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 #2
avatar+165 
+3
Aug 4, 2020
 #1
avatar+1084 
+5

Can you format your Latex correctly? I am having trouble understanding. 
 

refer to here for help with latex: https://web2.0calc.com/questions/help_31578

 

Also, you should maybe PM the link to this question to Melody. It is currently midnight in Australia (where she is from), and she might not see the questions when she comes online.

 

:)

Aug 4, 2020
 #2
avatar+1084 
+5

No, there is not a way to delete questions on this site. (read more here: https://web2.0calc.com/questions/c-2-0c-c_6)

 

If you feel super mortified by what your sister did, you could edit the post, and mark it as redacted. But, it's up for you to decide!

 

:)

Aug 4, 2020
 #5
avatar+118587 
+3

Thanks Heureka, 

Here is my contribution. 

It is not much different from yours, but the explanation is maybe a little more different.

 

 

If a 7-digit number 13ab45c is divisible by 792, what is b?

 

\(792=8*9*11\)

 

For this number to be dividable by 8, the last 3 digits must be divisible by 8

456/8=57  no other digits will work so  c=6

 

\(13ab45c\\ becomes\\ 13ab456\)

 

Now a number is divisible by 11 if

| (sum of even digits) - (sum of odd digits) | = multiple of 11

 

Sum of odd digits = 1+a+4+6 = 11+a

Sum of even digits = 3+b+5 = 8+b

absolute value of difference = | 11+a - 8-b | = |3+a-b |  

so 3+a+b can be negative but it must be a multiple of 11

3+a-b = 11k    where k is an integer.

\(-9\le a-b\le9\\ -6\le 3+a-b \le12\\ which\;\; means\\ 3+a-b=11\;\;or\;\;0\\ a-b=8\;\;or\;\;-3\\ a=8+b\;\;or \;\; a=-3+b\)

 

 

Now for a number to be divisible by 9, the sum of the digits must be a multiple of 9

so

\(1+3+a+b+4+5+6\\ =19+a+b\\\quad if\;\;a=8+b\;\;then\\ \\\quad =19+b+8+b\\\quad =27+2b\\ \quad b=0\;\;or\;\;9\\ b=9 \;\;doesn't \;\;work\\ so\;\; b=0,\;\;a=8 \; \text{is a possible solution}\\ ~\\ 1+3+a+b+4+5+6\\ =19+a+b\\\quad if\;\;a=-3+b\;\;then\\ \\\quad =19+b-3+b\\\quad =16+2b\\ \quad b=1,\;\;a=-2 \;\text{which is invalid}\)

 

If b=9 then a=17 or 6     17 is not good so

If b=9  a=-3+9=6

If b=0  a=8+0=8

 

So the number could be

1380456   or

1369456

 

So the number is

\(13ab45c= 13ab456 =1380456\\ a=8\\ b=0\\ c=6\)

 

 

 

 

 

LaTex:

-9\le a-b\le9\\ -6\le 3+a-b \le12\\ which\;\; means\\ 3+a-b=11\;\;or\;\;0\\ a-b=8\;\;or\;\;-3\\ 
a=8+b\;\;or \;\; a=-3+b

 

1+3+a+b+4+5+6\\ =19+a+b\\\quad if\;\;a=8+b\;\;then\\ \\\quad =19+b+8+b\\\quad  =27+2b\\ \quad b=0\;\;or\;\;9\\
b=9 \;\;doesn't \;\;work\\
so\;\;   b=0,\;\;a=8 \; \text{is a possible solution}\\
~\\

1+3+a+b+4+5+6\\ =19+a+b\\\quad if\;\;a=-3+b\;\;then\\ \\\quad =19+b-3+b\\\quad  =16+2b\\
 \quad b=1,\;\;a=-2 \;\text{which is invalid}

Aug 4, 2020
 #1
avatar+26364 
+1

A, B and C each represent a single digit with no two digits being the same.
Find the values of A, B and C so that the following is true: \(AB + C=38\) and \(BC+A=29\)

 

My attempt:

\(\begin{array}{|rcll|} \hline \mathbf{AB + C} &=& \mathbf{38} \\ 10A+B+C &=& 38 \qquad (1) \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{BC+A} &=& \mathbf{29} \\ 10B+C+A &=& 29 \qquad (2) \\ \hline \end{array} \\ \begin{array}{|lrcll|} \hline (1)-(2): & 10A+B+C -(10B+C+A) &=& 38-29 \\ & 10A+B+C - 10B -C -A &=& 38-29 \\ & 10A+B - 10B -A &=& 9 \\ & 9A-9B &=& 9 \quad | \quad :9 \\ & A-B &=& 1 \\ & \mathbf{B} &=& \mathbf{A-1} \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline (1): & \mathbf{10A+B+C} &=& \mathbf{38} \quad | \quad \mathbf{B=A-1} \\ & 10A+A-1+C &=& 38 \\ & 11A-1+C &=& 38 \\ & 11A +C &=& 39 \\ & \mathbf{ 11A } &=& \mathbf{39-C} \\ \hline \end{array}\)

 

\(\begin{array}{|c|c|c|} \hline C & 39-C & A = \dfrac{39-C}{11} \\ \hline 0 & 39 & \text{not integer} \\ \hline 1 & 38 & \text{not integer} \\ \hline 2 & 37 & \text{not integer} \\ \hline 3 & 36 & \text{not integer} \\ \hline 4 & 35 & \text{not integer} \\ \hline 5 & 34 & \text{not integer} \\ \hline \color{red}6 & 33 & \mathbf{A =3} \\ \hline 7 & 32 & \text{not integer} \\ \hline 8 & 31 & \text{not integer} \\ \hline 9 & 30 & \text{not integer} \\ \hline \end{array}\), so \(\mathbf{C=6}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{B} &=& \mathbf{A-1} \quad | \quad \mathbf{A =3} \\ B &=& 3-1 \\ \mathbf{B} &=& \mathbf{2} \\ \hline \end{array}\)

 

check:

\(\begin{array}{|rcll|} \hline 32+6 &=& 38 \qquad ( AB+C = 38) \\ 26+3 &=& 29 \qquad ( BC+A = 29) \\ \hline \end{array}\)

 

laugh

Aug 4, 2020

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