Numbers

[cinelli]

I have five numbers and their sums, taken two at a time, are as follows:

114 115 118 119 121 122 123 125 126 129

What are the numbers?

Cinelli

[jfgw]

Spoiler
55, 59, 60, 63, 66

If we sum the sums, we get 1212. Divide by 4 to get the sum of the original numbers = 303

If the numbers are a, b, c, d, and e in that order, with a being the smallest:

a+b = 114
d+e = 129
It is also clear (since the second sum is one greater than the first) that c = b+1.

c = 303 - 114 - 129 =60
b = 60 - 1 = 59
a = 114 - 59 = 55

I guessed that e - d = 3 which gives 2d = 129 - 3 =126
d = 63
e = 66


Julian F. G. W.

[modellingman]

Spoiler

Similar to jfgw but without any guessing.

Assume 5 numbers are A, B, C, D, E in ascending order.

Since the sums of all 10 possible pairs differ no two numbers can be the same.

Further, A+B is the smallest possible sum (so equal to 114) and D+E the largest (and equal to 129).

The sum of all 10 possible pairs is 4(A+B+C+D+E) and is equal to 1212 so A+B+C+D+E=303. Therefore C=303-[(A+B)+(C+D)]=303-(114+129)=303-243=60.

D and E must both exceed C and sum to 129 so there are 4 possibilities for the pair (D,E): (61,68), (62,67), (63,66) and (64,65). Examining each in turn:
(61,68) leads to C+E being 128 but this is not a valid sum;
(62, 67) leads to C+E being 129 which is also invalid since D+E takes this value;
(63, 66) leads to C+D being 123 and C+E being 126 both of which are valid possibilities;
(64,65) leads to C+D being 124 but this is not a valid sum.
Therefore, the only possibility leading to valid sums is (D,E)=(63,66).

The accounted for sums are
A+B=114
C+D=123
C+E=126
D+E=129

and the six unaccounted for sums
A+C, A+D, A+E, B+C, B+D and B+E
require allocation to the six values
115, 118, 119, 121, 122 and 125.

A+C must be the smallest of these (since A<B<C<D<E) so
A+C=115
and it follows that since A+B=114
B=C-1=60-1=59
and, further,
A=114-B=114-59=55

This completes the solution that
(A,B,C,D,E)=(55, 59,60,63,66).

For completeness, it can be verified that the (now) 5 unaccounted for sums of
A+D, A+E, B+C, B+D and B+E
equate, respectively, to the required values of
118, 121, 119, 122 and 125.


modellingman

[modellingman]

Small correction (underlined) to my spoiler...

The sum of all 10 possible pairs is 4(A+B+C+D+E) and is equal to 1212 so A+B+C+D+E=303. Therefore C=303-[(A+B)+(D+E)]=303-(114+129)=303-243=60.

modellingman

[cinelli]

Well solved, both. A relatively gentle puzzle for Christmas Day.

Cinelli

[9873210]

Spoiler

Similar to jfgw but without any guessing.


...

D and E must both exceed C and sum to 129 so there are 4 possibilities for the pair (D,E): (61,68), (62,67), (63,66) and (64,65).

modellingman

This assumes the numbers are integers. You can solve without this assumption, and incidentally prove that there are no real or complex solutions.

As before
a = 55, b = 59, c=60,

Then
d + a = d+a
d + b = d+a + 1
d + c = d+a + 4

e + a = e+a
e + b = e+a + 1
e + c = e+a + 4

Among the unassigned sums there are two series of m,m+1,m+4 and that gives d and e

[modellingman]

Spoiler

Similar to jfgw but without any guessing.


...

D and E must both exceed C and sum to 129 so there are 4 possibilities for the pair (D,E): (61,68), (62,67), (63,66) and (64,65).

modellingman

This assumes the numbers are integers. You can solve without this assumption, and incidentally prove that there are no real or complex solutions.

As before
a = 55, b = 59, c=60,

Then
d + a = d+a
d + b = d+a + 1
d + c = d+a + 4

e + a = e+a
e + b = e+a + 1
e + c = e+a + 4

Among the unassigned sums there are two series of m,m+1,m+4 and that gives d and e

You are correct, I rather lazily assumed that all solutions were integer. I could have avoided doing this by noting that since all summed pairs are integers (including those where c is a summand) the fact that c is an integer necessarily means a, b, d and e must also be integer.

Whilst I liked your approach to the solution for d and e, I had some initial difficulty following it.

After solving for a, b and c as 55, 59 and 60, respectively and assigning d+e as 129 (necessary for solving c) the unassigned sums are
118, 121, 122, 123, 125 and 126.

Whilst it is true that this contains two series of m, m+1, m+4 [these being (121, 122, 125) and (122, 123, 126)], the two series utilise the unassigned sum of 122 twice, and leave 118 unutilised. An error, surely?

Shouldn't the two series should be of the form m, m+4 and m+5? This would also give rise to two series - (118, 122, 123) and (121, 125, 126). These utilise all 6 unassigned sums just once and yield solutions for d and e of 63 and 66, respectively.


modellingman