~~Denise said that the date was an odd number; Kristell claimed it was greater than 13.
Wanda declared it was not a perfect square, while Linda swore it was a perfect cube. Finally, Anthony told me the date was less than one-fourth his age, which I know to be
68.
Yesterday I learned that only one of them had told the truth!
What is the date of the lunch?
There are 5 statements, only one of which is true.
-
Date is an odd number, i.e. D= 2n+1
-
Date is greater than 13, i.e. D > 13
-
D is not equal to a perfect square, i.e. D != n^2
-
D is equal to a perfect cube, i.e. D == n^2
-
D is less than 17 (68/4), i.e. D < 17
Table 1 - of squares and cubes
|
n
|
1
|
2
|
3
|
4
|
5
|
6
|
|
n^2
|
1
|
4
|
9
|
16
|
25
|
36
|
|
n^3
|
1
|
8
|
27
|
64
|
125
|
216
|
Creating a Table of Statements, true and false.
Table 2
|
|
S
|
nS
|
|
1
|
D = 2n + 1
|
D = 2n
|
|
2
|
D > 13
|
D < 13
|
|
3
|
D != n^2
|
D == n^2
|
|
4
|
D == n^3
|
D != n^3
|
|
5
|
D < 17
|
D > 17
|
We are told that only one of the original statements is true.
That means that in the column if not true statements (nS), only one is false, and all the rest are true.
i.e. we can’t have two false not-statements (the nS ones).
Since we can’t have two false statements, then both nS=2 and bS=5 can’t both be false.
They both can’t be true either since they are mutually exclusive.
Assume nS=2 is true. Then nS=5 is false.
This means that D is an even square number which is not a perfect cube, and is < 13.
Of all the squares from Table 1, only D = n^2 = 4 satisfies the conditions.
Assume nS=2 is false. Then nS=5 is true.
This means that D is an even square number which is not a perfect cube, and is > 17.
Of all the squares from Table 1, only D = n^2 = 36 satisfies the conditions.
But you can’t have 36 days in a month.
Solution is: all nS statements are true bar nS = 5, giving D = 4.
Answer: Lunch is on the 4th of the month.