Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The question is to find a thousand natural numbers such that their sum equals their product. Here's my approach :

I worked on this question for lesser cases :

\begin{align} &2 \times 2 = 2 + 2\\ &2 \times 3 \times 1 = 2 + 3 + 1\\ &3 \times 3 \times 1 \times 1 \times 1 = 3 + 3 + 1 + 1 + 1\\ &7 \times 7 \times 1 \times 1 \times \dots\times 1 \text{ (35 times) } = 7 + 7 + 1 + 1 .... \text{ (35 times) } \end{align}

Using this logic, I seemed to have reduced the problem in the following way.

$a \times b \times 1 \times 1 \times 1 \times\dots\times 1 = a + b + 1 + 1 +...$

This equality is satisfied whenever $ ab = a + b + (1000-n)$ Or $ abc\cdots n = a + b + \dots + n + ... + (1000 - n)$ In other words, I need to search for n numbers such that their product is greater by $1000-n$ than their sum. This allows the remaining spots to be filled by $1$'s. I feel like I'm close to the answer.

Note : I have got the answer thanks to Henning's help. It's $112 \times 10 \times 1 \times 1 \times 1 \times ...$ ($998$ times)$ = 10 + 112 + 1 + 1 + 1 + ...$ ($998$ times)

This is for the two variable case. Have any of you found answers to more than two variables ?

$abc...n = a + b + c + ... + n + (1000 - n) $

share|cite|improve this question
    
If you don't care how many numbers you have on each side, you can find solutions with any number of non-$1$ numbers. If you want five, choose any five positive integers $a,b,c,d,e$ then add $abcde-a-b-c-d-e\ 1$'s. – Ross Millikan Feb 2 at 4:55
    
@RossMillikan Well, not any number. We can't have the product of distinct 999 numbers equal to its sum. But, the problem is the more variables you introduce, the harder it is to find the combination. I wonder how many number choices are possible ... If 999 is the limit or if it comes much sooner than that. – user230452 Feb 2 at 11:53
    
I said any number of non-$1$s. You are right that it gets harder to find solutions with a given total number of numbers as the number of non-$1$'s gets large. – Ross Millikan Feb 2 at 14:30
    
The following pattern yields a finite sequence of natural numbers, whose sum equal its product: $A_1=k,A_2=2,A_3,\dots,A_k=1$. A few examples: $[2,2],[3,2,1],[4,2,1,1],[5,2,1,1,1]$. Following this pattern, a sequence of $1000$ numbers whose sum equals its product (equals $2000$) is $1000,2,\underbrace{1,\dots,1}_{998\text{ times}}$. – barak manos Jul 6 at 10:09

There's a sign error in your final equation; you want $$ a+b+998=ab $$ which simplifies to $$ (a-1)(b-1) = 999 $$ from which it should be easy to extract several integer solutions.

share|cite|improve this answer
1  
Your idea of adding one on both sides of the equation to factorise was brilliant. Well done – user230452 Feb 1 at 13:37

Do we have an alternative to fill up with ones?

Let $N=1000$, then \begin{align} \sum_{i=1}^N n_i &= \prod_{i=1}^N n_i \iff \\ N n_a &= n_g^N \end{align} where $n_a$ is the arithmetic mean and $n_g$ the geometric mean of the numbers $n_i$.

Those numbers on the left and right hand side of the equation drift apart very fast. Already for $n_a = n_g = 2$ we would have $2000$ vs. $2^{1000} \approx 10^{301}$.

Assuming $n_a \approx n_g$ we estimate: $$ N x = x^N \Rightarrow \\ N = x^{N-1} \Rightarrow \\ x = \sqrt[N-1]{N} $$ For $N=1000$ this gives $x=\sqrt[999]{1000}=1.0069\dotso$.

Such an estimated mean gives not much room for numbers $n_i > 1$. $$ 1.007 = \frac{1000+a}{1000} = 1 + a/1000 \Rightarrow \\ a = 7 $$ Spreading an excess of $7$ over a couple of numbers is too pessimistic. $$ 1.007 = \sqrt[1000]{b} \Rightarrow \\ b \approx 1070 $$ That looks better, the product has to be equal with a sum something above $1000$.

Have any of you found answers to more than two variables ?

A solution with three numbers different from $1$ is $x=67$, $y=z=4$. This gives $67+4+4+997=1072$. Also $67\times 4 \times 4 \times 1^{997}=1072$.

share|cite|improve this answer

A solution with four numbers different from 1 is:

$$16 \times 4 \times 4 \times 4 \times 1^{996} = 16 + 4 + 4 + 4 + (996 \times 1) = 1024$$

How was this found? $1024 = 2^{10}$ appeared to be a promising candidate for the sum and product because it's slightly larger than 1000 and has many factors. The problem then was to find a, b, c, d such that: $$a+b+c+d=10$$ $$2^a+2^b+2^c+2^d=1024 -(1000-4)=28$$ None of a-d can be more than 4 since $2^5=32 > 28$. But on trying $a=4$, reducing the problem to finding b,c,d such that $b+c+d=6$ and $2^b+2^c+2^d = 12$, the solution was apparent.

share|cite|improve this answer
    
Nice. How did you find this ? – user230452 Feb 11 at 22:54
    
@user230452 I've edited my answer to explain how the solution was found. – Adam Bailey Feb 12 at 10:23

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.