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
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?
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
To solve this problem, we need to find the values of a, b, x, and y that satisfy the given conditions. Let's break down the problem step by step.
1. Don chooses a positive integer, which he calls "a," less than 11. This implies that the possible values for a are 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10.
2. Don also chooses an integer, called "b," such that 11 <= b <= 20. This means that the possible values for b are 11, 12, 13, 14, 15, 16, 17, 18, 19, or 20.
3. Don adds all the positive integers between a and b, including both a and b, and the sum is equal to 90. Let's express this sum mathematically:
a + (a+1) + (a+2) + ... + (b-2) + (b-1) + b = 90.
4. Now, let's move on to Jim. He also chooses a positive integer, calling it "x," such that 1 <= x <= 10 (since he cannot choose the same number as Don).
5. Jim chooses an integer, called "y," such that 11 <= y <= 20 (also excluding the number chosen by Don).
6. Jim adds all the positive integers between x and y, including both x and y, and the sum is also 90. Again, let's express this sum mathematically:
x + (x+1) + (x+2) + ... + (y-2) + (y-1) + y = 90.
7. The sum of Don's chosen numbers (a + b) is greater than Jim's chosen numbers (x + y). We need to find the values of a, b, x, and y that satisfy all these conditions.
Now, let's find the solution by trying different combinations of values for a, b, x, and y until we find the correct one:
Let's start with a = 1 and b = 20 for Don:
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 = 210
Now, let's try different values for Jim:
Jim: x = 4, y = 17
4 + 5 + 6 + ... + 16 + 17 = 153
This shows that the sum of Don's chosen numbers (210) is indeed greater than Jim's chosen numbers (153).
Therefore, Don had numbers a = 1 and b = 20, while Jim had numbers x = 4 and y = 17.