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

 

Thank you so much to anyone who helps!!!

 Aug 4, 2020
 #1
avatar+1035 
+5

Try to find the prime factorization of 792. Then, use the prime factors, and see what restrictions the number puts.

 Aug 4, 2020
 #2
avatar+732 
+1

yeah i tried that

 

792=8*9*11

 

so c=6

 

then, 1+3+a+b+9+6=multiple of 9

so, a+b=8 or a+b =17

 

for it to be divisible by 11, 1-3+a-b+4-5+6= multiple of 11, so 3+a-b= multiple of 11, which means that a-b=8 or a-b = -3.

 

but then i tried to plug in the equations and i got x=7 and y=10 :/

 

can someone please tell me what I'm doing wrong?

lokiisnotdead  Aug 4, 2020
edited by lokiisnotdead  Aug 4, 2020
 #6
avatar+110835 
+1

Hi Loki,

 

Most of your logic looks spot on.

Have a look at mine and see if you can work out what is different (I have not looked that hard)

Melody  Aug 4, 2020
edited by Melody  Aug 4, 2020
 #7
avatar+1035 
+6

@lokiisnotdead, sorry for not seeing your post earlier, as I had to go to sleep. 

 

My way of doing it was similar to Melody's, so I will let her take the spotlight wink

 

:)

ilorty  Aug 4, 2020
 #3
avatar
0

1,380,456 mod 792 = 0

 Aug 4, 2020
 #4
avatar+25565 
+4

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

 

\(\mathbf{792=2^3*3^2*11}\)

 

1.

\(\mathbf{13ab45c}\) is divisible by \(\mathbf{2}\), so \(c=\{2,4,6,8\}\)

 

2.

\(\mathbf{13ab45c}\) is divisible by \(\mathbf{2^3=8}\), so \(45c \equiv 0 \pmod{8}\)

\(\begin{array}{|c|r|c|} \hline c & 45c & 45c \pmod{8} \equiv 0\ ?\\ \hline 2 & 452 & \text{no} \\ \hline 4 & 454 & \text{no} \\ \hline \mathbf{\color{red}6} & 456 & \text{yes} \\ \hline 8 & 458 & \text{no} \\ \hline \end{array} \), so \(\mathbf{c=6}\)

 

3.

\(\mathbf{13ab45c}\) is divisible by \(\mathbf{3^2=9}\), so \(13ab456 \equiv 0 \pmod{9}\)

\(\begin{array}{|rcll|} \hline 1+3+a+b+4+5+6 &\equiv& 0 \pmod{9} \\ a+b+19 &\equiv& 0 \pmod{9} \quad | \quad 19 \equiv 1 \pmod{9} \\ a+b+1 &\equiv& 0 \pmod{9} \quad | \quad - 1 \\ a+b &\equiv& -1 \pmod{9} \\ a+b &\equiv& -1+9 \pmod{9} \\ \mathbf{a+b} &\equiv& \mathbf{ 8 \pmod{9} } \\ \hline \end{array}\)

\(\begin{array}{|r|r|r|r|r|r|r|r|r|r|r|} \hline \mathbf{a+b}\pmod{9}= 8 \ ? \\ b \rightarrow & \mathbf{0} & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \mathbf{6} & \mathbf{7} & \mathbf{8} & \mathbf{9}\\ a \downarrow \\ \hline \mathbf{0} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & \color{red}8 & 0 \\ \hline \mathbf{1} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & \color{red}8 & 0 & 1 \\ \hline \mathbf{2} & 2 & 3 & 4 & 5 & 6 & 7 & \color{red}8 & 0 & 1 & 2 \\ \hline \mathbf{3} & 3 & 4 & 5 & 6 & 7 & \color{red}8 & 0 & 1 & 2 & 3 \\ \hline \mathbf{4} & 4 & 5 & 6 & 7 & \color{red}8 & 0 & 1 & 2 & 3 & 4 \\ \hline \mathbf{5} & 5 & 6 & 7 & \color{red}8 & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \mathbf{6} & 6 & 7 & \color{red}8 & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \mathbf{7} & 7 & \color{red}8 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \mathbf{8} & \color{red}8 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7& \color{red}8 \\ \hline \mathbf{9} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7& \color{red}8 & 0 \\ \hline \end{array}\)

