Question

If 7128 = 2

× 3

× 11 Find the value of p +q+ r​

Answer 1

Answer:

According to question,

7128 = {2}^{p} {3}^{q} {11}^{r}7128=2

p

3

q

11

r

Lets split 7128 into its factor

\begin{gathered}7128 = 2 \times 3564 \\ 3564 = 2 \times 1782 \\ 1782 = 2 \times 891 \\ 891 = 3 \times 297 \\ 297 = 3 \times 99 \\ 99 = 3 \times33 \\ 33 = 3 \times 11\end{gathered}

7128=2×3564

3564=2×1782

1782=2×891

891=3×297

297=3×99

99=3×33

33=3×11

So,

7128 = {2}^{3} {3}^{4} {11}^{1}7128=2

3

3

4

11

1

So,

p=3

q=4

r=1

so, pqr=12

Hope it helps you

Answer 2

Answer:

Ok I know it

Step-by-step explanation:

7128=2

7128/2=3564

3564×3=10692

3564×11/10692=11/3

p=11

Q=3

So,r =3564/3

=1782