Claude y el desafío matemático navideño de 2024
Dice Gaussianos:
Siguiendo la tradición desde 2012, vuelve el Desafío Matemático RSME-El País de Navidad. Este año, como hace ya tiempo, de nuevo es Adolfo Quirós (profesor de la Universidad Autónoma de Madrid (UAM) y director de La Gaceta de la Real Sociedad Matemática Española) quien nos lo presenta.
El problema es el siguiente:
El desafío comienza cuando elegimos dos números de la Lotería de Navidad (recordemos que tienen 5 cifras), con la única condición de que cumplan estos dos requisitos:
Entre los dos números, aparecen todos los dígitos del 0 al 9, necesariamente una vez cada uno.
La suma de los dos números vuelve a ser un número de lotería, es decir, tiene 5 cifras.
Por ejemplo, nos valen los números 01357 y 24689, que suman 26046. No nos valen 01357 y 24680, porque se repite el 0 y falta el 9. Tampoco nos valen 70135 y 42689, porque su suma, 112824, tiene 6 cifras.
Una vez elegidos los dos números, observamos que en su suma aparecen, en algún orden y alguna posición, los dígitos 1, 3, 5 y 7. Esto es, llamando X a la quinta cifra, la suma puede tener la forma 1357X, o 5X371, o 75X31, o…
El desafío consiste en decidir qué valores exactamente puede tomar el dígito X: ¿puede ser cualquiera entre 0 y 9? ¿pueden aparecer como X unos dígitos sí y otros no? ¿Puede ser que en realidad no aparezca ningún X (no hemos dicho que existan dos números que cumplan todas las condiciones que hemos dado)?
Se lo he pasado a Claude en estos términos:
Consider two 5 digit numbers such that their digits are the numbers 0, 1, … 9 without repetitions, obviously. Imagine you add them and the result is a 5 digit number (i.e, less than 100000) which contains digits 1, 3, 5 and 7. Which is the only option for the other digit? (Hint: think about some known results and theorems about the digits of numbers).
Y me ha contestado lo siguiente:
Let me solve this step by step by analyzing what we know. From the problem statement:
We have two 5-digit numbers using digits 0-9 without repetition Their sum is less than 100000 The sum contains digits 1, 3, 5, 7, and one unknown digit
Key insights:
If we add two 5-digit numbers whose sum is less than 100000, there must be a significant constraint on the leading digits A useful theorem here is the divisibility rule by 9: a number is divisible by 9 if and only if the sum of its digits is divisible by 9
Let’s analyze:
Looking at the given digits in the sum (1,3,5,7,x), we know their sum must be: 1+3+5+7+x = 16+x The sum of all digits 0-9 is 45 If we take any collection of these digits and split them into two groups (like our two numbers), their digits must sum to 45 By the divisibility rule of 9, the sum number must have the same remainder mod 9 as the sum of its individual digits
Therefore:
16+x ≡ (sum of two original numbers) mod 9 45 ≡ (sum of two original numbers) mod 9
This means: 16+x ≡ 45 mod 9 x ≡ 2 mod 9 Therefore, x must be 2. We can verify this is possible by finding two numbers that satisfy our conditions. For example: 98740 + 12356 = 11096 Therefore, 2 is the only possible value for the fifth digit.
Salvo por el hecho de que $98740 + 12356 \ne 11096$ (punto marrón menos para Claude), todo correcto. En menos de un segundo. Ha sido una sesión zero-shot: no ha habido interacciones previas. Le he pasado una pista porque previamente, con Gemini Experimental 1206, había tenido que indicarle que se remitiese a ese tipo de resultados después de una primera respuesta catastrófica.
En resumen: no vayáis diciendo que los LLMs no son sino loros estadísticos y que no son inteligentes, no vaya a ser que…
Coda
Le pasé el problema también a Qwen/QwQ-32B-Preview de la siguiente forma (sin la pista):
Let see if you can solve this maths problem. Consider two 5 digit numbers such that their digits are the numbers 0, 1, … 9 without repetitions, obviously. Imagine you add them and the result is a 5 digit number (i.e, less than 100000) which contains digits 1, 3, 5 and 7. Which is the only option for the other digit?
La respuesta, también correcta, aunque con varias idas y venidas, es más entretenida que la de Claude:
Let’s tackle this problem step by step. So, we have two 5-digit numbers, each using the digits 0 through 9 without repetition. When we add them together, the sum is another 5-digit number that contains only the digits 1, 3, 5, and 7, plus one other digit that we need to find.
First, let’s understand the constraints:
Each of the two numbers is a 5-digit number without repeating digits.
Their sum is also a 5-digit number, meaning it’s less than 100,000.
The sum contains only the digits 1, 3, 5, 7, and one other digit that we need to identify.
Since the sum is a 5-digit number and it’s less than 100,000, the first digit can’t be 0. Also, the digits in the sum are only from the set {1, 3, 5, 7, x}, where x is the unknown digit we’re trying to find.
Let me think about the possible range for the sum. The smallest possible sum would be if we add the smallest possible 5-digit numbers, like 10,234 + 34,567 = 44,801, but this is just a rough estimate. The actual smallest sum might be different, but it’s not crucial for now.
More importantly, since both numbers use digits without repetition and the sum also uses specific digits, there must be some constraints on the possible carries during addition.
