Problem of the Week
Problem C and Solution
Fractions to the Max
Problem
Suppose that six different numbers are selected from the set {1, 2, 3, 4, 5, 6, 7}. These six
numbers are then used to form three fractions. Each of the three numerators and three
denominators must contain a different digit from the selection. The three fractions are then
added together. Determine the largest possible sum that can be obtained.
Solution
Solution 1
We can start by observing that to get a fraction with the highest value we need a 7 in the
.
numerator. The choice of denominators is possibly obvious as well. 71 = 7, 72 = 3.5, 37 = 2.3,
.
7
= 1.75, 75 = 1.4 and 76 = 1.2. 71 is the largest fraction and any numerator other than 7 will
4
produce a lower value.
Now we have five numbers left to place: {2,3,4,5,6}.
Of these remaining numbers, since 6 is the largest it should go in the numerator. Then 62 = 3,
6
= 2, 46 = 1.5 and 65 = 1.2. 26 is the largest fraction and any numerator other than 6 will
3
produce a lower value.
Now we have three numbers left to place: {3,4,5}.
.
Of these remaining numbers, since 5 is the largest it should go in the numerator. Then 53 = 1.7
and 45 = 1.25. 53 is the largest fraction and any numerator other than 5 will produce a lower
value.
We can now determine the largest possible sum.
Largest Possible Sum =
∴ the largest possible sum is
35
3
7
1
+ 62 +
=
42
6
=
70
6
=
35
3
+
18
6
5
3
+
10
6
or 11 23 .
If you look closely, you will see that the largest number and the smallest are together in one
fraction, the second largest and the second smallest are together in another fraction, and the
third fraction uses the third largest and the third smallest numbers.
Solution 2
The largest fractions will be created by putting the three smallest numbers, 1, 2, and 3, in the
denominators and then placing the numbers 4, 5, 6, and 7 in the numerators. We will do this
in every possible way, determine the sums and choose the largest.
There are 24 different possible sums in which 1, 2, and 3 are the denominators and 4, 5, 6, 7
are the numerators. Only one sum will be worked out showing all of the steps. The other sums
would be determined in a similar fashion.
24 15 12
51
4 5 6
+ + =
+
+
=
1 2 3
6
6
6
6
Using 4,5,6
in the Numerators
Using 4,5,7
in the Numerators
Using 4,6,7
in the Numerators
Using 5,6,7
in the Numerators
4
1
+ 52 +
6
3
=
51
6
4
1
+ 52 +
7
3
=
53
6
4
1
+ 62 +
7
3
=
56
6
5
1
+ 62 +
7
3
=
62
6
4
1
+ 62 +
5
3
=
52
6
4
1
+ 72 +
5
3
=
55
6
4
1
+ 72 +
6
3
=
57
6
5
1
+ 72 +
6
3
=
63
6
5
1
+ 42 +
6
3
=
54
6
5
1
+ 42 +
7
3
=
56
6
6
1
+ 42 +
7
3
=
62
6
6
1
+ 52 +
7
3
=
65
6
5
1
+ 62 +
4
3
=
56
6
5
1
+ 72 +
4
3
=
59
6
6
1
+ 72 +
4
3
=
65
6
6
1
+ 72 +
5
3
=
67
6
6
1
+ 42 +
5
3
=
58
6
7
1
+ 42 +
5
3
=
64
6
7
1
+ 42 +
6
3
=
66
6
7
1
+ 52 +
6
3
=
69
6
6
1
+ 52 +
4
3
=
59
6
7
1
+ 52 +
4
3
=
65
6
7
1
+ 62 +
4
3
=
68
6
7
1
+ 62 +
5
3
=
70
6
∴ the largest possible sum is
7
1
+ 62 +
5
3
=
70
6
or 11 23 .
Algebraic Aside
Above we stated that the largest fractions will be created by putting the three smallest
numbers, 1, 2, and 3 in the denominators. Is there a way to justify why the 5, 6 and 7 are used
in the numerator and where they are?
a b c
We want to maximize the expression + + .
1 2 3
6a 3b 2c
6a + 3b + 2c
We can express this with a common denominator of 6 as follows:
+ +
=
.
6
2
3
6
If we are selecting values from 4, 5, 6 and 7, the largest value will be created when a = 7, b = 6
7 6 5
and c = 5. The largest sum is + + .
1 2 3