Paul and Kassey's Favorite Number :D

Kassey's favorite number is the integer n n . Paul's favorite number is the integer n + 1 n+1 .

Given that both n n and n + 1 n+1 have exactly four divisors and have the same sum of divisors. What is the number of divisors of their product?


The answer is 16.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Christopher Boo
Feb 23, 2015

n n and n + 1 n+1 are co-prime. That is, they don't share the same divisor except 1 1 . Hence,

d ( n ) × d ( n + 1 ) = 4 × 4 = 16 d(n) \times d(n+1)=4\times 4=16

Oh no I didn't thought of this, I found the numbers as 14 , 15 14,15 but not after wasting my 10 minutes.

Ronak Agarwal - 6 years, 3 months ago

Log in to reply

Can't the numbers be 27 and 28

Archit Boobna - 6 years, 2 months ago

Log in to reply

28 has 6 factors- 1,2,4,7,14,28 but 27 has 4

Kushagra Sahni - 5 years, 3 months ago

Great work @Christopher Boo ..Didn't think of this ..I worked out the numbers to be 14 and 15.

Ankit Kumar Jain - 4 years, 3 months ago
Paul Ryan Longhas
Feb 23, 2015

To say the numbers have exactly four divisors means that they are each a product of two distinct prime numbers, say p q pq and r s rs . To say the sum of their divisors is the same means 1 + p + q + p q = 1 + r + s + r s 1 + p + q + pq = 1 + r + s + rs , and to say they are consecutive means p q + 1 = r s pq + 1 = rs . Putting the two equations together gives p + q = 1 + r + s p+q = 1+r+s , which can happen only when one of the primes is even (hence 2 2 ). A little trial and error produces some primes that work: p = 2 , q = 7 , r = 3 p = 2, q = 7, r = 3 , and s = 5 s = 5 . So the two numbers could be 14 14 and 15 15 , both with sum of factors equal to 24 24 . The number of factors of 14 15 = 2 3 5 7 14 * 15 = 2 * 3 * 5 * 7 is 2 4 = 16 2^4 = 16 .

Given Christopher's answer, it probably would have been better to have just asked for the least possible sum of the favorite numbers, namely 29. 29. But this does raise an even more interesting question: how many more possibilities are there for the two favorite numbers? I'm finding that we must have that

n = 2 p + 2 ( p + 1 ) p 2 n = 2p + \dfrac{2(p + 1)}{p - 2}

for some prime p p , on the condition that, of course, n n must be a positive integer. Now both p = 3 p = 3 and p = 5 p = 5 yields n = 14 n = 14 , but I don't think that there are any more solutions, i.e., n = 14 n = 14 is the only solution. For p = 7 p = 7 we don't get an integer, and for primes p > 7 p \gt 7 we have that

2 < 2 ( p + 1 ) p 2 < 3 2 \lt \dfrac{2(p + 1)}{p - 2} \lt 3 ,

and so we can never obtain an integral value for n n other than 14 14 .

@Paul Ryan Longhas I have posted a rephrased version of your question that reflects this last calculation. I hope that's o.k. with you; I have made reference to your question in a comment to my solution of the new question.

Brian Charlesworth - 6 years, 3 months ago

Log in to reply

That's exciting. Could you post this version of the problem? Thanks!

Calvin Lin Staff - 6 years, 3 months ago

Log in to reply

O.k., here it is. :) I'll start playing around with variations of this problem, see if anything interesting comes up.

Brian Charlesworth - 6 years, 3 months ago

Lol! We posted same thing at same time. Given you have posted explanation, I can delete my comment.

Pranjal Jain - 6 years, 3 months ago

Log in to reply

Haha. Yeah, I guess once is enough. :) It is an interesting question though to show that there is a unique solution to what the favorite numbers are.

Brian Charlesworth - 6 years, 3 months ago

Yes, I can feel that the problem maker wasn't expecting my type of solution. The problem should be stated in a different way such that the answer is non-trivial.

Christopher Boo - 6 years, 3 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...