Finding the values $A$ can take. [closed]

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$$A2B2 equiv 0 space space (textmod space 12)$$

$$A-B>4$$

Find the values $A$ can take.

Could you help me out?

Regards

modular-arithmetic divisibility

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asked Aug 6 at 12:59

Hamilton

967

closed as off-topic by JavaMan, Shailesh, Brahadeesh, gammatester, amWhy Aug 6 at 13:35

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – JavaMan, Shailesh, Brahadeesh, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Should that be $A^2B^2$? What are your thoughts on the problem? What you have tried? Where are you stuck?
  – JavaMan
  Aug 6 at 13:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What have you tried so far? Where do you get stuck? What does it mean $A2B2$?
  – Taroccoesbrocco
  Aug 6 at 13:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JavaMan That shouldn't. I'm beginner at modular arithmetics. Hence I couldn't show my attempt.
  – Hamilton
  Aug 6 at 13:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So then what is $A2B2$?
  – JavaMan
  Aug 6 at 13:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JavaMan $A2B2$ is a number
  – Hamilton
  Aug 6 at 13:03
  

  
  
  
  

 | 
show 1 more comment

up vote
-3
down vote

favorite

1

$$A2B2 equiv 0 space space (textmod space 12)$$

$$A-B>4$$

Find the values $A$ can take.

Could you help me out?

Regards

modular-arithmetic divisibility

share|cite|improve this question

asked Aug 6 at 12:59

Hamilton

967

closed as off-topic by JavaMan, Shailesh, Brahadeesh, gammatester, amWhy Aug 6 at 13:35

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – JavaMan, Shailesh, Brahadeesh, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Should that be $A^2B^2$? What are your thoughts on the problem? What you have tried? Where are you stuck?
  – JavaMan
  Aug 6 at 13:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What have you tried so far? Where do you get stuck? What does it mean $A2B2$?
  – Taroccoesbrocco
  Aug 6 at 13:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JavaMan That shouldn't. I'm beginner at modular arithmetics. Hence I couldn't show my attempt.
  – Hamilton
  Aug 6 at 13:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So then what is $A2B2$?
  – JavaMan
  Aug 6 at 13:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JavaMan $A2B2$ is a number
  – Hamilton
  Aug 6 at 13:03
  

  
  
  
  

 | 
show 1 more comment

up vote
-3
down vote

favorite

1

up vote
-3
down vote

favorite

1

1

$$A2B2 equiv 0 space space (textmod space 12)$$

$$A-B>4$$

Find the values $A$ can take.

Could you help me out?

Regards

modular-arithmetic divisibility

share|cite|improve this question

asked Aug 6 at 12:59

Hamilton

967

$$A2B2 equiv 0 space space (textmod space 12)$$

$$A-B>4$$

Find the values $A$ can take.

Could you help me out?

Regards

modular-arithmetic divisibility

share|cite|improve this question

asked Aug 6 at 12:59

Hamilton

967

share|cite|improve this question

share|cite|improve this question

asked Aug 6 at 12:59

Hamilton

967

asked Aug 6 at 12:59

Hamilton

967

asked Aug 6 at 12:59

Hamilton

967

967

closed as off-topic by JavaMan, Shailesh, Brahadeesh, gammatester, amWhy Aug 6 at 13:35

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – JavaMan, Shailesh, Brahadeesh, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by JavaMan, Shailesh, Brahadeesh, gammatester, amWhy Aug 6 at 13:35

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – JavaMan, Shailesh, Brahadeesh, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Should that be $A^2B^2$? What are your thoughts on the problem? What you have tried? Where are you stuck?
  – JavaMan
  Aug 6 at 13:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What have you tried so far? Where do you get stuck? What does it mean $A2B2$?
  – Taroccoesbrocco
  Aug 6 at 13:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JavaMan That shouldn't. I'm beginner at modular arithmetics. Hence I couldn't show my attempt.
  – Hamilton
  Aug 6 at 13:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So then what is $A2B2$?
  – JavaMan
  Aug 6 at 13:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JavaMan $A2B2$ is a number
  – Hamilton
  Aug 6 at 13:03
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Should that be $A^2B^2$? What are your thoughts on the problem? What you have tried? Where are you stuck?
  – JavaMan
  Aug 6 at 13:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What have you tried so far? Where do you get stuck? What does it mean $A2B2$?
  – Taroccoesbrocco
  Aug 6 at 13:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JavaMan That shouldn't. I'm beginner at modular arithmetics. Hence I couldn't show my attempt.
  – Hamilton
  Aug 6 at 13:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So then what is $A2B2$?
  – JavaMan
  Aug 6 at 13:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JavaMan $A2B2$ is a number
  – Hamilton
  Aug 6 at 13:03
  

