Brain Teasers
Ten Numbers
There are ten natural numbers, two of which are equal. Adding any nine of them, we can get the following sums:
82, 83, 84, 85, 87, 89, 90, 91, 92.
Find the largest number of the ten.
82, 83, 84, 85, 87, 89, 90, 91, 92.
Find the largest number of the ten.
Answer
There are ten possible sums, arrived at by alternately excluding each one of the ten numbers. Two of the sums are equal. If we add the ten sums, it should equal nine times the sum of the ten numbers, as each number will be present within all of the sums nine times.82+83+84+85+87+89+90+91+92=87*9=783, which is divisible by 9, therefore the repeated sum must be divisible by 9. The only sum which meets this criteria is 90.
The sum of the ten numbers is (783+90)/9=97. The largest number is found by subtracting the lowest sum of nine numbers (which excluded the largest of the ten numbers) from the sum of all ten numbers. Thus: 97-82=15. 15 is the largest number of the ten.
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Comments
I think there is a problem here....
The problem states that of the 10 natural numbers, 2 are repeated. The proposed solution however, seems to imply that 2 of the sums of 9 numbers are repeated. Perhaps I am mistaken.....
The problem states that of the 10 natural numbers, 2 are repeated. The proposed solution however, seems to imply that 2 of the sums of 9 numbers are repeated. Perhaps I am mistaken.....
Oh boy....I made a mistake.
I almost had them by trial and error, until I went to check, and saw that the numbers were not continuous
I like your solution, it's more elegant than mine.
I came at this from a different angle. I knew that the total of the numbers minus the largest was 82. I also knew that the gaps in the sequence of the numbers when moving from low to high would match the gaps in the sums when moving from high to low.
Assigning the values A..I to the nine different values and R to the repeated value, I get the following equations:
B = A+1
C = A+2
D = A+3
E = A+5
F = A+7
G = A+8
H = A+9
I = A+10
A+B+C+D+E+F+G+H+R = 82
Substituting the values in the last equation gives:
8A+35+R = 82
R = 47-8A and R >= A and R = 4.111...
Since I know A is an integer, A must be 5, which makes:
R = 47-8*5 = 7
and the 10 numbers are:
5, 6, 7, 7, 8, 10, 12, 13, 14, 15
I came at this from a different angle. I knew that the total of the numbers minus the largest was 82. I also knew that the gaps in the sequence of the numbers when moving from low to high would match the gaps in the sums when moving from high to low.
Assigning the values A..I to the nine different values and R to the repeated value, I get the following equations:
B = A+1
C = A+2
D = A+3
E = A+5
F = A+7
G = A+8
H = A+9
I = A+10
A+B+C+D+E+F+G+H+R = 82
Substituting the values in the last equation gives:
8A+35+R = 82
R = 47-8A and R >= A and R = 4.111...
Since I know A is an integer, A must be 5, which makes:
R = 47-8*5 = 7
and the 10 numbers are:
5, 6, 7, 7, 8, 10, 12, 13, 14, 15
Apparently the site doesn't handle greater than and less than signs in the post too well because it sure mangled my last post. Here it is again with .ge. for "greater than or equal to" and .le. for "less than or equal to" instead of the symbols:
I like your solution, it's more elegant than mine.
I came at this from a different angle. I knew that the total of the numbers minus the largest was 82. I also knew that the gaps in the sequence of the numbers when moving from low to high would match the gaps in the sums when moving from high to low.
Assigning the values A..I to the nine different values and R to the repeated value, I get the following equations:
B = A+1
C = A+2
D = A+3
E = A+5
F = A+7
G = A+8
H = A+9
I = A+10
A+B+C+D+E+F+G+H+R = 82
Substituting the values in the last equation gives:
8A+35+R = 82
R = 47-8A
I also know that
R .ge. A
and
R .le. A+10
Plugging in the min and max values for R gives:
A .le. 47-8A
9A .le. 47
A .le. 47/9
A .le. 5.222...
and
A+10 .ge. 47-8A
9A+10 .ge. 47
9A .ge. 37
A .ge. 37/9
A .ge. 4.111...
Since I know A is an integer, A must be 5, which makes:
R = 47-8*5 = 7
and the 10 numbers are:
5, 6, 7, 7, 8, 10, 12, 13, 14, 15
I like your solution, it's more elegant than mine.
I came at this from a different angle. I knew that the total of the numbers minus the largest was 82. I also knew that the gaps in the sequence of the numbers when moving from low to high would match the gaps in the sums when moving from high to low.
Assigning the values A..I to the nine different values and R to the repeated value, I get the following equations:
B = A+1
C = A+2
D = A+3
E = A+5
F = A+7
G = A+8
H = A+9
I = A+10
A+B+C+D+E+F+G+H+R = 82
Substituting the values in the last equation gives:
8A+35+R = 82
R = 47-8A
I also know that
R .ge. A
and
R .le. A+10
Plugging in the min and max values for R gives:
A .le. 47-8A
9A .le. 47
A .le. 47/9
A .le. 5.222...
and
A+10 .ge. 47-8A
9A+10 .ge. 47
9A .ge. 37
A .ge. 37/9
A .ge. 4.111...
Since I know A is an integer, A must be 5, which makes:
R = 47-8*5 = 7
and the 10 numbers are:
5, 6, 7, 7, 8, 10, 12, 13, 14, 15
Thanks for your puzzle.
I managed to solve it with scip (a linear, mixed integer and nonlinear programming solver).
I just uploaded the model file that I used to pastebin for anyone who might be interested to play with it.
https://pastebin.com/DkJ1XPQW
I managed to solve it with scip (a linear, mixed integer and nonlinear programming solver).
I just uploaded the model file that I used to pastebin for anyone who might be interested to play with it.
https://pastebin.com/DkJ1XPQW
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