Number Theory 1

• Jul 20th 2012, 10:26 PM
timkuc
Number Theory 1
I add first N natural numbers and finds the sum to be 1850 . But actually one number was added twice by mistake .
Find the difference between N and that number ?
• Jul 20th 2012, 10:29 PM
Prove It
Re: Number Theory 1
Quote:

Originally Posted by timkuc
I add first N natural numbers and finds the sum to be 1850 . But actually one number was added twice by mistake .
Find the difference between N and that number ?

The sum of the first N natural numbers is equal to \displaystyle \begin{align*} \frac{N(N+1)}{2} \end{align*}.
• Jul 20th 2012, 10:31 PM
richard1234
Re: Number Theory 1
$1 + 2 + \dots + N + k = 1850$, where k is the duplicated number.

$\frac{N(N+1)}{2} + k = 1850$.

Note that if N = 60, 60*61/2 = 1830. This means that k = 20, and the difference between N and k is 40.
• Jul 20th 2012, 10:33 PM
timkuc
Re: Number Theory 1
so we have to do it intuitively (means by the guessing ) ? We can't get value by calculation right ?

hmmm got it . thanks
• Jul 20th 2012, 11:19 PM
richard1234
Re: Number Theory 1
Yeah, guess-and-check seems to be the best solution.
• Jul 21st 2012, 12:40 AM
Deveno
Re: Number Theory 1
well we know that N(N+1)/2 < 1850 so that:

N(N+1) < 3700.

that is: N2 + N - 3700 < 0

using the quadratic equation, we have:

N2 + N - 3700 = (N - (1/2)(19√41 - 1))(N + (1/2)(19√41 + 1)).

for this to be < 0, the factors must be of different signs, and since N is assumed positive,

we must have N - (1/2)(19√41 - 1) < 0, that is N < (1/2)(19√41 - 1) ~ 60.33

so 60 is the largest natural number N could be.

how small could N be? since k (the repeated number) must be ≤ N, we have:

1850 ≤ N(N+1)/2 + N, so

0 ≤ N2 + 3N - 3700, and thus:

0 ≤ (N - (1/2)(√14809 - 3))(N + (1/2)(√14809 + 3)), which means N ≥ (1/2)(√14809 - 3) ~ 59.35

this means that 60 is the smallest natural number N can be.

therefore we conclude N = 60, whereupon the rest follows by richard1234's post.

no guessing required.