Checking for $343$ cases

Question

Problem:

Let $_1, _2, _3$ be integers. Prove that there exist integers $_1, _2, _3$ such that

•$_i$ is equal to −, or for all ∈ {1,2,3}

• all numbers can’t be at the same time.

• the number = $_1_1 + _2_2 + _3_3$ is perfectly divisible by .

My approach:

let $a_1 \equiv {n_1}\pmod{7}$ where $0\leq{n_1}<7$

let $a_2 \equiv {n_2}\pmod{7}$ where $0\leq{n_2}<7$

let $a_3 \equiv {n_3}\pmod{7}$ where $0\leq{n_3}<7$

I have to prove that $n_1_1 + n_2_2 + n_3_3\equiv 0\pmod{7}$

There are a total of $7^3$ or 343 possible triple of $(n_1,n_2,n_3)$

Sadly, I checked all 343 cases and prove it.

Conclusion:

This took a long time to check all 343 cases. I want a simple solution where we don't have to check all 343 cases.This question appeared in BDMO 2024 national(secondary). So,there was a time limit.Feel free to help me.

Source: BDMO(Bangladesh Mathematical Olympiad) 2024 national secondary question(problem no-4)

Answer 1

We may replace $a_i$ with $a_i\bmod 7$, so $0 \le a_i\le 6$.

• If any $a_i$ is zero, we can set $b_i=1$; so we may assume all $a_i$ are non-zero.
• If any $a_i$ is $> 3$, we can replace it with $7-a_i$, and then change the sign of $b_i$ at the end. So we may assume $1\le a_i\le 3$.
• If $a_i=a_j$ for some $i\ne j$, we can set $b_i=1,b_j=-1$.

Therefore we may assume $\{a_1,a_2,a_3\}=\{1,2,3\}$ (in some order)! And now it's easy.

Answer 2

This is just PGH principle.

Pigeons: Consider the $8$ values of the form $ c_1 a_1 + c_2 a_2 + c_3 a_3 $, where $ c_i \in \{ 0, 1 \}$.

Holes: The remainder mod $7$.

PGH: There are $2$ values with the same remainder mod $7$ (not necessarily $0$).
Let's call them $c_1' a_1 + c_2' a_2 + c_3' a_3$ and $c_1^* a_1 + c_2^* a_2 + c_3^* a_3$.

Corollary: Setting $b_i = c_i' - c_i^*$, conclude that $b_1 a_1 + b_2 a_2 + b_3 a_3$ is a multiple of $7$.
(For completeness, explain why $b_i \in \{-1, 0, 1\}$ and why some $b_i \neq 0$.)

Bonus Generalization: Given any $n$ integers, there are some integers $b_i \in \{-1, 0, 1 \}$, not all zero, such that $\sum b_i a_i$ is a multiple of $2^n-1$.
The other solutions require non-trivial work to extend to the general case, as their case work explodes. EG Try $n=5$.

Answer 3

Assume for a contradiction that no such $b_i$ exist.

$a_1,a_2,a_3\in\{0,\dots,6\}$ must be distinct (or $a_i-a_j=0$ yields a solution) and non-zero (or $a_i=0$ yields a solution).

So, $a_1+a_2$ and $a_1+a_3$ are also distinct from each of $a_1,a_2,a_3$.

If $a_1+a_2+a_3$ was equal to $a_1$ then $a_2+a_3=0$ gives a solution for $b_i$, and the same applies for $a_2$ and $a_3$. It also can't be equal to $a_1+a_2$ as $a_3\neq 0$ and the same for $a_1+a_3$.

Therefore, these are each six distinct values, none of which are 0 (mod 7): $a_1,a_2,a_3,a_1+a_2,a_1+a_3,a_1+a_2+a_3$.

