Asked by Lone
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
Answers
Answered by
Steve
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?
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?
Answered by
Reiny
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