We are given that has only two prime factors that have two digits. Find the sum of these two prime factors.
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It is known that if n is a positive odd integer ⇒ a + b ∣ a n + b n
Write 2 3 0 + 3 3 0 as an expression with odd exponents.
2 3 0 + 3 3 0 = 4 1 5 + 9 1 5 = ( 4 3 ) 5 + ( 9 3 ) 5 = ( 4 5 ) 3 + ( 9 5 ) 3 ⇒ 4 + 9 , 4 3 + 9 3 , 4 5 + 9 5 divide 2 3 0 + 3 3 0 , so one of the two prime factors searched is 1 3 . To find the other apply divisibility rules to:
4 3 + 9 3 = 7 9 3 = 1 3 × 6 1
4 5 + 9 5 = 6 0 0 7 3 = 1 3 × 4 6 2 1
tTe two prime factors of two digits are 1 3 and 6 1 , ∴ its sum is 7 4