Find the number of different primes p > 2 such that p is a factor of 71^2 – 37^2 – 51.
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Find the number of different primes p > 2 such that p is a factor of 71^2 – 37^2 – 51.
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Question:
Find the number of different primes p > 2 such that p is a factor of 71² – 37² – 51.
ANSWER:
3, 17, 71
SOLUTION:
Step 1: solve for the value of expression,
( 71)² – (37)² – 51
= 5041 - 1369 - 51
= 3621
Step 2: Let's list down first few prime numbers in a table for our reference,
2 3 5 7 11 13 17 19 23 29
31 37 41 43 47 53 59 61 67 71
73 79 83 89 97 101 103 107 109 113
Step 3: Let's look for prime factors of the number 3621
Between the digits 2 to 9, 3621 can only be divided by 3, so
3621 ÷ 3 = 1207
Hence we already have one of our prime factors which is 3
Step 4: Now we try to look for remaining prime factors of 1207
As can be seen from 1207, it ends in a 7, so that means looking at our prime numbers table above, let's look for two numbers whose product will end in a 7.
see examples below,
7 × 11 will end in a 7
17 × 11 will end in a 7
37 × 11 will end in a 7
and by trial and error of a few combinations, we arrived at
17 × 71 = 1207
Hence our number 3621 has 3 prime factors as follows,
3, 17, 71
PROOF:
3 x 17 x 71 = 3621
3621 = 3621 ✔✔✔ TRUE
* as always, double check my answers for errors or carelessness.
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