Thread: 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 ?

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 .

, where k is the duplicated number.

.

Note that if N = 60, 60*61/2 = 1830. This means that k = 20, and the difference between N and k is 40.

so we have to do it intuitively (means by the guessing ) ? We can't get value by calculation right ?


hmmm got it . thanks

Yeah, guess-and-check seems to be the best solution.

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.