Is the integer ## (447836)_{9} ## divisible by ## 3 ## and ## 8 ##?

[Math100]

Homework Statement

Is the integer $(447836)_{9}$ divisible by $3$ and $8$?

Relevant Equations

None.

Observe that $(447836)_{9}=6+3\cdot 9+8\cdot 9^{2}+7\cdot 9^{3}+4\cdot 9^{4}+4\cdot 9^{5}=268224$.
Then $2+6+8+2+2+4=24$.
Thus $3\mid (2+6+8+2+2+4)\implies 3\mid (447836)_{9}$ and $8\mid (2+6+8+2+2+4)\implies 8\mid (447836)_{9}$.
Therefore, the integer $(447836)_{9}$ is divisible by $3$ and $8$.

 

[fresh_42]

Homework Statement:: Is the integer $(447836)_{9}$ divisible by $3$ and $8$?
Relevant Equations:: None.

Observe that $(447836)_{9}=6+3\cdot 9+8\cdot 9^{2}+7\cdot 9^{3}+4\cdot 9^{4}+4\cdot 9^{5}=268224$.
Then $2+6+8+2+2+4=24$.
Thus $3\mid (2+6+8+2+2+4)\implies 3\mid (447836)_{9}$ and $8\mid (2+6+8+2+2+4)\implies 8\mid (447836)_{9}$.
Therefore, the integer $(447836)_{9}$ is divisible by $3$ and $8$.

Correct. How do you check that a number is divisible by eight?

 

[Mark44]

Homework Statement:: Is the integer $(447836)_{9}$ divisible by $3$ and $8$?
Relevant Equations:: None.

Observe that $(447836)_{9}=6+3\cdot 9+8\cdot 9^{2}+7\cdot 9^{3}+4\cdot 9^{4}+4\cdot 9^{5}=268224$.
Then $2+6+8+2+2+4=24$.

It's a well-known fact that if the digits of a decimal number (i.e., in base-10) add up to 3 or a multiple of 3, then the number itself is divisible by 3. It's also a fact that if the digits of the same number add up to 9 or a multiple of 9, then the number is divisible by 9.

Thus $3\mid (2+6+8+2+2+4)\implies 3\mid (447836)_{9}$ and $8\mid (2+6+8+2+2+4)\implies 8\mid (447836)_{9}$.

You wrote $3\mid (2+6+8+2+2+4)\implies 3\mid (447836)_{9}$ twice. You don't need to do that. Also, and fresh_42 beat me to it, why is it the case that 8 divides (2+6+8+2+2+4)? This is not at all obvious.

Therefore, the integer $(447836)_{9}$ is divisible by $3$ and $8$.

 

[Math100]

Correct. How do you check that a number is divisible by eight?

I haven't thought about that. Do you know how? In this given question/problem, it's obvious since $24$ is divisible by both $3$ and $8$. But when it comes to large integers, do you know any rule?

 

[Math100]

It's a well-known fact that if the digits of a decimal number (i.e., in base-10) add up to 3 or a multiple of 3, then the number itself is divisible by 3. It's also a fact that if the digits of the same number add up to 9 or a multiple of 9, then the number is divisible by 9.
You wrote $3\mid (2+6+8+2+2+4)\implies 3\mid (447836)_{9}$ twice. You don't need to do that. Also, and fresh_42 beat me to it, why is it the case that 8 divides (2+6+8+2+2+4)? This is not at all obvious.

$8$ divides $(2+6+8+2+2+4)$ because $8\mid 24$.

 

[Mark44]

$8$ divides $(2+6+8+2+2+4)$ because $8\mid 24$.

Clearly, but does 8 divide 268,224?

 

[fresh_42]

$8$ divides $(2+6+8+2+2+4)$ because $8\mid 24$.

This is wrong. We know that $8\,|\,1000$ and any number $a_na_{n-1}\ldots a_1=a_na_{n-1}\ldots a_4\cdot 1000 + a_3a_2a_1.$ So what is the rule?

 

[Mark44]

There are two other facts that I didn't mention:
If 4 divides that last two digits of a decimal number (i.e., the digits in the ten's and one's places), 4 divides the number.
If 8 divides that last three digits of a decimal number, 8 divides the number.

 

[Math100]

Clearly, but does 8 divide 268,224?

Yes, since $224$, the last three digits of $268,224$ is divisible by $8$.

 

[Math100]

This is wrong. We know that $8\,|\,1000$ and any number $a_na_{n-1}\ldots a_1=a_na_{n-1}\ldots a_4\cdot 1000 + a_3a_2a_1.$ So what is the rule?

I've never seen this before. What is this rule?

 

[fresh_42]

I've never seen this before. What is this rule?

$8$ divides all thousands, so we have to check whether $8$ divides the last number formed by the last three digits. If $n\,|\,a$ then $n\,|\,(a+b) \Longleftrightarrow n\,|\,b.$

 

[Delta2]

Perhaps you could argue in a different way that 3 divides the given number in base 9. Because 3 divides 9 and 3 divides the last digit of the number which is 6.

 

[SammyS]

Homework Statement:: Is the integer $(447836)_{9}$ divisible by $3$ and $8$?
Relevant Equations:: None.

Observe that $(447836)_{9}=6+3\cdot 9+8\cdot 9^{2}+7\cdot 9^{3}+4\cdot 9^{4}+4\cdot 9^{5}=268224$.
Then $2+6+8+2+2+4=24$.
Thus $3\mid (2+6+8+2+2+4)\implies 3\mid (447836)_{9}$ and $8\mid (2+6+8+2+2+4)\implies 8\mid (447836)_{9}$.
Therefore, the integer $(447836)_{9}$ is divisible by $3$ and $8$.

Yes, $(447836)_{9}$ which is 268224 in base ten, is divisible by $8$. Notice that $4+4+7+8+3+6=32$ which is divisible by $8$. That is a valid test of divisibility by 8 for a base 9 number. The reason this works is that $9\equiv 1\pmod 8$.
The fact that $8$ divides $2+6+8+2+2+4$ is merely a coincidence.
Take the decimal number $267496$. It's divisible by $8$, but the sum of those digits is not divisible by $8$.
If I did my calculation correctly, that is $(446837)_{9}$ in base 9.

As for testing for divisibility by $3$ of a number written in base 9 : simply check the one's digit. After all, $9\equiv 0\pmod 3$ .
No need to convert to decimal representation.
Observe that $(447836)_{9}=6+3\cdot 9+8\cdot 9^{2}+7\cdot 9^{3}+4\cdot 9^{4}+4\cdot 9^{5}\equiv 6\pmod 9$.
And 6 is equivalent to 0 .

Added in Edit:
I should have said: $(447836)_{9}=6+3\cdot 9+8\cdot 9^{2}+7\cdot 9^{3}+4\cdot 9^{4}+4\cdot 9^{5}\equiv 6\equiv 0\pmod 3$.

We only needed to check the one's digit, base 9. This is similar to the test for divisibility by 5 or by ten in the case of decimal representation.