###### Asked in School Subjects, Ancient Egypt, Paper

# Gene has 2n pieces of paper numbered 1 through 2n He removes n pieces of paper that are numbered consecutively The sum of the numbers on the remaining pieces of paper is 1615 Find all possible value n?

## Answer

###### Wiki User

###### 01/16/2008

I am fairly sure the only possible values are 34 and 38 ..........

From observations made on small series' of 1 to 2n, the amount remaining on the paper after the numbers are crossed out is equal to (1 + 2+ .... +n ) + kn where k is any integer between 0 and n. ( this formula is not too hard to find, but would take a while to explain) The sum of the numbers 1 to n is Â½ (n2 +n). ( this can be proven by adding up pairs of numbers, starting from the outside of the series and going in, e.g. if n= 100, you add up (100 +1) + (99 +2) + (98 +3).......) Therefore: Â½ (n2 + n) +nk = 1615 So: (n2 + n) + 2nk = 3230 And: n + 1 + 2k = 3230/n As we want an integer value for n + 1 + k, we look at the factors of 3230. 3230 = 2 x 5 x 17 x 19 n must be a combination of these factors There are many combinations of these factors , but some e.g. 2 are going to be way off. Instead of using tonnes of trial and error, lets find a range. Min value: if n is low, k is high, therefore min value for n would occur when k is n. Subbing in values, we get 3n2 + n - 3230 = 0 using a quadratics function on graphics calculator (if you dont have one, go look up the quadratic formula) this equals approx 32.6 max value for n would have k = 0 n2 + n = 3230 = 56.33 approx. Combining the factors the values found within this range are: 34 (2x17), 38 (2x19) If n = 34 34 + 1 + 2k = 3230/34 2k = 95-35 2k= 60 k = 30 this means it works as k is less than n and is an integer. If n= 38 38 +1 +2k = 3230/38 2k = 85 - 39 2k =46 k= 23

i hope this is right - Cat : )