Question

the total number of prime factors in (25)^10x (24)^20 x (26)^25 is?​

Answer 1

Answer:

150

Step-by-step explanation:

(25)^10∗(24)^20∗(26)^25

(5*5)^10∗(2*12)^20∗(2*13)^25

Note: (x∗y)^k=x^k∗y^k

(5^10)*(5^10)*(2^20)*(12^20)*(2^25)*(13^25)

(5^10)*(5^10)*(2^20)*((2*3*2)^20)*(2^25)*(13^25)

(5^10)*(5^10)*(2^20)*((2^20)*(3^20)*(2^20)*(2^25)*(13^25)

add all the powers

10+10+20+20+20+20+25+25

150

Answer 2

The total number of prime factors

$\bf{(25)}^{10} \times ( {24)}^{20} \times {(26)}^{25}$ are 150.

Given:

• ${(25)}^{10} \times ( {24)}^{20} \times {(26)}^{25}$

To find:

• Find the total number of prime factors.

Solution:

Step 1:

Write the prime factors of 25, 24 and 26.

We know that

$25 = 5 \times 5$

$24 = 2 \times 2 \times 2\times 3$

$26 = 2 \times 13$

Step 2:

Simplify the expression in terms of prime factors.

${(25)}^{10} \times ( {24)}^{20} \times {(26)}^{25} = \\ {(5 \times 5)}^{10} \times ( {2 \times 2 \times 2 \times 3)}^{20} \times {(13 \times 2)}^{25}$

$\implies{( {5}^{2} )}^{10} \times ( { {2}^{3} \times 3)}^{20} \times {(13 \times 2)}^{25}$

we know that

$\bf ( { {a}^{n} })^{m} = {a}^{nm}$

$\bf {a}^{n} \times {a}^{m} = {a}^{n + m}$

$\implies{{5}}^{20} \times { {2}^{60} \times 3}^{20} \times {13^{25} \times 2}^{25}$

Add the powers of a similar prime number.

$\implies \bf {{5}}^{20} \times { {2}^{85} \times 3}^{20} \times 13^{25}$

To find the total number of prime factors, add all the powers.

$= 20 + 85 + 20 + 25$

$\bf = 150$

Thus,

Total number of prime factors are 150.

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