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 ?
Printable View
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 toQuote:
Originally Posted by timkuc
.
, 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.