How can both sums be 90, yet Don's sum is greater?
I think you mean that a > x
If Don's numbers cover a range of n values, beginning with a, then the sum is
a(a+2n-1)/2
Hmmm. The only solution I can find starting with a number from 1-10 is the set of numbers from 6-14.
Any others out there?
Don chooses a positive integer less than 11 which he calls a and an integer b such that 11<=b<=20. He adds all the positive integers between a and b, including a and b. His sum is 90. Jim does the same as Don but chooses different numbers from Don. He calls them x and y. He is surprised to find that his sum is also 90. The sum of Don's chosen numbers is greater than Jim's. Who had what numbers
2 answers
There are actually 2 such solutions,
I made up a nifty primitive computer program (GW-Basic from the 1980's)
and it gave me the following results:
The first column is a, the second is b
10 FOR A = 1 TO 10
20 FOR B = 11 TO 20
40 IF (A+B)*(B-A+1)/2 = 90 THEN PRINT A,B
60 NEXT B
70 NEXT A
Ok
run
2 13
6 14
Ok
Steve had the 2nd set
6 and 14 for 9 terms
sum = (6+14(9/2) = 90
my additional was 2 and 13 for 12 terms
so sum = (2+13)(12/2) = 90
Since the sum of Don's chosen numbers is greater than Jim's,
Don had picked 6 and 14, Jim had picked 2 and 13
I made up a nifty primitive computer program (GW-Basic from the 1980's)
and it gave me the following results:
The first column is a, the second is b
10 FOR A = 1 TO 10
20 FOR B = 11 TO 20
40 IF (A+B)*(B-A+1)/2 = 90 THEN PRINT A,B
60 NEXT B
70 NEXT A
Ok
run
2 13
6 14
Ok
Steve had the 2nd set
6 and 14 for 9 terms
sum = (6+14(9/2) = 90
my additional was 2 and 13 for 12 terms
so sum = (2+13)(12/2) = 90
Since the sum of Don's chosen numbers is greater than Jim's,
Don had picked 6 and 14, Jim had picked 2 and 13
0 answers