Another GMAT Prep Question
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H(n) = 2*4*6*8.....n ( all are even integers)winny wrote:someone please explain...
lets take example:
H(8)+1 = 2.4.6.8 +1 = 2^6(1*2*3)+1 = 385 ; so, least prime factor H(8)+1 is 5 (i.e 5*7*11)
similarly..H(10)+1 = 2*4*6*8*10 +1 = 2^7*(1*2*3*5)+1= 3841;
so, least prime factor of H(10)+1 is 23 (i.e. 23*167)
now come to H(100)+1 = 2^50*(1*2*3*...*50) + 1, will not have prime factors < = 50.
Hence the least prime factor must be >50
Answer E!
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h(100) + 1 = (2*4*6*8*10*...*100) + 1 = 2*(1*2*3*4*5*...*50) + 1winny wrote:someone please explain...
Thus when [h(100) + 1] is divided by any integer (including all the primes) less than or equal to 50, it'll leave a remainder of 1. Thus p must be greater than 50 in turn p > 40.
The correct answer is E.
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I am unable to find any relation between your examples and the solution of the question.n@resh wrote:H(n) = 2*4*6*8.....n ( all are even integers)winny wrote:someone please explain...
lets take example:
H(8)+1 = 2.4.6.8 +1 = 2^6(1*2*3)+1 = 385 ; so, least prime factor H(8)+1 is 5 (i.e 5*7*11)
similarly..H(10)+1 = 2*4*6*8*10 +1 = 2^7*(1*2*3*5)+1= 3841;
so, least prime factor of H(10)+1 is 23 (i.e. 23*167)
now come to H(100)+1 = 2^50*(1*2*3*...*50) + 1, will not have prime factors < = 50.
Hence the least prime factor must be >50
Answer E!
Looks like you jumped to the solution directly (please don't take me wrong).
let's do it with one of your example i.e. h(10)+1 = 2*4*6*8*10+1 = 2^4(1*2*3*4*5)+1
Now as per the remainder theorem if you divide (p+1)by p you will get remainder = 1.
The same logic is applicable here as well.
h(10)+1 = 2^4(1*2*3*4*5)+1 -- means if you divide this number by any integer till 5 you will get remainder = 1.
The same logic is applicable to this question (you don't need to multiply the whole thing and then find the prime factor because that will take lot of time)
h(100)+1 = 2^50 (1*2*.......*50) +1
this number when divided by any number till 50 will give 1 as the remainder. Hence your answer is greater than 50.
Option E