Finding the values $A$ can take.

Question

$$A2B2 \equiv 0 \space \space (\text{mod} \space 12)$$

$$A-B>4$$

Find the values $A$ can take.

Could you help me out?

Regards

Answer 1

Note that my answer is based off the comment the OP said: "$A2B2$ is a number". If this is not what they are looking for, then I will edit my answer later.

What we want is a four digit number $A2B2$ that satisfies the conditions given above.

Condition 1: $A2B2 \equiv 0 \mod{12}$

This means that our number must be divisible by $12 = 4 \times 3$. For a four digit number to be divisible by $4$, then the last two digits must either be a multiple of $4$ or both be zero. For one to be divisible by $3$, we need each digit added up to be divisible by $3$ (if you didn't know of this trick, it can be proved if you have some understanding of proofs and divisibility). A number that is divisible by both $4$ and $3$ will be divisible by $12$ as well.

Condition 2: $A-B \gt 4$ this is a straightforward thing to check after we check for condition 1.

Using condition 1, we can find all possible values for $B$:

• $A202 \not\equiv 0 \mod 4 \Rightarrow B \neq 0$
• $A212 \equiv 0 \mod 4 \Rightarrow B = 1$
• $A222 \not\equiv 0 \mod 4 \Rightarrow B \neq 2$
• $A232 \equiv 0 \mod 4 \Rightarrow B=3$
• $A242 \not\equiv 0 \mod 4 \Rightarrow B \neq 4$
• $A252 \equiv 0 \mod 4 \Rightarrow B=5$
• $A262 \not\equiv 0 \mod 4 \Rightarrow B \neq 6$
• $A272 \equiv 0 \mod 4 \Rightarrow B = 7$
• $A282 \not\equiv 0 \mod 4 \Rightarrow B \neq 8$
• $A292 \equiv 0 \mod 4 \Rightarrow B=9$

So $B$ could be either $1,3,5,7,9$, but take note that if $B$ were to be equal to $5,7, or 9$, then $A$ would need to be larger than $9$ by condition 2 which wouldn't make sense using the assumption that $A2B2$ is a four digit number. Therefore we now know for sure that $B = 1$ or $B=3$ and our number is either of the form: $$A212$$ Or $$A232$$

Now using condition 2, we can deduce that $A$ can be any of the following: $6,7,8,9$. Remember that our number must be divisible by three, so we need $A + 2 + 1 + 2$ to be a multiple of $3$, and this can only happen when $A = 7$.

If we let $B=3$ then we need $A+2+3+2$ to be a multiple of three and $A=8$ works.

So our only allowed value for $A$ is $7$ if $B=1$ or $8$ if $B=3$, and we can check with a calculator that $$7212 ÷ 12 = 601$$ $$7-1 \gt 4$$ Or $$8232÷12 = 686$$ $$8-3 \gt 4$$

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If in fact you did mean to say $A^2B^2$ then see that we only need one of the perfect squares to be divisible by $12$. The smallest perfect square that $12$ divides into is $36 = 6^2$. If we let $A$ be $6$ then we simply set $B$ to be either $0$ or $\pm1$. Otherwise set $B=6$ and let $A \ge 11$.

So either $|A|=6$ or $|A| \ge 11$

Answer 2

If I were you, and admitting $$A2B2$$ is $$A^2B^2$$, I would take a look at the different values that a square number can take modulo 12.