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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

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 \(\frac{w}{9}\):

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) 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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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..

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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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.

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]

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07 Jun 2013, 00:20

1

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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

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]

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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

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

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05 Nov 2015, 06:25

1 and 2 gives the same information knowing the sum of all digits, we clearly see that N is not divisible by 9, or that N+5 would be divisible by 9. This is sufficient. 2 - gives directly N+5 divisible by 9. both statements sufficient.

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

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05 Jan 2016, 22:54

timothyhenman1 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

Nice question. Answer is (D) - Each alone is sufficient. Here's why:

(1) SUFFICIENT.

The rule for divisibility by 9 is that you add up all the digits of the given number. If this addition is divisible by nine, the number too is divisible by nine. In case the addition works out to be a number > 9, add the digits again. The same works for remainders - for 9, the remainder you get when you add digits and divide the result by 9 is the same remainder you get after dividing the original number by 9.

In our case, w+x+y+z = 13 Add digits again - 1+3 = 4 So, remainder after dividing by 9 is exactly 4 => SUFFICIENT.

Now for the analytical part - WHY?

A four digit number can be represented as 1000w + 100x + 10y + z.

Let's take the Remainder operation on "number" and call it Rem(number/9). Note that we can add remainders as long as we make the final adjustment to make them < original number (which is 9 here).

So, overall remainder is = Rem(1000w/9) + Rem (100x/9) + Rem(10y/9) +z

For any multiple of 10, you'll notice that 10^n - 1 is always divisible by 9 (e.g. 9, 99, 999 and so on). So, Rem(1000w/9) becomes Rem(1000/9)*Rem(w/9) = 1*Rem(w/9) As w is a single digit, Rem(w/9) = w (even if w is 9, remainder can be technically 9. we will make the adjustment in the final phase)

Similarly, Rem(100x/9) = x, Rem (10y/9) = y and Rem(z/9) = z

So, overall remainder = Rem{(w+x+y+z)/9} That's what we did above. We took the remainder after adding all the digits. This is complicated to remember, so just focus on the part before the incurably curious mind asks "WHY". It's a rule worth remembering.

(2) SUFFICIENT

If N+5 is divisible by 9, it means N+5 = 9k (where k is any integer). You can just stop here and say that this is sufficient, as if you know this then intuitively you know that N+4 would leave a remainder of 8, N+3 would leave 7 and so on.. with N/9 leaving a remainder of 4. Hence, SUFFICIENT.

Here's a more analytical treatment (and unnecessary at that - you MUST save time on DS if you can. If you are sure with the above approach, don't even do this) -

So, N = 9k - 5.

i.e. N/9 = k -5/9

compare this with Dividend / Divisor = Quotient + Remainder/Divisor

We get Remainder = -5.

For negative remainders, we add back the divisor to get the proper positive remainder. So, remainder = -5 +9 = 4. Therefore, SUFFICIENT.
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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

When you divide some integer n with 9, the remainder is same as the remainder which is the sum of all digits of the integer n and divided by 9. In the original condition, there is 1 variable(n), which should match with the number of equations. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer. In 1), the sum of all digits of n is 13. From 13=9*1+4, 4 is the remainder as divided by 9, which is unique and sufficient. In 2), when dividing n+5 with 9, from n+5=9m, n=9m-5=9(m-1)+4, 4 is also the remainder as dividing n with 9, which is unique and sufficient. Therefore, the answer is D.

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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