Maths Olympiad Prep

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, 2016

Number theory Difficulty 7.7 National olympiad, round 2 Prove it Asia Pacific Mathematics Olympiad (APMO)

A positive integer is called fancy if it can be expressed in the form
2a1+2a2++2a100 2^{a_{1}} + 2^{a_{2}} + \cdots + 2^{a_{100}}
where a1,a2,,a100a_{1}, a_{2}, \ldots, a_{100} are non-negative integers that are not necessarily distinct.
Find the smallest positive integer nn such that no multiple of nn is a fancy number.

Solution

Let kk be any positive integer less than 210112^{101}-1. Then kk can be expressed in binary notation using at most 100 ones, and therefore there exists a positive integer rr and non-negative integers a1,a2,,ara_{1}, a_{2}, \ldots, a_{r} such that r100r \leq 100 and k=2a1++2ark = 2^{a_{1}} + \cdots + 2^{a_{r}}. Notice that for a positive integer ss we have:
2sk=2a1+s+2a2+s++2ar1+s+(1+1+2++2s1)2ar=2a1+s+2a2+s++2ar1+s+2ar+2ar++2ar+s1 \begin{aligned} 2^{s} k & = 2^{a_{1}+s} + 2^{a_{2}+s} + \cdots + 2^{a_{r-1}+s} + \left(1 + 1 + 2 + \cdots + 2^{s-1}\right) 2^{a_{r}} \\ & = 2^{a_{1}+s} + 2^{a_{2}+s} + \cdots + 2^{a_{r-1}+s} + 2^{a_{r}} + 2^{a_{r}} + \cdots + 2^{a_{r}+s-1} \end{aligned}
This shows that kk has a multiple that is a sum of r+sr+s powers of two. In particular, we may take s=100r0s = 100 - r \geq 0, which shows that kk has a multiple that is a fancy number.

We will now prove that no multiple of n=21011n = 2^{101} - 1 is a fancy number. In fact we will prove a stronger statement, namely, that no multiple of nn can be expressed as the sum of at most 100 powers of 2.

For the sake of contradiction, suppose that there exists a positive integer cc such that cnc n is the sum of at most 100 powers of 2. We may assume that cc is the smallest such integer. By repeatedly merging equal powers of two in the representation of cnc n we may assume that
cn=2a1+2a2++2ar c n = 2^{a_{1}} + 2^{a_{2}} + \cdots + 2^{a_{r}}
where r100r \leq 100 and a1<a2<<ara_{1} < a_{2} < \ldots < a_{r} are distinct non-negative integers. Consider the following two cases:

- If ar101a_{r} \geq 101, then 2ar2ar101=2ar101n2^{a_{r}} - 2^{a_{r}-101} = 2^{a_{r}-101} n. It follows that 2a1+2a2++2ar1+2ar1012^{a_{1}} + 2^{a_{2}} + \cdots + 2^{a_{r-1}} + 2^{a_{r}-101} would be a multiple of nn that is smaller than cnc n. This contradicts the minimality of cc.

- If ar100a_{r} \leq 100, then {a1,,ar}\{a_{1}, \ldots, a_{r}\} is a proper subset of {0,1,,100}\{0, 1, \ldots, 100\}. Then
ncn<20+21++2100=n. n \leq c n < 2^{0} + 2^{1} + \cdots + 2^{100} = n.
This is also a contradiction.

From these contradictions we conclude that it is impossible for cnc n to be the sum of at most 100 powers of 2. In particular, no multiple of nn is a fancy number.

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