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Why is can number be divided by 3(ofc I mean we get an integer result) if its sum of digits can be divided by 3 [duplicate]

0

$begingroup$

This question already has an answer here:

• 
  Prove that if the sum of digits of a number is divisible by 3, so is the number itself.
  
  2 answers
  
  

Why is can number be divided by 3(ofc I mean we get an integer result) if its sum of digits can be divided by 3?
Please post an easy proof since Im in grade 9:)
I know there are already proofs but can someone show me a proof for the level of a grade 9 student?

proof-explanation

share|cite|improve this question

edited Apr 2 at 15:06

Selim Jean Ellieh

asked Apr 2 at 14:58

Selim Jean ElliehSelim Jean Ellieh

22118

$endgroup$

marked as duplicate by Dietrich Burde, callculus, Lord Shark the Unknown, Leucippus, José Carlos Santos Apr 3 at 8:35

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But I cannot understand this proof I am in grade 9 :(
  $endgroup$
  – Selim Jean Ellieh
  Apr 2 at 15:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You only need the definition $aequiv bbmod 3$, i.e., that $3$ divides $b-a$. This one can understand in grade $9$, I think. Also, there are several other answers.
  $endgroup$
  – Dietrich Burde
  Apr 2 at 15:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You probably mean the sum of the digits ?
  $endgroup$
  – Yves Daoust
  Apr 2 at 15:04
  
  

  
  
  
  


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  1
  

  
  
  
  
  
  

  
  $begingroup$
  It has to do with the remainders. When you divide $10$ by $3$, what is the remainder? How about $100$, $1000$. Notice a pattern?
  $endgroup$
  – Vasya
  Apr 2 at 15:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  the remainder is always 1 . and so?
  $endgroup$
  – Selim Jean Ellieh
  Apr 2 at 15:07
  
  

  
  
  
  

add a comment | 

0

$begingroup$

This question already has an answer here:

• 
  Prove that if the sum of digits of a number is divisible by 3, so is the number itself.
  
  2 answers
  
  

Why is can number be divided by 3(ofc I mean we get an integer result) if its sum of digits can be divided by 3?
Please post an easy proof since Im in grade 9:)
I know there are already proofs but can someone show me a proof for the level of a grade 9 student?

proof-explanation

share|cite|improve this question

edited Apr 2 at 15:06

Selim Jean Ellieh

asked Apr 2 at 14:58

Selim Jean ElliehSelim Jean Ellieh

22118

$endgroup$

marked as duplicate by Dietrich Burde, callculus, Lord Shark the Unknown, Leucippus, José Carlos Santos Apr 3 at 8:35

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But I cannot understand this proof I am in grade 9 :(
  $endgroup$
  – Selim Jean Ellieh
  Apr 2 at 15:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You only need the definition $aequiv bbmod 3$, i.e., that $3$ divides $b-a$. This one can understand in grade $9$, I think. Also, there are several other answers.
  $endgroup$
  – Dietrich Burde
  Apr 2 at 15:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You probably mean the sum of the digits ?
  $endgroup$
  – Yves Daoust
  Apr 2 at 15:04
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It has to do with the remainders. When you divide $10$ by $3$, what is the remainder? How about $100$, $1000$. Notice a pattern?
  $endgroup$
  – Vasya
  Apr 2 at 15:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  the remainder is always 1 . and so?
  $endgroup$
  – Selim Jean Ellieh
  Apr 2 at 15:07
  
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

• 
  Prove that if the sum of digits of a number is divisible by 3, so is the number itself.
  
  2 answers
  
  

Why is can number be divided by 3(ofc I mean we get an integer result) if its sum of digits can be divided by 3?
Please post an easy proof since Im in grade 9:)
I know there are already proofs but can someone show me a proof for the level of a grade 9 student?

proof-explanation

share|cite|improve this question

edited Apr 2 at 15:06

Selim Jean Ellieh

asked Apr 2 at 14:58

Selim Jean ElliehSelim Jean Ellieh

22118

$endgroup$

share|cite|improve this question

edited Apr 2 at 15:06

Selim Jean Ellieh

asked Apr 2 at 14:58

Selim Jean ElliehSelim Jean Ellieh

22118

share|cite|improve this question

edited Apr 2 at 15:06

Selim Jean Ellieh

asked Apr 2 at 14:58

Selim Jean ElliehSelim Jean Ellieh

22118

asked Apr 2 at 14:58

Selim Jean ElliehSelim Jean Ellieh

22118

asked Apr 2 at 14:58

Selim Jean ElliehSelim Jean Ellieh

22118

3 Answers
3

active

oldest

votes

1

$begingroup$

A number such as $2437$ is understood as the sum $2cdot1000+4cdot100+3cdot10+7$. But this can be written

$$2cdot(9cdot111+1)+4cdot(9cdot11+1)+3cdot(9+1)+7
\=9cdot(2cdot111+4cdot11+3)+2+4+3+7.$$

