the number of prime factors of 2 ^ 202 × 3 ^ 211 ×70 ^8
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Solution :
Given
2^202 × 3^211 × 70^8
= 2^202 × 3^211 × ( 7 × 2 × 5 )^8
= 2^202 × 3^211 × 7^8×2^8×5^8
= 2^202+8 × 3^211 × 5^8 × 7^8
= 2^210 × 3^211 × 5^8 × 7^8
Therefore ,
Number of factors
= ( 210+1)(211+1)(8+1)(8+1)
= 211 × 212 × 9 × 9
•••••
Given
2^202 × 3^211 × 70^8
= 2^202 × 3^211 × ( 7 × 2 × 5 )^8
= 2^202 × 3^211 × 7^8×2^8×5^8
= 2^202+8 × 3^211 × 5^8 × 7^8
= 2^210 × 3^211 × 5^8 × 7^8
Therefore ,
Number of factors
= ( 210+1)(211+1)(8+1)(8+1)
= 211 × 212 × 9 × 9
•••••
Answered by
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Answer:
Number of prime factors = 437
Step-by-step explanation:
2^202 × 3^211 × 70^8
= 2^202 × 3^211 × (2×5×7)^8
= 2^202 × 3^211 × 2^8 × 5^8 × 7^8
= 2^(202 + 8) × 3^211 × 5^8 × 7^8
= 2^210 × 3^211 × 5^8 × 7^8
So, the prime factors of 2^202 × 3^211 × 70^8 are 2, 3, 5 and 7.
The number of prime factors of 2^202 × 3^211 × 70^8
=210 + 211 + 8 + 8 (Add the powers of prime numbers of the number.)
=437
So, the number of prime factors of 2^202 × 3^211 × 70^8 is 437.
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