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# If w, x, y, and z are the digits of the four-digit number N,

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If w, x, y, and z are the digits of the four-digit number N, [#permalink]  08 Nov 2009, 02:43
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 13 Aug 2012, 06:39, edited 1 time in total.
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Re: Remainder when divided by 9? [#permalink]  08 Nov 2009, 04:17
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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 $$\frac{w}{9}$$:

$$\frac{3000}{9}$$ remainder $$3$$, $$\frac{3}{9}$$remainder $$3$$.

The same with $$100x$$ and $$10y$$.

So, the remainder when $$1000w+100x+10y+z$$ is divided by 9 would be:

$$\frac{w}{9}+\frac{x}{9}+\frac{y}{9}+\frac{z}{9}=\frac{w+x+y+z}{9}$$

(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.

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Kudos [?]: 130 [1] , given: 9

Re: Remainder when divided by 9? [#permalink]  08 Nov 2009, 05:21
1
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 $$\frac{w}{9}$$:

$$\frac{3000}{9}$$ remainder $$3$$, $$\frac{3}{9}$$remainder $$3$$.

The same with $$100x$$ and $$10y$$.

So, the remainder when $$1000w+100x+10y+z$$ is divided by 9 would be:

$$\frac{w}{9}+\frac{x}{9}+\frac{y}{9}+\frac{z}{9}=\frac{w+x+y+z}{9}$$

(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.

kool !! and that's the OA
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Kudos [?]: 2300 [6] , given: 359

Re: Number Properties [#permalink]  25 Mar 2010, 11:55
6
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
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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..
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Kudos [?]: 2300 [0], given: 359

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.
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Kudos [?]: 49 [0], given: 7

Re: Remainder [#permalink]  01 Jun 2011, 06:29
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.

Ans D
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Re: Number Properties [#permalink]  03 Jun 2011, 17:22
1. Sufficient

sum of the digits is not divisible by 9

which is enough to tell that number is not divisible by 9.

2. Sufficient

N+5 is divisible by 9
=> N is not divisible by 9

or we can try the following
=>N+5 = 9k
=> N = 9k-5
=> N = 9k-5-4 +4
=9(k-1)+4

=> N is not divisible by 9.

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Re: Number Properties [#permalink]  05 Jun 2011, 23:44
13-9 = 4

a and b both giving 4 each.

D it is.
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Re: If w, x, y, and z are the digits of the four-digit number N, [#permalink]  06 Jun 2013, 05:10
Expert's post
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Kudos [?]: 23 [0], given: 99

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 $$\frac{w}{9}$$:

$$\frac{3000}{9}$$ remainder $$3$$, $$\frac{3}{9}$$remainder $$3$$.

The same with $$100x$$ and $$10y$$.

So, the remainder when $$1000w+100x+10y+z$$ is divided by 9 would be:

$$\frac{w}{9}+\frac{x}{9}+\frac{y}{9}+\frac{z}{9}=\frac{w+x+y+z}{9}$$

(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.

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?

eg -
1345/8 > r=1.
5431/8 > r=7.

KR,
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Re: If w, x, y, and z are the digits of the four-digit number N, [#permalink]  07 Jun 2013, 00:20
1
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.
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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

Hence D
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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.

thought exactly the same way.
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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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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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