If 2352-2 p - 3 q - 7 r then find the values of p - q and r-A- 87-8 - 10-25 KgB- 8-78 - 1025 KgC- 8-78 - 1023 KgD- 878 - 10-23 m

The correct option is B 8.78 × Kg Given 2352 = 2^p × 3^q × 7^r we know 2352 = 2^4 × 3^1 × 7^2 (by prime factorisation) ⇒ 2^4 × 3^1 × 7^2 = 2^p × 3^q ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard VIII

Mathematics

Powers and Exponents

If 2352=2 p ×...

Question

If $2352 = 2^{p} \times 3^{q} \times 7^{r}$ then find the values of $p, q a n d r$ .

878 × m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

8.78 × Kg

No worries! We‘ve got your back. Try BYJU‘S free classes today!

87.8 × Kg

No worries! We‘ve got your back. Try BYJU‘S free classes today!

8.78 × Kg

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D
8.78 × Kg

Given

2352 = $2^{p} \times 3^{q} \times 7^{r}$

we know $2352 = 2^{4} \times 3^{1} \times 7^{2}$ (by prime factorisation)

$\Rightarrow 2^{4} \times 3^{1} \times 7^{2} = 2^{p} \times 3^{q} \times 7^{r}$

$\Rightarrow p = 4, q = 1 a n d r = 2$

Suggest Corrections

Similar questions

If $2352 = 2^{p} \times 3^{q} \times 7^{r}$ , frind the values of $p, q a n d r$

The quantity $87, 800, 000, 000, 000, 000, 000, 000, 000 K g$ equals to _________.

If 2160 = $2^{p} \times 3^{q} \times 5^{r}$ then find the values of $p, q a n d r$ .

Q. Find the value of

2 p - 3 q + 4 r - 8 s

, when

p = 10, q = 8, r = 6

and

s = 2

Q. Find the value of

2 p^{2} q - 3 p q^{2} + 2 p - 3 q + 1

when

p = 3, q = - 3

Join BYJU'S Learning Program

If 2352=2p×3q×7r then find the values of p,q and r.

The correct option is D 8.78 × Kg Given 2352 = 2p×3q×7r we know 2352= 24×31×72 (by prime factorisation) ⇒ 24×31×72 = 2p×3q×7r ⇒ p=4,q=1 and r=2

If $2352 = 2^{p} \times 3^{q} \times 7^{r}$ then find the values of $p, q a n d r$ .

The correct option is D
8.78 × Kg

Given

2352 = $2^{p} \times 3^{q} \times 7^{r}$

we know $2352 = 2^{4} \times 3^{1} \times 7^{2}$ (by prime factorisation)

$\Rightarrow 2^{4} \times 3^{1} \times 7^{2} = 2^{p} \times 3^{q} \times 7^{r}$

$\Rightarrow p = 4, q = 1 a n d r = 2$