An integer $n\geq 2$ having exactly $s$ positive divisors $1=d_1<d_2<\cdots<d_s=n$ is said to be good if there exists an integer $k$, with $2\leq k\leq s$, such that $d_k>1+d_1+\cdots+d_{k-1}$. An integer $n\geq 2$ is said to be bad if it is not good. (a) Show that there are infinitely many bad integers. (b) Prove that, among any seven consecutive integers all greater than $2$, there are always at least four good integers. (c) Show that there are infinitely many sequences of seven consecutive good integers.
Problem
Source: BxMO 2018, Problem 4
Tags: BxMO, Benelux, number theory
Tanb
28.04.2018 19:30
For a) just choose a power of 2.
Superguy
01.05.2018 17:14
Solution A) Choose $n=2^m$ $\boxed{Proof}$ If we choose any such k Then sum of divisors of n less than $2^k$ is $d_1+d_2...........d_{k-1}=1+2+2^2+2^3+......2^{k-1}\implies 2^k-1$ Now as per the condition for good number We must have $d_k>1+d_1+\cdots+d_{k-1}$ $\implies 2^k>2^k$ which is clearly a contradiction. We conclude that all such n is bad$\square$ B) We can easily conclude that all odd numbers are good. As $d_2>1+d_1$ So if among seven consecutive integers we have Three multiples of 2 we are done Lets see the case when we have 4 multiples of 2
When we have three multiples of 3
In this case we must have
First and last number multiples of 6
Consider the sequence of consecutive integers
$\boxed{a_1,a_2,a_3,a_4,a_5,a_6,a_7}$
Here we have
$a_1\equiv 0\pmod{6}$
$a_2\equiv 1\pmod{2}$
$a_3\equiv 2,4\pmod{6}$
$a_4\equiv 3\pmod{6}$
$a_5\equiv 2,4\pmod{6}$
$a_6\equiv 1\pmod{2}$
$a_7\equiv 0\pmod{6}$
Now either $a_1\equiv a_5\equiv 0\pmod{4}-(1)$
Or $a_3\equiv a_7\equiv 0\pmod{4}-(2)$
From $(1)$
We get $\boxed{a_2,a_3,a_4,a_6}$ are good.
From $(2)$
We get $\boxed{a_2,a_4,a_5,a_6}$ are good.
When we have 2 multiples of 3
We must have an odd multiple of 3
Similar arguement as above gives four good numbers.
C) Consider a sequence of consecutive integers $\boxed{a_1,a_2,a_3,a_4,a_5,a_6,a_7}$ Where $a_1\equiv 3\pmod{6}$ $a_2\equiv 2\pmod{4}$ $a_3\equiv 1\pmod{2}$ $a_4=288l+12$ $a_5\equiv 1\pmod{2}$ $a_6\equiv 2\pmod{4}$ $a_7\equiv 3\pmod{6}$ Now for $a_4$ We choose $l=5\cdot7\cdot11\cdot13\cdot17\cdot19\cdot23\cdot29\cdot(31b+11)$ We get $31|a_4$ Thus $31=d_7>1+d_1+d_2+d_3+d_4+d_5+d_6=1+1+2+3+4+6+12=29$ which is true hence $a_4$ is good.Similarly we can get all seven numbers are good.We also have an infinite sequence of such integers.$\square$