2011 N1 = 2024 A2

I am always harping on my students to write solutions well rather than aiming for just mathematically correct, and now I have a pair of problems to illustrate why.

Shortlist 2011 N1

Here is Shortlist 2011 N1, proposed by Suhaimi Ramly:

For any integer ${d > 0}$ , let ${f(d)}$ be the smallest positive integer that has exactly ${d}$ positive divisors (for example, ${f(1)=1}$ , ${f(5)=16,}$ and ${f(6)=12}$ ). Prove that for every integer ${k \geq 0}$ , ${f(2^k)}$ divides ${f(2^{k+1})}$ .

I like this problem, so try it out if you haven’t. This is a problem for which the correct idea is quite nice, but it’s easy to get lost in notation and formalism and miss it.

(Actually. Do the problem, or you won’t get anything out of this blog post.)

I’m actually willing to claim that basically every solution to 2011 N1 is essentially identical in terms of mathematical content. But depending on how you wrote it, the idea may be more or less obscured. In other words, Shortlist 2011 N1 is a problem for which it’s unusually common that a student can write a 7-point paper while lacking a complete understanding. I won’t pick on anyone in particular, but it’s not hard to find examples of write-ups that are worth 7 points but do not make it easy to understand the key idea.

Shortlist 2024 A2

This problem came up again in my memory because on the 2024 IMO Shortlist there was a problem whose key idea is secretly the same as 2011 N1. It left an impression on me because when I worked on the problem, I had the revelation that I had seen the same idea before. It’s Shortlist 2024 A2:

Let ${n}$ be a fixed positive integer. Find the minimum possible value of ${S = 2^0x_0^2 + 2^1x_1^2 + \dots + 2^nx_n^2}$ , where ${x_0}$ , ${\dots}$ , ${x_n}$ are nonnegative integers such that ${x_0+\dots+x_n=n}$ .

The reason I like this pair so much is that normally, when I try to talk about “full understanding”, I have to be pretty nebulous and philosophical. But for this particular pair of problems, I can be precise: a full understanding of 2011 N1 means you should easily find the elegant solution to 2024 A2.

And that’s why if you only focus on writing technically-mathematically-correct 7-point solutions in practice, instead of writing well, you are going to lose out.

(The 2024 A2 problem has two solution paths, one that I think is elegant, and one that uses an annoying calculation via induction on ${n}$ . Try to find the non-induction solution.)

4 thoughts on “2011 N1 = 2024 A2”

The elegant solution also has the advantage that it generalizes to the continuous case where xi, n do not have to be integers.

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Would APMO 2017/3 lie on the same line of problems?

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You can actually determine f(2^k) exactly in N1. For k>=3, it’s 4*(the product of the first k-1) primes. Then the result is obvious.

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@above this is incorrect. f(64)=2^3*3^3*5*7, which is less than 2^3*3*5*7*11.

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presheaf 27 July, 202507:48 Reply

The elegant solution also has the advantage that it generalizes to the continuous case where xi, n do not have to be integers.

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superb2970ac623e 4 August, 202504:31 Reply

Would APMO 2017/3 lie on the same line of problems?

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math fan 4 August, 202518:25 Reply

You can actually determine f(2^k) exactly in N1. For k>=3, it’s 4*(the product of the first k-1) primes. Then the result is obvious.

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Vincent Wang 6 September, 202511:40 Reply

@above this is incorrect. f(64)=2^3*3^3*5*7, which is less than 2^3*3*5*7*11.

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