How to get the n if I know the result of 1+2+3+..+n=n*(n+1)/2?

Question

If you know the result of 1+2+3+..+n, which is n*(n+1)/2.
For example, if the result is 5050, then I can know n is 100. How can I obtain the n. However, you can only calculate the n by only addition and substraction.

And I know how I can get the n i is that I can traverse the natural number from 1 to n, by computing per different 1+2+3+...+n, like 1, 1+2, 1+2+3,...1+2+..+n, then I can check every result to 5050, then I can find the n is 100. But I find that the calculating steps will over 2000 steps, so there is a good algorithm to find the n?
Thanks!.

Adding the numbers from 1 to 100 should only take you 100 steps, not 2000. — melpomene, Jan 29 '17 at 13:26
Edit your **tag** and add what kind of **programming language** — Shakiba Moshiri, Jan 29 '17 at 13:26
@melpomene, not from 1 to 100, but from 1 to 2, then 1 to 3, until 1 to 100, then you can find the correct n. — zhenguoli, Jan 29 '17 at 13:27
@k-five, there is no specific programming language, but the language can only support *addition* and *substraction*. And I want a good algorithm to the problem. — zhenguoli, Jan 29 '17 at 13:29
@zhenguoli Why restart from 1 every time? If you just computed 1+2+...+n, you can get 1+2+...+n+(n+1) by simply adding n+1 to your previous result. — melpomene, Jan 29 '17 at 13:29
@adamwy, I have said that the programming language just support the *addtion* and *substraction*. — zhenguoli, Jan 29 '17 at 13:31

score 3 · Answer 1 · answered Jan 29 '17 at 14:45

3

Note that:

1 + 2 + ... + N = V = N(N+1)/2

Can be rewritten as:

N^2 + N - 2V = 0

The value you are looking for is thus:

(-1+sqrt(1+4*2*value))/2;

That is one of the zeros of the given quadratic equation.

In C++ you can use a function like this:

int rev(int value) {
    return (-1+std::sqrt(1+4*2*value))/2;
}

answered Jan 29 '17 at 14:45

skypjack

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"However, you can only calculate the n by only addition and substraction." – Grzegorz Górkiewicz Jan 30 '17 at 09:48

Grzegorz Górkiewicz · Accepted Answer · 2017-01-29T14:31:28.147

2

Only by addition and subtraction... Simply add numbers from 1 and check whether the sum is the given number.

In pseudocode. Given 5050.

number := 5050
next, sum := 0
while sum <= number
  next := next + 1
  sum := sum + next
return next

edited Jan 29 '17 at 14:31

answered Jan 29 '17 at 13:29

Grzegorz Górkiewicz

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Thanks, this is what I want. – zhenguoli Jan 29 '17 at 13:32
1

you should change the condition in the loop to `while sum <= number`. Otherwise this code will produce an infinite loop, if `number` isn't of the form `1 + 2 + ... + n`. – Jan 29 '17 at 14:30
@Paul, yes. Edited, but tbh I only cared about the algorithm being correct for allowed input `number`. – Grzegorz Górkiewicz Jan 29 '17 at 14:32

score 0 · Answer 3 · edited Jun 20 '20 at 09:12

Suppose 1 + 2 + .. + n = n*(n+1)/2 = r (r is known n and is unknown).

Now we want to know the positive value of n for which the above equation is satisfied.

Theoretical method

We have, n^2+n=2*r => (n+1/2)^2 = 2*r + 1/4 => n + 1/2 = sqrt(2*r+1/4) (taking only the +ve sqrt since n>0) => n = sqrt(2*r+1/4) - 1/2 For example, r=55 => n =10, which can be found in O(1) time by plugging the value of r.

Numerical method

Now, we may fidn the solution numerically as well, by finding the positive root of the equation (n^2+n)/2-r=0. If r=55, the graph of the quadratic equation looks like the following, the root can be found out numerically by using bisection or newton-raphson like numerical methods. The following graph of the quadratic polynomial shows that n=10 for r=55.

How to get the n if I know the result of 1+2+3+..+n=n*(n+1)/2?

3 Answers3

Theoretical method

Numerical method