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How do you know whether such a large number is prime as non-calculator exam question is to determine which two numbers between 2010-2020 are prime?
Wiki User
∙ 7y ago
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Anibal Gusikowski
Wiki User
∙ 7y agoTry eliminating the non-primes by using division tests for 2, 3, 5, etc.
If there are two primes between 2010 and 2020 then:
All the even numbers (multiples of 2) cannot be prime, so that excludes 2010, 2012, 2014, 2016, 2018 & 2020, leaving: 2011, 2013, 2015, 2017, 2019 as the possible primes.
To be a multiple of 3, the sum of the digits of a number must be a multiple of 3:
2011 : 2 + 0 +1 + 1 = 4; not a multiple of 3
2013 : 2 + 0 + 1 + 3 = 6; multiple of 3
2015 : 2 + 0 + 1 + 5 = 8; not a multiple of 3
2017 : 2 + 0 + 1 + 7 = 10; not a multiple of 3
2019 : 2 + 0 + 1 + 9 = 12; multiple of 3.
Thus 2013 & 2019 are multiples of 3 and can be eliminated, leaving 2011, 2015, 2017 are possible primes
All multiples of 5 end in 0 or 5, so 2015 cannot be a prime, leaving 2011, 2017 as possible primes.
As we are told there are two primes, then 2011 and 2017 must be the primes.
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In any set of 6 consecutive numbers: 6n, 6n+1, 6n+2, 6n+3, 6n+4, 6n+5 (for n = 1, 2, 3, ...):
6n is divisible by 2 and 3
6n+2 is divisible by 2
6n+3 is divisible by 3
6n+4 is divisible by 2
Leaving only 6n+1 and 6n+5 as the possible candidates for being prime (they may not be prime, eg when n is a multiple of 5, 6n+5 is divisible by 5 since n = 5k (for some integer k) → 6n+5 = 6(5k)+5 = 5(6k)+5 = 5(6k+1) which is a multiple of 5).
Let m = n+1, then 6n+5 = 6n+6-6+5 = 6(n+1)-1 = 6m-1
Thus all primes are 6n±1 for some integer n.
In the range 2010 to 2020:
2010 ÷ 6 = 335
2020 ÷ 6 = 336 r 4 (these can be worked out by long division by hand without a calculator)
→ the possible primes are: 6×335+1 = 2011, 6×336-1 = 2015 and 6×336+1 = 2017
2015 is divisible by 5, which means the only possible candidates are 2011 and 2017.
As we are told there are two, they must be 2011 and 2017.
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