Question

if 1800= p³xq²xr² where p, q and r are prime numbers then teh value of (p-q+r)​

Answer 1

Answer:

If 1800=$p^{3}q^{2} r^{2}$, where p,q and r are prime numbers then p-q+r=4

Step-by-step explanation:

In the question it is given that

$1800=p^{3}q^{2} r^{2}$

We need to find the factors of 1800 and choose three prime numbers from them that will satisfy the equation.

Among all the factors of 1800 we have 8,9,25

We can see that $8=2^{3} , 9=3^{2} , 25=5^{2}$

where 2,3 and 5 are prime numbers.

We can write the following from the above observations:

$1800=8.9.25=2^{3}3^{2} 5^{2}$

We observe that p=2,q=3 and r=5

Thus, p-q+r=2-3+5=4