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Re: Remainder when divided by 9? [#permalink]
08 Nov 2009, 04:17

8

This post received KUDOS

Expert's post

6

This post was BOOKMARKED

gmattokyo wrote:

If w, x, y, and z are the digits of the four-digit number N, a positive integer, what is the remainder when N is divided by 9? 1. w + x + y + z = 13 2. N + 5 is divisible by 9

Remainder when a number is divided by 9 is the same as remainder when the sum of its digits is divided by 9:

\(Remainder \frac{N}{9}=Remainder \frac{w + x + y + z}{9}\)

Let's show this on our example:

Our 4 digit number is \(1000w+100x+10y+z\). what is the remainder when it's divided by 9?

When 1000w is divided by 9 the remainder is \(\frac{w}{9}\):

Re: Remainder when divided by 9? [#permalink]
08 Nov 2009, 05:21

1

This post received KUDOS

Bunuel wrote:

gmattokyo wrote:

If w, x, y, and z are the digits of the four-digit number N, a positive integer, what is the remainder when N is divided by 9? 1. w + x + y + z = 13 2. N + 5 is divisible by 9

Remainder when a number is divided by 9 is the same as remainder when the sum of its digits is divided by 9:

\(Remainder \frac{N}{9}=Remainder \frac{w + x + y + z}{9}\)

Let's show this on our example:

Our 4 digit number is \(1000w+100x+10y+z\). what is the remainder when it's divided by 9?

When 1000w is divided by 9 the remainder is \(\frac{w}{9}\):

Re: Number Properties [#permalink]
25 Mar 2010, 11:55

6

This post received KUDOS

Expert's post

D

1) N = w*1000 + x*100 + y*10 + z --> N = (w + w*999) + (x + x*99) + (y + y*9) + z --> N = (w + x + y + z) + 9*(w*111+x*11+y) ---> N = 13 + 9*(w*111+x*11+y) = 9 + 4 + 9*(w*111+x*11+y) --> remainder is 4. Sufficient

2) N + 5 = 9k --> N = 9k - 5 --> N = 9(k-1) + 4 --> remainder is 4. Sufficient _________________

Re: Number Properties [#permalink]
25 Mar 2010, 11:59

walker wrote:

D

1) N = w*1000 + x*100 + y*10 + z --> N = (w + w*999) + (x + x*99) + (y + y*9) + z --> N = (w + x + y + z) + 9*(w*111+x*11+y) ---> N = 13 + 9*(w*111+x*11+y) = 9 + 4 + 9*(w*111+x*11+y) --> remainder is 4. Sufficient

2) N + 5 = 9k --> N = 9k - 5 --> N = 9(k-1) + 4 --> remainder is 4. Sufficient

Hi Walker, nice explanation.. i really had to try with couple of actual numbers to come up with answer D. Thank you! learned a new strategy. On a side note I have sent you a PM..

Re: Number Properties [#permalink]
25 Mar 2010, 12:05

Expert's post

My first thought was it is B but then I recalled the rule for divisibility by 3 and by 9: divisibility of sum of digits. So, that was a clue to start thinking about the first statement. _________________

Divisibility rule for 9: sum of digits must be divisible by 9 from 1) we know sum of digits = 13, which is 5 less than 18. If the sum had been 18 then remainder would be 0. So for any number whose sum of digits is 13 remainder will be 4 when divided by 9. Therefore sufficient. From 2) we know that N+5 = multiple of 9. So remainder of N will be 4. Hence sufficient.

Re: Remainder when divided by 9? [#permalink]
06 Jun 2013, 20:40

Bunuel wrote:

gmattokyo wrote:

If w, x, y, and z are the digits of the four-digit number N, a positive integer, what is the remainder when N is divided by 9? 1. w + x + y + z = 13 2. N + 5 is divisible by 9

Remainder when a number is divided by 9 is the same as remainder when the sum of its digits is divided by 9:

\(Remainder \frac{N}{9}=Remainder \frac{w + x + y + z}{9}\)

Let's show this on our example:

Our 4 digit number is \(1000w+100x+10y+z\). what is the remainder when it's divided by 9?

When 1000w is divided by 9 the remainder is \(\frac{w}{9}\):

(1) w + x + y + z = 13 --> remainder 13/9=4, remainder N/9=4. Sufficient.

(2) N+5 is divisible by 9 --> N+5=9k --> N=9k-5=4, 13, 22, ... --> remainder upon dividing this numbers by 9 is 4. Sufficient.

Answer: D.

Hi Bunnel,

Is this rule applicable to only 9. I tried to check the divisibility with 8 for a 4 digit number with digits adding up to 13. Will it work for 8 some other sum of the digits?

Re: If w, x, y, and z are the digits of the four-digit number N, [#permalink]
07 Jun 2013, 00:20

1

This post received KUDOS

gmattokyo wrote:

If w, x, y, and z are the digits of the four-digit number N, a positive integer, what is the remainder when N is divided by 9?

(1) w + x + y + z = 13 (2) N + 5 is divisible by 9

There is a divisibility rule for 9 - if the sum of the digits of the number is divisible to 9 then the whole number is divisible to 9.

(1) the sum of the digits is 13 which is 4 more than 9, it means in order to be divisible to 9 one of the digits or the sum of some digits within the number should be 4 less. Otherwise there will be 4 extra when we divide to 9. So the remainder is 4. Sufficient.

(2) the same rule as in the first statement applies here as well. Moreover it does not really matter how big is the number whether four digits or two. For example lets take possible two digits numbers for N: 13, 22, 31 etc. in all case the remainder when divided by 9 will be 4. Or four digit numbers: 1129, or 1138, we will have the same remainder. Sufficient. _________________

If you found my post useful and/or interesting - you are welcome to give kudos!

Re: If w, x, y, and z are the digits of the four-digit number N, [#permalink]
27 Jun 2013, 08:07

gmattokyo wrote:

If w, x, y, and z are the digits of the four-digit number N, a positive integer, what is the remainder when N is divided by 9?

(1) w + x + y + z = 13 (2) N + 5 is divisible by 9

A tip: whenever you get a question in which digits of a no. are given in terms of x,y .. always write it down in form 100x+10y+z ... according to no. of digits in that number. you'll automatically start proceeding with the solution and eventually reach it

st. 1: N= 1000w+100x+10y+z = (999w+99x+9y)+(w+x+y+z). First part of this is divisible by 9 and value of 2nd part is given. we can calculate the remainder.

st. 2: N+5 is divisible by 9 .. [(N+5)-9]=[N-4] will also be divisible by 9. To reach to N we need to add 4. hence 4 is the remainder

Re: Number Properties [#permalink]
02 Oct 2013, 22:24

walker wrote:

My first thought was it is B but then I recalled the rule for divisibility by 3 and by 9: divisibility of sum of digits. So, that was a clue to start thinking about the first statement.

Re: If w, x, y, and z are the digits of the four-digit number N, [#permalink]
01 Dec 2014, 12:21

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