, so \((a,b)=\{(0,8),(1,7),(2,6),(3,5),(4,4),(5,3),(6,2),(8,0),(8,9),(9,8)\}\)

 

4.
\(\mathbf{13ab45c}\) is divisible by \(\mathbf{11}\), so \(13ab456 \equiv 0 \pmod{11}\)

\(\begin{array}{|rcll|} \hline 1-3+a-b+4-5+6 &\equiv& 0 \pmod{11} \\ a-b+3 &\equiv& 0 \pmod{11} \quad | \quad - 3 \\ a-b &\equiv& -3 \pmod{11} \\ a-b &\equiv& -3+11 \pmod{11} \\ \mathbf{a-b} &\equiv& \mathbf{ 8 \pmod{11} } \\ \hline \end{array}\)

\(\begin{array}{|r|r|r|r|r|r|r|r|r|r|r|} \hline \mathbf{a-b}\pmod{11}= 8 \ ? \\ b \rightarrow & \mathbf{0} & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \mathbf{6} & \mathbf{7} & \mathbf{8} & \mathbf{9}\\ a \downarrow \\ \hline \mathbf{0} & 0 & 10 & 9 & \color{red}8 & 7 & 6 & 5 & 4 & 3 & 2 \\ \hline \mathbf{1} & 1 & 0 & 10 & 9 & \color{red}8 & 7 & 6 & 5 & 4 & 3 \\ \hline \mathbf{2} & 2 & 1 & 0 & 10 & 9 & \color{red}8 & 7 & 6 & 5 & 4 \\ \hline \mathbf{3} & 3 & 2 & 1 & 0 & 10 & 9 & \color{red}8 & 7 & 6 & 5 \\ \hline \mathbf{4} & 4 & 3 & 2 & 1 & 0 & 10 & 9 & \color{red}8 & 7 & 6 \\ \hline \mathbf{5} & 5 & 4 & 3 & 2 & 1 & 0 & 10 & 9 & \color{red}8 & 7 \\ \hline \mathbf{6} & 6 & 5 & 4 & 3 & 2 & 1 & 0 & 10 & 9 & \color{red}8 \\ \hline \mathbf{7} & 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0 & 10 & 9 \\ \hline \mathbf{8} & \color{red}8 & 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0 & 10 \\ \hline \mathbf{9} & 9 & \color{red}8 & 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0 \\ \hline \end{array}\)

, so \((a,b)=\{(0,3),(1,4),(2,5),(3,6),(4,7),(5,8),(6,9),(8,0),(9,1)\}\)

 

compare:

\(\mathbf{a+b}\pmod{9}= 8 : \quad (a,b)=\{(0,8),(1,7),(2,6),(3,5),(4,4),(5,3),(6,2),{\color{red}(8,0)},(8,9),(9,8)\} \\ \mathbf{a-b}\pmod{11}= 8 :\quad (a,b)=\{(0,3),(1,4),(2,5),(3,6),(4,7),(5,8),(6,9),{\color{red}(8,0)},(9,1)\}\)

,so \(a=8\) and \(b=0\)

 

\(\begin{array}{|lr|} \hline & \mathbf{13ab45c} = \mathbf{1380456} \\ \hline \text{check}: \\ & 1380456 : 792 = 1743 \\ \hline \end{array}\)

 

laugh

 Aug 4, 2020
 #8
avatar+732 
+1

thank you so much!!! i see what i did wrong now. thank you for the great and clear solution!!! :)

lokiisnotdead  Aug 5, 2020
 #5
avatar+110835 
+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
edited by Melody  Aug 4, 2020
 #9
avatar+732 
+1

thank you! yeah I forgot about adding a-b=8 and a+b=8 together. thank you so much for the work you put into the clear solution! :)

lokiisnotdead  Aug 5, 2020

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