So the remainder of the division of $2437$ by $9$ is the same as the remainder of the division of $2+4+3+7$.

You can generalize.

(Also note that $2+4+3+7=16$, so that the remainder is $1+6$: you may iterate.)

share|cite|improve this answer

answered Apr 2 at 15:08

Yves DaoustYves Daoust

133k676232

$endgroup$

add a comment | 

1

$begingroup$

Any number can be written in powers of $10$, i.e. $2472=2 cdot 1000 + 4 cdot 100 + 7 cdot 10 + 2$. As you noticed, if we divide power of ten by three, the remainder is one. If we divide each member of the sum by $3$ and add remainders, we will get $2+4+7+2=15$ (notice that this is the sum of digits) as the remainder. Thus we can write $2472=k cdot 3 +15=3(k+5)$ which is divisible by $3$.

share|cite|improve this answer

answered Apr 2 at 15:14

VasyaVasya

4,5441619

$endgroup$

add a comment | 

0

$begingroup$

If you have a multiple of $3$, and you add or subtract another multiple of $3$ then you will get a multiple of $3$.

So notice that $9$, and $99$ and $999$, etc are all multiples of $3.$ So if you take a number like $456$ you have:

$$456 = 4times 100 + 5times 10 + 6 $$

$$ = 4 times (99+1) + 5times (9+1) +6$$

$$=4+5+6+mbox a multiple of 3.$$

So you have that, if $4+5+6$ is a multiple of 3, then so is $456.$

share|cite|improve this answer

answered Apr 2 at 15:11

B. GoddardB. Goddard

20.2k21543

$endgroup$

add a comment | 

1

$begingroup$

A number such as $2437$ is understood as the sum $2cdot1000+4cdot100+3cdot10+7$. But this can be written

$$2cdot(9cdot111+1)+4cdot(9cdot11+1)+3cdot(9+1)+7
\=9cdot(2cdot111+4cdot11+3)+2+4+3+7.$$

So the remainder of the division of $2437$ by $9$ is the same as the remainder of the division of $2+4+3+7$.

You can generalize.

(Also note that $2+4+3+7=16$, so that the remainder is $1+6$: you may iterate.)

share|cite|improve this answer

answered Apr 2 at 15:08

Yves DaoustYves Daoust

133k676232

$endgroup$

add a comment | 

1

$begingroup$

A number such as $2437$ is understood as the sum $2cdot1000+4cdot100+3cdot10+7$. But this can be written

$$2cdot(9cdot111+1)+4cdot(9cdot11+1)+3cdot(9+1)+7
\=9cdot(2cdot111+4cdot11+3)+2+4+3+7.$$

So the remainder of the division of $2437$ by $9$ is the same as the remainder of the division of $2+4+3+7$.

You can generalize.

(Also note that $2+4+3+7=16$, so that the remainder is $1+6$: you may iterate.)

share|cite|improve this answer

answered Apr 2 at 15:08

Yves DaoustYves Daoust

133k676232

$endgroup$

add a comment | 

$begingroup$

A number such as $2437$ is understood as the sum $2cdot1000+4cdot100+3cdot10+7$. But this can be written

$$2cdot(9cdot111+1)+4cdot(9cdot11+1)+3cdot(9+1)+7
\=9cdot(2cdot111+4cdot11+3)+2+4+3+7.$$

So the remainder of the division of $2437$ by $9$ is the same as the remainder of the division of $2+4+3+7$.

You can generalize.

