Brain buster

Sum on positive integers from 1 to n is equal to n(n+1)/2

Let m be the double counted number.

n >= m

n, m are positive integers.

n(n+1)/2 + m = 1000

m > 1

so n(n+1)/2 < 1000

n^2 + n - 2000 < 0

=> n < 45

Let's assume that n = 44

44(44+1)/2 + m = 1000

=> m = 10

if n < 44

m = 10 + 990 - n(n+1)/2 = 1000 - n(n+1)/2

1000 - n(n+1)/2 < n

2000 - n^2 - n < 2n

2000 - n^2 - 3n < 0

n^2 + 3n - 2000 > 0

n > (sqrt(8009) - 3)/2

n >= 44

so if n < 44 thenn >= 44 which is not possible so

n = 44 is the only possible

so m = 10 which the only solution

Result: 10

Sum of consecutive integers=n ((n+1))/2 1000=n ((n+1))/2 2000=n^2+n 0=n^2+n-2000 Use quadratic formula to solve

n= -1±√(1^2-4(1)(-2000))/2a

n= -45.222,or 44.224 Therefore, the boy added from 1 to 44, but that sum of 1-44 is only 990, so he had to have counted 10 twice. Double counting any number less than 10 won’t yield enough to reach 1000. Double counting any value over 10 yields a result over 1000.
Additional proof can be found in attempting to sum 1-43, which only yields 946. In order to reach 1000, you would need an extra 54. Since 54 is outside of the consecutive 1-43, that can’t be the solution. Or, consider the sum of 1-45. This is 1035, well over the 1000 without any double counting. The only solution is that the boy counted from 1-44 and counted 10 twice.

Sum of 1 to 44 is 990. To reach 1000 must have added 10, so 10 was double counted.