Find the missing digits in the calculation below

[Math100]

Homework Statement

Working modulo $9$ or $11$, find the missing digits in the calculation below:
$2\chi 99561=[3(523+\chi)]^{2}$.

Relevant Equations

None.

Observe that $2\chi 99561=[3(523+\chi)]^{2}=9(523+\chi)^{2}$.
This means $9\mid [9(523+\chi)^{2}]\implies 9\mid 2\chi 99561$.
Thus $9\mid (2+\chi+9+9+5+6+1)\implies 9\mid (32+\chi)$.
Therefore, $\chi=4$.

 

[Mark44]

Homework Statement:: Working modulo $9$ or $11$, find the missing digits in the calculation below:
$2\chi 99561=[3(523+\chi)]^{2}$.
Relevant Equations:: None.

Observe that $2\chi 99561=[3(523+\chi)]^{2}=9(523+\chi)^{2}$.
This means $9\mid [9(523+\chi)^{2}]\implies 9\mid 2\chi 99561$.
Thus $9\mid (2+\chi+9+9+5+6+1)\implies 9\mid (32+\chi)$.
Therefore, $\chi=4$.

I don't understand what "Working modulo 9 or 11 ... " means in the context of this problem. Is the above the exact statement of the problem? Could you have mistakenly mixed in part of one problem with another?

The work you did seems to assume a problem statement of "If $2\chi 99561$ is divisible by 9, find the missing digit from the equation ##$2\chi 99561 = [3(523 + \chi)]^2$.

How does 11 figure into this problem?

 

[SammyS]

I don't understand what "Working modulo 9 or 11 ... " means in the context of this problem. Is the above the exact statement of the problem? Could you have mistakenly mixed in part of one problem with another?

The work you did seems to assume a problem statement of "If $2\chi 99561$ is divisible by 9, find the missing digit from the equation $2\chi 99561 = [3(523 + \chi)]^2$.

How does 11 figure into this problem?

Seems it would have been better for OP to start out with the observation that
$(3(523+\chi))^{2}=9(523+\chi)^2$ , thus $9$ divides $(3(523+\chi))^{2}$ .

Therefore, ...

 

[Math100]

I don't understand what "Working modulo 9 or 11 ... " means in the context of this problem. Is the above the exact statement of the problem? Could you have mistakenly mixed in part of one problem with another?

The work you did seems to assume a problem statement of "If $2\chi 99561$ is divisible by 9, find the missing digit from the equation ##$2\chi 99561 = [3(523 + \chi)]^2$.

How does 11 figure into this problem?

There's no mistake in the problem statement. And do I also have to include/show the modulo $11$ part? I thought with only working modulo $9$, it is sufficient since the solution/missing digit is $4$, which I've found. Do I still have to work modulo $11$?

 

[SammyS]

There's no mistake in the problem statement. And do I also have to include/show the modulo $11$ part? I thought with only working modulo $9$, it is sufficient since the solution/missing digit is $4$, which I've found. Do I still have to work modulo $11$?

I think you're right. The way I read the problem, it seems to be giving you the option.

 

[Mark44]

There's no mistake in the problem statement. And do I also have to include/show the modulo $11$ part? I thought with only working modulo $9$, it is sufficient since the solution/missing digit is $4$, which I've found. Do I still have to work modulo $11$?

I would think so, since the problem author mentioned it. But like I said, I don't quite get what the author wants you to do with the modulo 11 part.

 

[Math100]

I would think so, since the problem author mentioned it. But like I said, I don't quite get what the author wants you to do with the modulo 11 part.

Me neither. I don't get it either.

 

[Delta2]

Someone give me a proof of the lemma :
"9 divides a number if and only if divides the sum of its digits in base 10" cause I can't prove it lol.

Oh wait something clicked, do we use binomial expansion and the fact that $10^n=(9+1)^n$?

I think this lemma can be generalized as
"k divides a number n if and only if it divides the sum of digits of n in base k+1".

 

[Delta2]

BTW problem says "working with modulo 9 OR modulo 11", it doesn't say "AND" modulo 11.

 

[Mark44]

"9 divides a number if and only if divides the sum of its digits in base 10" cause I can't prove it lol.

Here's an explanation that could be extended to a proof by induction.
Let $N = a_0*10^2 + a_1*10^1 + a_2$, where $a_0 + a_1 + a_2 = 9k$ for some integer k.
$N = a_010^2 + a_1*10^1 + a_2 = 10[a_0 * 10 + a_1] + a_2 = 9[a_0 * 10 + a_1] + [a_0 * 10 + a_1] + a_2$
$= 9[a_0 * 10 + a_1] + 9 * a_0 + a_0 + a_1 + a_2$
$= 9[a_0 * 10 + a_1] + 9 * a_0 + 9k = 9[a_0 * 10 + a_1 + a_0 + k]$
Therefore, N is divisible by 9 provided that the decimal digits of N add up to 9 or a multiple thereof.

BTW problem says "working with modulo 9 OR modulo 11", it doesn't say "AND" modulo 11.

So why was modulo 11 added to the problem? Could the problem be worked if it said only "working with modulo 11"? I've spent a fair amount of time on this problem, but don't see any way that modulo 11 plays any part.

 

[Delta2]

There is another lemma for modulo 11:
"11 divides a number if and only if it divides the alternating sum/difference of its digits in base 10"
By alternating I mean the sum $$\sum (-1)^n a_n$$ where $a_n$ the digits of the number (n=0 the units digit).

 

[Delta2]

NVM now I see that that number is not divisible by 11

 

[Mark44]

There is another lemma for modulo 11:
"11 divides a number if and only if it divides the alternating sum/difference of its digits in base 10"

Yes, I know this rule. I don't think it has much to do with modulo 11, but rather can be used to determine whether a decimal (i.e., base-10) number is divisible by 11.
If $2\chi99561$ is divisible by 11, it must be the case that $\chi = 2$. If so, the number on the right side is $9 \cdot 525 = 3^2 \cdot 5^4$, which is not divisible by 11. This is something I checked earlier.

Relative to modulo 9 and modulo 11, I also checked whether the numbers were written in base-9 (no, since $2\chi99561$ is not a base-9 number) or base-11, neither of which yielded any enlightenment.

 

[SammyS]

Seems it would have been better for OP to start out with the observation that
$(3(523+\chi))^{2}=9(523+\chi)^2$ , thus $9$ divides $(3(523+\chi))^{2}$ .

Therefore, ...

So, do we need to do this by explicitly working modulo 9 ? - or is is it enough to say that the above shows that the right hand side of the given equation is divisible by 9, therefore, the left hand side must also be divisible by 9, which allows us to use the divisibility test for division by 9 on LHS ?
( For a whole number written as a decimal, the test OP is using for divisibility by 9 is based on equivalence modulo 9 . )

As for the instruction: "Working modulo $9$ or $11$" :
It would not be unusual to have a single instruction for a whole set of exercises.