Which number is a factor of 1001^32-1

Hi,

I tried calculating it from the link provided by you and the no gets perfectly divided by 768.

The explanation provided is perfect. That is how we attack such question. Two consecutive numbers do not have any common factor. I am also for A.

Excellent explanation. I heard that 1001 is a number to be remembered as it is a product of 3 consecutive prime numbers 7, 11, 13 and is sometimes used in exams.

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Given \(1001^{32}-1\)--> The next consecutive integer is \(1001^{32}\) which can be factorized as \((7*11*13)^{32}\). Note that any 2 consecutive integers are co-prime, i.e. they don't have any common factor other than 1. As 7,11, and 13 are all factors of \(1001^{32}\), thus \(1001^{32}-1\) will have NO common factors. From the given options, we have to look for a number, which is not a factor of any of the three primes. Only A satisfies the criteria.
A.

+1 to DDB1981

Zarrolou,

Shouldnt we actually be concerned about \(1001^{32}-1\). Why have you calculated for \(1001^{19}-1\) ? Thank you.

_________
It's a typo.

Hi Bunnel,

Could you please comment if my understanding of the problem is correct.

a^n+b^n is divisible by a+b when n is even,and a^n-b^n is always divisible by a-b

〖1001〗^32-1=〖1001〗^32-1^32 is divisible by 1001+1=1002=2.3.167 and 1001-1=1000=2^3.5^3

From this limited information we know that 〖1001〗^32-1 is definitely a factor of 〖2.3.167.2〗^3.5^3 or any combinations of these integers

By implementing this information we can see that option B,C,D,E doesnot contain 167 as prime factor,there fore A must be an option as 768=(2^4.3).2^4

Thanks in advance

using a^2 - b^2 = (a+b) (a-b)
we get 1000* 1002^5 = 2^3 * 5^3 * 2^5 * 3^5 * 167^5
now check which options are divisible by 2,3,5 or 167

A. 768
B. 819
C. 826
D. 858
E. 924

A has 3 * 2^8 clear winner
for B-E it will take 20 seconds more...hope this approach is clear

A id the correct option as per above explation

The key to answering this question is to recognize that 1001^32 − 1 is a difference of squares
And so it 1001^16 - 1
And 1001^18 - 1
etc

1001^32 − 1 = (1001^16 + 1)(1001^16 - 1)
= (1001^16 + 1)(1001^8 + 1)(1001^8 - 1)
= (1001^16 + 1)(1001^8 + 1)(1001^4 + 1)(1001^4 - 1)
= (1001^16 + 1)(1001^8 + 1)(1001^4 + 1)(1001^2 + 1)(1001^2 - 1)
= (1001^16 + 1)(1001^8 + 1)(1001^4 + 1)(1001^2 + 1)(1001 + 1)(1001 - 1)

Now let's evaluate some of the NICE parts.
= (1001^16 + 1)(1001^8 + 1)(1001^4 + 1)(1001^2 + 1)(1002)(1000
= (1001^16 + 1)(1001^8 + 1)(1001^4 + 1)(1001^2 + 1)((2)(3)(167))((2)(2)(2)(3)(3)(3))

Now check the answer choices....
A. 768 = (2)(2)(2)(2)(2)(2)(2)(2)(3) = (2^8)(3)
Hmmm, it looks like we might not have enough 2's in the factorization of 1001^16 - 1 in order for 768 to be a factor.
However, if we recognize that (1001^16 + 1), (1001^8 + 1), (1001^4 + 1), and (1001of ^2 + 1) are all EVEN numbers, we can see that we have enough two's in the factorization of 1001^16 - 1 for 768 to be a factor.

Answer: A

Cheers,
Brent

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