

210xypq=529200

xypq= 529200÷210

xypq= 2520

Therefore HCF of x, y, p, q is

1

and HCF of x+y+p+q= 2+5+2+0 = 9 is 1