  
  
  
  

1

1

Should that be $A^2B^2$? What are your thoughts on the problem? What you have tried? Where are you stuck?
– JavaMan
Aug 6 at 13:00

Should that be $A^2B^2$? What are your thoughts on the problem? What you have tried? Where are you stuck?
– JavaMan
Aug 6 at 13:00

What have you tried so far? Where do you get stuck? What does it mean $A2B2$?
– Taroccoesbrocco
Aug 6 at 13:01

What have you tried so far? Where do you get stuck? What does it mean $A2B2$?
– Taroccoesbrocco
Aug 6 at 13:01

@JavaMan That shouldn't. I'm beginner at modular arithmetics. Hence I couldn't show my attempt.
– Hamilton
Aug 6 at 13:01

@JavaMan That shouldn't. I'm beginner at modular arithmetics. Hence I couldn't show my attempt.
– Hamilton
Aug 6 at 13:01

So then what is $A2B2$?
– JavaMan
Aug 6 at 13:02

So then what is $A2B2$?
– JavaMan
Aug 6 at 13:02

@JavaMan $A2B2$ is a number
– Hamilton
Aug 6 at 13:03

@JavaMan $A2B2$ is a number
– Hamilton
Aug 6 at 13:03

 | 
show 1 more comment

2 Answers
2

active

oldest

votes

up vote
0
down vote



accepted

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: $A2B2 equiv 0 mod12$

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 $B$:

• $A202 notequiv 0 mod 4 Rightarrow B neq 0$


• $A212 equiv 0 mod 4 Rightarrow B = 1$


• $A222 notequiv 0 mod 4 Rightarrow B neq 2$


• $A232 equiv 0 mod 4 Rightarrow B=3$


• $A242 notequiv 0 mod 4 Rightarrow B neq 4$


• $A252 equiv 0 mod 4 Rightarrow B=5$


• $A262 notequiv 0 mod 4 Rightarrow B neq 6$


• $A272 equiv 0 mod 4 Rightarrow B = 7$


• $A282 notequiv 0 mod 4 Rightarrow B neq 8$


• $A292 equiv 0 mod 4 Rightarrow B=9$

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 $$8232÷12 = 686$$
$$8-3 gt 4$$

---

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 $|A| ge 11$

share|cite|improve this answer

edited Aug 6 at 14:08

answered Aug 6 at 13:54

WaveX

1,8211616

add a comment | 

up vote
0
down vote

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.

share|cite|improve this answer

answered Aug 6 at 13:21

Pjonin

3206

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

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: $A2B2 equiv 0 mod12$

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 $B$:

• $A202 notequiv 0 mod 4 Rightarrow B neq 0$


• $A212 equiv 0 mod 4 Rightarrow B = 1$


• $A222 notequiv 0 mod 4 Rightarrow B neq 2$


• $A232 equiv 0 mod 4 Rightarrow B=3$


• $A242 notequiv 0 mod 4 Rightarrow B neq 4$


• $A252 equiv 0 mod 4 Rightarrow B=5$


• $A262 notequiv 0 mod 4 Rightarrow B neq 6$


• $A272 equiv 0 mod 4 Rightarrow B = 7$


• $A282 notequiv 0 mod 4 Rightarrow B neq 8$


• $A292 equiv 0 mod 4 Rightarrow B=9$

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 $$8232÷12 = 686$$
$$8-3 gt 4$$

---

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 $|A| ge 11$

share|cite|improve this answer

edited Aug 6 at 14:08

answered Aug 6 at 13:54

WaveX

1,8211616

add a comment | 

up vote
0
down vote



accepted

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: $A2B2 equiv 0 mod12$

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 $B$:

