A positive integer N is interoceanic if its prime factorization N=px11px22⋯pxkk satisfies x1+x2+⋯+xk=p1+p2+⋯+pk. Find all interoceanic numbers less than 2020.
Problem
Source: Centroamerican 2020, problem 6
Tags: number theory, prime factorization
29.10.2020 01:00
We have that 28⋅3⋅5>2020, so N is divisible only by 2 and 3 and it should be easy to look at all the cases from here. @below Yeah, this was just off the top of my head.
29.10.2020 01:06
Math00954 wrote: We have that 28⋅3⋅5>2020, so N is divisible only by 2 and 3 and it should be easy to look at all the cases from here. The prime factorization does not necessarily have to contain all of the first k primes. 27−n5n works for 1≤n≤3 and 28⋅7 does as well.
29.10.2020 09:15
I proposed this problem. This is more or less the solution I sent First, let's prove that k≤3. If there are at least 4 prime numbers, we have that p1+p2+p3+p4≥2+3+5+7=17. x1+x2+x3+x4=p1+p2+p3+p4≥17→N>2x1⋅2x2⋅2x3⋅2x4=2x1+x2+x3+x4=217>2020, which is a contradiction. Now we have three cases: k=3 Assume WLOG that p1<p2<p3. N=px11⋅px22⋅px33>2x1⋅2x2⋅4x3=2x1+x2+2x3=2p1+p2+p3+x3≥22+3+5+x3=210+x3≥211>2020. We have a contradiction, so there are not solutions for k=3. k=2 We have that p1+p2=x1+x2≤10. We are limited to use the prime numbers 2,3,5,7. We have to do some casework with the pairs (2,3),(2,5),(2,7),(3,5),(3,7) Case 1: x1+x2=2+3=5 We have the solutions N=34⋅2,33⋅22,32⋅23,3⋅24 Case 2: x1+x2=2+5=7 54⋅23>2020, the solutions for this case are N=53⋅24,52⋅25,5⋅26 Case 3: x1+x2=2+7 The only solution is N=28⋅7 Case 4:x1+x2=3+5=8 Since 37⋅5>2020, there are not solutions in this case. Case 5:x1+x2=3+7=10 Since 39⋅7>2020, there are not solutions in this case. k=1 We take N=22,33. The interoceanic numbers are: 4,27,48,72,108,162,320,800,1792,2000