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01 Nov 2016, 06:56
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There exists a positive prime number n, such that 16n+1 equals the cube of a positive integers. What is the value of n?
This is AIME question. Logic used in solving this question can come in handy for tackling “hard” gmat problems. Sorry I don’t have answer choices.
Kudos for correct solution.
This is AIME question. Logic used in solving this question can come in handy for tackling “hard” gmat problems. Sorry I don’t have answer choices.
Kudos for correct solution.
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01 Nov 2016, 07:46
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It's 307 but I have no idea how you're supposed to this fast. The dumb way I did it was plugging in 999 to n and figuring out that that the highest cube number will be less than 15985 so it's gonna be a cube between 1-25. Also you can eliminate all the even numbers since n is odd and multiplied by 16 + 1 will always get you an odd number. So then you cube every odd number from 1-25, subtract 1 from the result, and find out which one is divisible by 16. Eventually you get 17 cubed which is just 16 x 307 + 1. But that took me like 15-20 minutes and I'm sure there's a much easier way to figure it out
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01 Nov 2016, 08:39
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Best tool for tackling such kind of questions is “parity”.
Let’s represent question as following expression.
\(16n+1=n^3\)
\(n^3-1=16n\)
\((n-1)(n2+n+1)=2^4n\)
Now at left side \(n^2+n+1\). Let’s check it’s parity.
If n is even => \((2n)^2+2n+1\) this expression is odd
If n is odd => \((2n+1)^2+2n+1+1=4n^2+4n+1+2n+1+1=4n^2+6n+3=e+e+odd=odd\) So this expression is always odd.
n-1 can be either odd o even.
That’s why all factors of 2 could be absorbed only by n-1. And we have n-1=16
n=17.
Now what’s left is to put it into \(n^2+n+1\) and find the product. The result will be 307.
You’ve got you kudos the hard way.
Let’s represent question as following expression.
\(16n+1=n^3\)
\(n^3-1=16n\)
\((n-1)(n2+n+1)=2^4n\)
Now at left side \(n^2+n+1\). Let’s check it’s parity.
If n is even => \((2n)^2+2n+1\) this expression is odd
If n is odd => \((2n+1)^2+2n+1+1=4n^2+4n+1+2n+1+1=4n^2+6n+3=e+e+odd=odd\) So this expression is always odd.
n-1 can be either odd o even.
That’s why all factors of 2 could be absorbed only by n-1. And we have n-1=16
n=17.
Now what’s left is to put it into \(n^2+n+1\) and find the product. The result will be 307.
You’ve got you kudos the hard way.
