Question

challenging question please solve.
If p ,q and r are the positive integers such that , p + q + r + pq + qr + pr+pqr = 1000 , then find p^2+q^2+r^2 ? a)349 b)351 c)280 d)365

Answer 1

Answer:280

Step-by-step explanation: 1+p+q+r+pq+pr+qr+pqr=1000+1

(1+p) +(q+r+pq+qr+pr+pqr) =1001

(1+p) +1(q+r)+p(q+r)+qr(1+p)=1001

(1+p) +(1+p)(1+r)+qr(1+p)=1001

(1+p)(1+1+r+qr)=1001

(1+p)(1+p)(1+r) =1001

Now factors of 1001 is 7*11*13

1+p=7, p=6

1+q=11, q=10

1+r=13, r=12

p²+q²+r²=6²+10²+12²=36+100+144=280