Random problem
Number Theory
Level
3
This is not an original problem
The number 27000001 has 4 prime factors Find their sum
The answer is 652.
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2 7 0 0 0 0 0 1 = 3 0 0 3 + 1
⇒ 2 7 0 0 0 0 0 1 = ( 3 0 0 + 1 ) ( 3 0 0 2 − 3 0 0 + 1 )
⇒ 2 7 0 0 0 0 0 1 = 3 0 1 ∗ ( 3 0 0 2 + 2 ∗ 3 0 0 + 1 − 3 ∗ 3 0 0 )
⇒ 2 7 0 0 0 0 0 1 = 3 0 1 ∗ ( ( 3 0 0 + 1 ) 2 − 3 0 2 )
⇒ 2 7 0 0 0 0 0 1 = 3 0 1 ∗ ( 3 0 1 − 3 0 ) ∗ ( 3 0 1 + 3 0 )
We thus have 3 0 1 , 2 7 1 , 3 3 1 as factors of 2 7 0 0 0 0 0 1 . Since 3 0 1 can be further factorised into 3 0 1 = 7 ∗ 4 3 , we have the 4 prime factors 7 , 4 3 , 2 7 1 , 3 3 1 with their sum being 7 + 4 3 + 2 7 1 + 3 3 1 = 6 5 2 .