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How many different answers can be obtained by adding 


two different integers chosen from the set below ?


-13, -12, -11, ..., 6, 7, 8 ]

Guest Jul 14, 2014

Best Answer 

 #1
avatar+90001 
+5

Not as many as you think  !!!

Notice that the -13 can be combined with any of the other numbers. So this = 21 different answers.

But, look at -12.....We can make every sum we made in the first example, except the sum of -12 and 8. = -4.  (The largest sum we can make with -13 and any of the other numbers is -13 + 8 = -5.)

The same thing with -11. We can make every sum we've had so far except -11 + 8 = -3. (Again, the largest sum so far is -4)

If you were to continue this process, you would find that the number of different sums would be equal to (n-1) + (n-2)  = (22-1) + (22-2) = (21) + (20)  = 41, where n is the number of integers in the set, i.e., 22.

To get a feel for this, look at the six numbers  [1 2 3 4 5 6]

Note that there are nine possible sums:

1+2, 1+3, 1+4, 1+5, 1+6, 2+6, 3+6, 4+6, 5+6

Note that the first number is summed with all the others, but all the other numbers - 2 through 5 - are just summed with the last one. And there are 9 different sums = (n-1) + (n-2) = (6-1) + (6-2) = (5) + (4) = 9  !!!

P.S. - We could express (n-1) + (n-2) as (2n-3) if we wanted a more "compact" answer !!

 

CPhill  Jul 14, 2014
 #1
avatar+90001 
+5
Best Answer

Not as many as you think  !!!

Notice that the -13 can be combined with any of the other numbers. So this = 21 different answers.

But, look at -12.....We can make every sum we made in the first example, except the sum of -12 and 8. = -4.  (The largest sum we can make with -13 and any of the other numbers is -13 + 8 = -5.)

The same thing with -11. We can make every sum we've had so far except -11 + 8 = -3. (Again, the largest sum so far is -4)

If you were to continue this process, you would find that the number of different sums would be equal to (n-1) + (n-2)  = (22-1) + (22-2) = (21) + (20)  = 41, where n is the number of integers in the set, i.e., 22.

To get a feel for this, look at the six numbers  [1 2 3 4 5 6]

Note that there are nine possible sums:

1+2, 1+3, 1+4, 1+5, 1+6, 2+6, 3+6, 4+6, 5+6

Note that the first number is summed with all the others, but all the other numbers - 2 through 5 - are just summed with the last one. And there are 9 different sums = (n-1) + (n-2) = (6-1) + (6-2) = (5) + (4) = 9  !!!

P.S. - We could express (n-1) + (n-2) as (2n-3) if we wanted a more "compact" answer !!

 

CPhill  Jul 14, 2014

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