Let’s consider that each column (units, tens, hundreds, etc.) adds up to a digit in the sum, possibly with a carryover to the next column.
Given that the sum contains only 1, 3, 5, 7, and one other digit, and considering that the digits in the two numbers are from 0 to 9 without repetition, there must be some unique property that determines the possible values for the fifth digit in the sum.
Maybe I should look at the possible digit sums and see what constraints that imposes.
Let’s denote the two numbers as A and B, and their sum as S.
A and B are both 5-digit numbers with unique digits.
S is a 5-digit number with digits only from {1, 3, 5, 7, x}.
Since A and B together use 10 distinct digits (from 0 to 9), and their sum S has digits from the specified set, there must be a relationship between the digits of A and B and the digits of S.
Perhaps I can consider the properties of addition with carries. For each column, the sum of the digits plus any carry from the previous column determines the digit in S and any carry to the next column.
This seems complicated. Maybe there’s a better approach.
Let me think about the total sum of all digits from 0 to 9. The sum of digits from 0 to 9 is 45.
Since A and B together use all digits from 0 to 9 exactly once, the sum of their digits is 45.
Therefore, the sum S should have digits that add up to 45, considering any carryover in the addition process.
Wait, no. The sum of the digits of S is not necessarily 45 because carryover affects the digit sum.
Let me think differently. Maybe I can consider the possible carries in each column.
In a 5-digit addition, there are five columns (units to ten-thousands). Each column can have a carryover to the next column.
Let me denote the carryover from column i to column i+1 as c_i, where c_i is either 0 or 1.
Then, for each column, the sum of the digits plus any carry from the previous column equals the digit in S plus 10 times the carry to the next column.
Mathematically, for column i:
a_i + b_i + c_{i-1} = s_i + 10 * c_i
Where a_i and b_i are the digits in A and B for column i, s_i is the digit in S for column i, and c_{i-1} and c_i are the carryover from the previous and to the next column, respectively.
This seems too general. Maybe I need to look for a different approach.
Let me consider the possible values for the fifth digit in S.
The sum S has digits from {1, 3, 5, 7, x}, and we need to find x.
Perhaps x must be a digit that can be obtained through the addition process given the constraints.
Alternatively, maybe there’s only one digit that can satisfy the condition that S is less than 100,000 and has only these digits.
Wait, maybe I can consider the possible ranges for S.
Since S is a 5-digit number less than 100,000, it must be between 10,000 and 99,999.
Given that S contains only the digits 1, 3, 5, 7, and x, and x is a digit from 0 to 9 not used in A or B, but since A and B use all digits from 0 to 9, x must be one of the digits in S that isn’t 1, 3, 5, or 7.
But wait, A and B together use all digits from 0 to 9 exactly once, but S is only 5 digits, so S can have digits from A and B, but with the addition process and possible carries, it’s not straightforward.
This is getting complicated. Maybe I should look for an example or try to find a property that uniquely determines x.
Let me consider the possible digit sums again.
The sum of the digits of S is s_1 + s_2 + s_3 + s_4 + s_5, where each s_i is in {1, 3, 5, 7, x}.
But I don’t know the carries, so this might not help directly.
Alternatively, maybe I can consider the possible values for S and see what x must be.
Given that S is a 5-digit number with digits only from {1, 3, 5, 7, x}, and x is a digit from 0 to 9 not already used in S, but since A and B use all digits from 0 to 9, x must be determined by the addition process.
This is still confusing. Maybe I need to think about the properties of the digits in S.
Wait a minute, perhaps I can consider the possible digit sums modulo 9, since the sum of digits modulo 9 is preserved in the sum S.
The sum of digits of A and B is 45, which is 0 modulo 9.
Therefore, the sum S should also have digits that sum to a multiple of 9.
So, s_1 + s_2 + s_3 + s_4 + s_5 ≡ 0 mod 9.
Given that s_i are from {1, 3, 5, 7, x}, and x is a digit from 0 to 9, we can use this to find possible values for x.
Let me calculate the sum of the known digits: 1 + 3 + 5 + 7 = 16.
So, 16 + x ≡ 0 mod 9 ⇒ x ≡ -16 mod 9 ⇒ x ≡ 2 mod 9.
Since x is a digit from 0 to 9, the possible values for x are 2 and 11, but 11 is not a digit, so x must be 2.
Wait, but 11 mod 9 is 2, so x ≡ 2 mod 9, thus x = 2.
But is this the only possible option?
Let me verify this.
If x = 2, then the sum of digits in S is 16 + 2 = 18, which is divisible by 9.
Is there any other digit x that can satisfy this condition?
If x = 11, it’s not a valid digit, so x must be 2.
Therefore, the only option for the other digit is 2.
But wait, the problem says “which is the only option for the other digit?” and according to this, it’s 2.
But I need to make sure that this is indeed the case and that there are no other possibilities.
Let me consider if there could be any other x that satisfies the condition.
Given that s_1 + s_2 + s_3 + s_4 + s_5 ≡ 0 mod 9, and s_1 + s_2 + s_3 + s_4 = 16, then x must be such that 16 + x ≡ 0 mod 9.
This implies x ≡ -16 mod 9 ≡ -16 + 18 = 2 mod 9.
Therefore, x must be 2 or 11, but since x is a single digit, x = 2 is the only possibility.
Thus, the only option for the other digit is 2.
Final Answer
[ \boxed{2} ]