(Also note that $2+4+3+7=16$, so that the remainder is $1+6$: you may iterate.)

share|cite|improve this answer

answered Apr 2 at 15:08

Yves DaoustYves Daoust

133k676232

$endgroup$

share|cite|improve this answer

answered Apr 2 at 15:08

Yves DaoustYves Daoust

133k676232

answered Apr 2 at 15:08

Yves DaoustYves Daoust

133k676232

answered Apr 2 at 15:08

Yves DaoustYves Daoust

133k676232

1

$begingroup$

Any number can be written in powers of $10$, i.e. $2472=2 cdot 1000 + 4 cdot 100 + 7 cdot 10 + 2$. As you noticed, if we divide power of ten by three, the remainder is one. If we divide each member of the sum by $3$ and add remainders, we will get $2+4+7+2=15$ (notice that this is the sum of digits) as the remainder. Thus we can write $2472=k cdot 3 +15=3(k+5)$ which is divisible by $3$.

share|cite|improve this answer

answered Apr 2 at 15:14

VasyaVasya

4,5441619

$endgroup$

add a comment | 

1

$begingroup$

Any number can be written in powers of $10$, i.e. $2472=2 cdot 1000 + 4 cdot 100 + 7 cdot 10 + 2$. As you noticed, if we divide power of ten by three, the remainder is one. If we divide each member of the sum by $3$ and add remainders, we will get $2+4+7+2=15$ (notice that this is the sum of digits) as the remainder. Thus we can write $2472=k cdot 3 +15=3(k+5)$ which is divisible by $3$.

share|cite|improve this answer

answered Apr 2 at 15:14

VasyaVasya

4,5441619

$endgroup$

add a comment | 

$begingroup$

Any number can be written in powers of $10$, i.e. $2472=2 cdot 1000 + 4 cdot 100 + 7 cdot 10 + 2$. As you noticed, if we divide power of ten by three, the remainder is one. If we divide each member of the sum by $3$ and add remainders, we will get $2+4+7+2=15$ (notice that this is the sum of digits) as the remainder. Thus we can write $2472=k cdot 3 +15=3(k+5)$ which is divisible by $3$.

share|cite|improve this answer

answered Apr 2 at 15:14

VasyaVasya

4,5441619

$endgroup$

share|cite|improve this answer

answered Apr 2 at 15:14

VasyaVasya

4,5441619

answered Apr 2 at 15:14

VasyaVasya

4,5441619

answered Apr 2 at 15:14

VasyaVasya

4,5441619

0

$begingroup$

If you have a multiple of $3$, and you add or subtract another multiple of $3$ then you will get a multiple of $3$.

So notice that $9$, and $99$ and $999$, etc are all multiples of $3.$ So if you take a number like $456$ you have:

$$456 = 4times 100 + 5times 10 + 6 $$

$$ = 4 times (99+1) + 5times (9+1) +6$$

$$=4+5+6+mbox a multiple of 3.$$

So you have that, if $4+5+6$ is a multiple of 3, then so is $456.$

share|cite|improve this answer

answered Apr 2 at 15:11

B. GoddardB. Goddard

20.2k21543

$endgroup$

add a comment | 

0

$begingroup$

If you have a multiple of $3$, and you add or subtract another multiple of $3$ then you will get a multiple of $3$.

So notice that $9$, and $99$ and $999$, etc are all multiples of $3.$ So if you take a number like $456$ you have:

$$456 = 4times 100 + 5times 10 + 6 $$

$$ = 4 times (99+1) + 5times (9+1) +6$$

$$=4+5+6+mbox a multiple of 3.$$

So you have that, if $4+5+6$ is a multiple of 3, then so is $456.$

share|cite|improve this answer

answered Apr 2 at 15:11

B. GoddardB. Goddard

20.2k21543

$endgroup$

add a comment | 

$begingroup$

If you have a multiple of $3$, and you add or subtract another multiple of $3$ then you will get a multiple of $3$.

So notice that $9$, and $99$ and $999$, etc are all multiples of $3.$ So if you take a number like $456$ you have:

$$456 = 4times 100 + 5times 10 + 6 $$

$$ = 4 times (99+1) + 5times (9+1) +6$$

$$=4+5+6+mbox a multiple of 3.$$

So you have that, if $4+5+6$ is a multiple of 3, then so is $456.$

share|cite|improve this answer

answered Apr 2 at 15:11

B. GoddardB. Goddard

20.2k21543

$endgroup$

share|cite|improve this answer

answered Apr 2 at 15:11

B. GoddardB. Goddard

20.2k21543

answered Apr 2 at 15:11

B. GoddardB. Goddard

20.2k21543

answered Apr 2 at 15:11

B. GoddardB. Goddard

20.2k21543