Question

Find the no of prime factors in 2²²²×3³³³×5⁵⁵⁵

Answer 1

To find :-

• Find the no of prime factors in 2²²²×3³³³×5⁵⁵⁵ ?

Concept used :-

Fundamental theorem of arithmetic states that every composite number can be expressed as a product of two or more prime numbers.

Let N be a composite number and a,b & c are its prime factors. Then :

N = a^p * b^q * c^r

Than, we Have :-

• Number of factors = (p+1)(q+1)(r+1)
• Number of unique factors = 3
• Number of prime factors = p+q+r
• Sum of factors = (a^0+a^1+..+a^p)(b^0+b^1+..+b^q)(c^0+c^1+..+c^r)
• Product of factors = N^(Number of factors/2)

Solution :-

N = 2²²²×3³³³×5⁵⁵⁵

Comparing it with a^p * b^q * c^r we get,

→ a = 2

→ b = 3

→ c = 5

→ p = 222

→ q = 333

→ r = 555

since a , b & c all are Prime Numbers.

Than,

→ Number of prime factors = p + q + r = 222 + 333 + 555 = 1110 (Ans.)

Answer 2

SOME POINTS TO NOTICE:-

1. IN QUESTIONS RELATED TO SOLVING PROBLEMS ON NO. OF PRIME FACTORS.
2. IF QUESTIONS ARE OF FORM IN POWERS X^T × Y^Q × Z^R
3. THEN THE CONCEPT USED HERE IS THAT THE FUNDAMENTAL THEOREM OF ARITHMETIC STATES THAT THE EVERY COMPOSITE NO. CAN BE EXPRESSED OF THE FORM TWO OR MORE PRIME NO.S

• CONSIDER A TERM P SUCH THAT

P = X^T × Y^Q × Z^R

THEN IT'S FACTORS CAN BE CONCLUDED BY:-

1. NO. OF FACTORS CAN BE WRITTEN AS (T + 1)(Q + 1)(R + 1)
2. ALSO NO. OF UNIQUE FACTORS HERE IS 3
3. SUM OF FACTORS ARE WRITTEN AS (X^0 + X^1 ......X^N)(Y^0 + Y^1 .......Y^N)(Z^0 + Z^1 .......Z^N)

ALSO NO. OF FACTORS = N^no. of factors/2

NO. OF PRIME FACTORS = p + q + r

GIVEN:- 2²²²×3³³³×5⁵⁵⁵

TO FIND:- THE PRIME FACTORS

SOLUTION:-

AS WE HAVE TAKEN THE STANDARD EQUATION ABOVE ,

P = X^T × Y^Q × Z^R

THEN COMPARING IT WE GET,

T = 222

Q = 333

R = 555

X = 2

Y = 3

Z = 5

NOW WE ALSO KNOW ,

THAT 2 , 3, 5 ARE PRIME NO.S

THEREFORE BY USING THE FORMULA OF NO. OF PRIME FACTORS

NO. OF PRIME FACTORS = p + q + r

NO. OF PRIME FACTORS = 222 + 333 + 555

=> 1110 (YOUR ANSWER)