• $A202 notequiv 0 mod 4 Rightarrow B neq 0$


• $A212 equiv 0 mod 4 Rightarrow B = 1$


• $A222 notequiv 0 mod 4 Rightarrow B neq 2$


• $A232 equiv 0 mod 4 Rightarrow B=3$


• $A242 notequiv 0 mod 4 Rightarrow B neq 4$


• $A252 equiv 0 mod 4 Rightarrow B=5$


• $A262 notequiv 0 mod 4 Rightarrow B neq 6$


• $A272 equiv 0 mod 4 Rightarrow B = 7$


• $A282 notequiv 0 mod 4 Rightarrow B neq 8$


• $A292 equiv 0 mod 4 Rightarrow B=9$

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 $$8232÷12 = 686$$
$$8-3 gt 4$$

---

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 $|A| ge 11$

share|cite|improve this answer

edited Aug 6 at 14:08

answered Aug 6 at 13:54

WaveX

1,8211616

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

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: $A2B2 equiv 0 mod12$

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 $B$:

• $A202 notequiv 0 mod 4 Rightarrow B neq 0$


• $A212 equiv 0 mod 4 Rightarrow B = 1$


• $A222 notequiv 0 mod 4 Rightarrow B neq 2$


• $A232 equiv 0 mod 4 Rightarrow B=3$


• $A242 notequiv 0 mod 4 Rightarrow B neq 4$


• $A252 equiv 0 mod 4 Rightarrow B=5$


• $A262 notequiv 0 mod 4 Rightarrow B neq 6$


• $A272 equiv 0 mod 4 Rightarrow B = 7$


• $A282 notequiv 0 mod 4 Rightarrow B neq 8$


• $A292 equiv 0 mod 4 Rightarrow B=9$

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 $$8232÷12 = 686$$
$$8-3 gt 4$$

---

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 $|A| ge 11$

share|cite|improve this answer

edited Aug 6 at 14:08

answered Aug 6 at 13:54

WaveX

1,8211616

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: $A2B2 equiv 0 mod12$

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 $B$:

• $A202 notequiv 0 mod 4 Rightarrow B neq 0$


• $A212 equiv 0 mod 4 Rightarrow B = 1$


• $A222 notequiv 0 mod 4 Rightarrow B neq 2$


• $A232 equiv 0 mod 4 Rightarrow B=3$


• $A242 notequiv 0 mod 4 Rightarrow B neq 4$


• $A252 equiv 0 mod 4 Rightarrow B=5$


• $A262 notequiv 0 mod 4 Rightarrow B neq 6$


• $A272 equiv 0 mod 4 Rightarrow B = 7$


• $A282 notequiv 0 mod 4 Rightarrow B neq 8$


• $A292 equiv 0 mod 4 Rightarrow B=9$

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 $$8232÷12 = 686$$
$$8-3 gt 4$$

---

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 $|A| ge 11$

share|cite|improve this answer

edited Aug 6 at 14:08

answered Aug 6 at 13:54

WaveX

1,8211616

share|cite|improve this answer

share|cite|improve this answer

edited Aug 6 at 14:08

edited Aug 6 at 14:08

edited Aug 6 at 14:08

answered Aug 6 at 13:54

WaveX

1,8211616

answered Aug 6 at 13:54

WaveX

1,8211616

answered Aug 6 at 13:54

WaveX

1,8211616

1,8211616

add a comment | 

add a comment | 

up vote
0
down vote

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.

share|cite|improve this answer

answered Aug 6 at 13:21

Pjonin

3206

add a comment | 

up vote
0
down vote

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.

share|cite|improve this answer

answered Aug 6 at 13:21

Pjonin

3206

add a comment | 

up vote
0
down vote

up vote
0
down vote

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.

share|cite|improve this answer

answered Aug 6 at 13:21

Pjonin

3206

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.

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answered Aug 6 at 13:21

Pjonin

3206

share|cite|improve this answer

share|cite|improve this answer

answered Aug 6 at 13:21

Pjonin

3206

answered Aug 6 at 13:21

Pjonin

3206

answered Aug 6 at 13:21

Pjonin

3206

3206

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