Now consider $a_2+a_3$. If it is not in the above list then it must be 0 (mod 7), by the pigeonhole principle (all other values mod 7 appear). If it is in the above list then we get some contradiction:

• $a_2+a_3=a_1$ implies $a_1-a_2-a_3=0$ (solution is $b_1=1,b_2=-1,b_3=-1$)
• $a_2+a_3=a_2$ implies $a_3=0$
• $a_2+a_3=a_3$ implies $a_2=0$
• $a_2+a_3=a_1+a_2$ implies $a_1=a_3$
• $a_2+a_3=a_1+a_3$ implies $a_1=a_2$
• $a_2+a_3=a_1+a_2+a_3$ implies $a_1=0$

Answer 4

First of all, it is sufficient to restrict $~a_1, ~a_2, ~a_3~$ to $~\{0,1,2,\cdots,6\},~$ since all that you are interested in is the congruence $~\pmod{7}~$ of $~[ ~(a_1 ~b_1) + ~(a_2 ~b_2) + ~(a_3 ~b_3) ~].$

Second, you can assume, without loss of generality, that none of $~a_1, ~a_2, ~a_3~$ are equal to $~0.$ That is, if $~a_1 = 0,~$ then you could set $~(b_1, b_2, b_3) = (1, 0, 0).$

Further, you can assume, without loss of generality that $~a_1, a_2, a_3,~$ are all distinct elements in $~\{1,2,3,4,5,6\}.~$ That is, if $~a_1 = a_2,~$ then you could set $~(b_1, b_2, b_3) = (1,-1,0).$

Further, you can not have any two of the elements $~a_1, a_2, a_3~$ such that they sum to $~7.~$ That is, if $~(a_1 + a_2) = 7,~$ then you could have $~(b_1,b_2,b_3) = (1,1,0).$

Then, just to ease the discussion, you can assume, without loss of generality, that $~a_1 < a_2 < a_3.~$

At this point, the case work is not that bad.

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$\underline{\text{Case 1:} ~a_1 = 1}$

Suppose that $~a_3 = 6. ~$ Then,

• Then $~a_1 + a_3 = 7.$

Suppose that $~a_3 = 5. ~$ Then,

• You can't have $~a_2 = 2, ~$ because then $~a_2 + a_3 = 7.$

• You can't have $~a_2 = 3,~$ because then $~a_2 + a_3 - a_1 = 7.~$

• You can't have $~a_2 = 4,~$ because then $~a_1 + a_2 - a_3 = 0.~$

Suppose that $~a_3 = 4. ~$ Then,

• You can't have $~a_2 = 2, ~$ because then $~a_1 + a_2 + a_3 = 7.$

• You can't have $~a_2 = 3,~$ because then $~a_2 + a_3 = 7.~$

Suppose that $~a_3 = 3. ~$ Then,

• You can't have $~a_2 = 2, ~$ because then $~a_1 + a_2 - a_3 = 0.$

Therefore, $~a_1 = 1~$ can not produce a counter-example.

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$\underline{\text{Case 2:} ~a_1 = 2}$

Suppose that $~a_3 = 6. ~$ Then,

• You can't have $~a_2 = 3, ~$ because then $~a_2 + a_3 - a_1 = 7.$

• You can't have $~a_2 = 4, ~$ because then $~a_1 + a_2 - a_3 = 0.$

• You can't have $~a_2 = 5, ~$ because then $~a_1 + a_2 = 7.$

Suppose that $~a_3 = 5. ~$ Then,

• Then $~a_1 + a_3 = 7.$

Suppose that $~a_3 = 4. ~$ Then,

• You can't have $~a_2 = 3, ~$ because then $~a_2 + a_3 = 7.$

Therefore, $~a_1 = 2~$ can not produce a counter-example.

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$\underline{\text{Case 3:} ~a_1 = 3}$

Suppose that $~a_3 = 6. ~$ Then,

• You can't have $~a_2 = 4, ~$ because then $~a_1 + a_2 = 7.$

• You can't have $~a_2 = 5, ~$ because then $~a_1 + a_2 + a_3 = 14.$

Suppose that $~a_3 = 5. ~$ Then,

• You can't have $~a_2 = 4, ~$ because then $~a_1 + a_2 = 7.$

Therefore, $~a_1 = 3~$ can not produce a counter-example.

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$\underline{\text{Case 4:} ~a_1 = 4}$

This forces $~a_2 = 5, ~a_3 = 6 \implies a_2 + a_3 - a_1 = 7.$

Therefore, $~a_1 = 4~$ can not produce a counter-example.