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Kwenyao
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Prime factorisation 16 Feb 2016, 18:18
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For each of the following numbers, determine if the number is prime. If not, factorize
it into product of primes if it is not prime.
A)1747
B)139^3 + 1
C)47^8 - 37^8

How do i go about doing these?



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ENGRTOMBA2018
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Prime factorisation 16 Feb 2016, 18:32
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Kwenyao
For each of the following numbers, determine if the number is prime. If not, factorize
it into product of primes if it is not prime.
A)1747
B)139^3 + 1
C)47^8 - 37^8

How do i go about doing these?
Show more

By definition, a prime number 'n' is a number that is not divisible by anything else but 1 and n.

Let us analyse each of the 3 options:

A)1747 : try the usual way. 41^2= 1681, so you need to check for all numbers from 1 to 41 to see if there are any divisors of 1747. You wont find any. Thus 1747 is a prime number.

B)139^3 + 1: no need to solve. Recognise that 9^3 ends in 9 and thus the last digit 9+1 = last digit of 0. As 0 is an even number (divisible by 2)---> the resulting number 139^3+1 is NOT a prime.

C)47^8 - 37^8 ---> apply a^2-b^2=(a+b)(a-b) ---> 47^8 - 37^8 = (47^4+37^4)(47^4-37^4) = (odd+odd)(odd+odd) = even*even = even . NOT a prime number.

Hope this helps.
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Prime factorisation 16 Feb 2016, 18:36
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Engr2012
Kwenyao
For each of the following numbers, determine if the number is prime. If not, factorize
it into product of primes if it is not prime.
A)1747
B)139^3 + 1
C)47^8 - 37^8

How do i go about doing these?

By definition, a prime number 'n' is a number that is not divisible by anything else but 1 and n.

Let us analyse each of the 3 options:

A)1747 : try the usual way. 41^2= 1681, so you need to check for all numbers from 1 to 41 to see if there are any divisors of 1747. You wont find any. Thus 1747 is a prime number.

B)139^3 + 1: no need to solve. Recognise that 9^3 ends in 9 and thus the last digit 9+1 = last digit of 0. As 0 is an even number (divisible by 2)---> the resulting number 139^3+1 is NOT a prime.

C)47^8 - 37^8 ---> apply a^2-b^2=(a+b)(a-b) ---> 47^8 - 37^8 = (47^4+37^4)(47^4-37^4) = (odd+odd)(odd+odd) = even*even = even . NOT a prime number.

Hope this helps.
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Hi, thanks for the reply. However how do I go about finding the prime factors for the last 2 parts?
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Prime factorisation 16 Feb 2016, 18:38
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Kwenyao
Engr2012
Kwenyao
For each of the following numbers, determine if the number is prime. If not, factorize
it into product of primes if it is not prime.
A)1747
B)139^3 + 1
C)47^8 - 37^8

How do i go about doing these?

By definition, a prime number 'n' is a number that is not divisible by anything else but 1 and n.

Let us analyse each of the 3 options:

A)1747 : try the usual way. 41^2= 1681, so you need to check for all numbers from 1 to 41 to see if there are any divisors of 1747. You wont find any. Thus 1747 is a prime number.

B)139^3 + 1: no need to solve. Recognise that 9^3 ends in 9 and thus the last digit 9+1 = last digit of 0. As 0 is an even number (divisible by 2)---> the resulting number 139^3+1 is NOT a prime.

C)47^8 - 37^8 ---> apply a^2-b^2=(a+b)(a-b) ---> 47^8 - 37^8 = (47^4+37^4)(47^4-37^4) = (odd+odd)(odd+odd) = even*even = even . NOT a prime number.

Hope this helps.

Hi, thanks for the reply. However how do I go about finding the prime factors for the last 2 parts?
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You should not even think about prime factorizing such big numbers. It wont make sense. As shown above, you dont really have to go into prime factorization to check whether some of the larger numbers actually prime or not.
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Kwenyao
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Prime factorisation 16 Feb 2016, 18:46
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Yes, but the second part of the question states that i am supposed to factorize it into a product of primes if the number is not a prime
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Prime factorisation 19 Feb 2016, 16:27
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The standard way to approach (C) would be to do what Engr2012 suggests above. It's a classic 'hidden special quadratic' problem: whenever you see a difference of even exponents, like 100^6-8^10 or x^8-y^8, use the difference of squares rule to factor it. That said, (C) isn't really a fair problem. You can quickly factor down differences of squares, but when you're looking at a sum of squares, like 47^4 + 37^4, there isn't a general rule. You wouldn't be asked to do something like this on the GMAT unless there was some hidden trick to it, which in this case there isn't. Basically, the problem just isn't very well designed.



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