SOLUTION: The number 42 can be written in three different ways as the sum of two or more consecutive positive integers.
42 = 13 + 14 +15
= 9 + 10 + 11 + 12
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Question 1165637: The number 42 can be written in three different ways as the sum of two or more consecutive positive integers.
42 = 13 + 14 +15
= 9 + 10 + 11 + 12
= 3 + 4 + 5 + 6 + 7 + 8 + 9
Is it possible to write 105 in eight different ways as the sum of two or more consecutive positive numbers?
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!
The number 42 can be written in three different ways as the sum of two or more consecutive positive integers.
42 = 13 + 14 +15
= 9 + 10 + 11 + 12
= 3 + 4 + 5 + 6 + 7 + 8 + 9
Is it possible to write 105 in eight different ways as the sum of two or more consecutive positive numbers?
ANSWER: No
That is, in fact, the answer to your question; but it is probably not what you wanted to know or learn from the problem.
In any set of consecutive integers, the sum is the number of terms, multiplied by the average of the terms.
If the number of terms is odd, then the average of the terms is an integer; if the number of terms is even, then the average of the terms is halfway between two integers.
So the sum of a set of consecutive integers can be either
(a) an odd number, multiplied by an integer; or
(b) an even number, multiplied by a number halfway between two integers.
To find the different ways of writing a given number as the sum of consecutive integers, write the number as the product of any two integers and try to use those two integers to fit either pattern (a) or pattern (b).
For the number 105, the analysis goes like this....
(1) 105 = 1*105
1 number with an average of 105: 105 (doesn't satisfy the requirement of being the sum of 2 or more numbers)
2 numbers with an average of 105/2 = 52.5: 52+53 ** Solution #1
(2) 105 = 3*35
3 numbers with an average of 35: 34+35+36 ** Solution #2
6 numbers with an average of 35/2 = 17.5 (3 integers less than 17.5 and 3 greater): 15+16+...+19+20 ** Solution #3
35 numbers with an average of 3: 3, and the 17 integers closest to 3 on the either side: (-14)+(-13)+...+0+...+13+14+15+16+...+19+20
That is the same as the previous answer, with the integers from -14 to +14 added. Since that sequence includes negative integers, it is not a solution we are looking for.
In fact, every solution we find will have a corresponding solution with additional positive and negative integers that cancel each other in the sum.
(3) 105 = 5*21
5 numbers with an average of 21: 19+20+21+22+23 ** Solution #4
10 numbers with an average of 21/2 = 10.5: 6+7+...+14+15 ** Solution #5
21 numbers with an average of 5 (will need negative integers)
42 numbers with an average of 2.5 (will use negative integers)
(4) 105 = 7*15
7 numbers with an average of 15: 12+13+...+17+18 ** Solution #6
14 numbers with an average of 7.5: 1+2+...+13+14 ** Solution #7
15 numbers with an average of 7: 0+1+2+...+13+14 (includes 0, which is not a positive integer)
30 numbers with an average of 3.5 (will use negative integers)
That's all the possibilities. We have 7 solutions in positive integers -- not 8.
Note we would get 8 solutions if 0 were allowed -- that is, if the integers are non-negative instead of positive.
(1) 52+53
(2) 34+35+36
(3) 15+16+...+19+20
(4) 19+20+21+22+23
(5) 6+7+...+14+15
(6) 12+13+...+17+18
(7) 1+2+...+13+14
Note you can find the same set of 7 solutions simply by dividing 105 by the integers 2, 3, ... and finding the ones in which the quotient is either an integer or halfway between two integers.
105/2 = 52.5 --> 52+53
105/3 = 35 --> 34+35+36
...
105/5 = 21 --> 19+20+21+22+23
105/6 = 17.5 --> 15+16+17+18+19+20
105/7 = 15 --> 12+13+...+17+18
...
105/10 = 10.5 --> 6+7+...+14+15
105/14 = 7.5 --> 1+2+...+13+14
In finding the solutions by that method, you know you are finished with the last solution, because it already starts with the smallest positive integer, 1. Any sequence of more than 14 consecutive integers with a sum of 105 would have to include 0 and/or negative integers.
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