Data Sufficiency

If W Y X Z are the digits of the 4 digit number N,a positive integer, what is the remainder when N is divided by 9?
1. W+Y+Z+X= 13
2. N+5 is divisible by 9


D

If W Y X Z are the digits of the 4 digit number N,a positive integer, what is the remainder when N is divided by 9?

1. W+Y+Z+X= 13

N = 1000W + 100Y + 10Z + X
N = 999W + 99Y + 9Z + W + Y + Z + X
N = 9*K + W + Y + Z + X
So, the remainder of N/9 is same as the remainder of W + Y + Z + X/9.
The remainder of 4! So, statement I is sufficient to answer the question.

2. N+5 is divisible by 9

N + 5 is divisible by 9
N + 5 = 9K
N = 9K - 5
N = 9K - 9 + 4
N = 9*Integer + 4
The remainder of 4! So, statement I is sufficient to answer the question.

Answer : D

avada wrote:If W Y X Z are the digits of the 4 digit number N,a positive integer, what is the remainder when N is divided by 9?
1. W+Y+Z+X= 13
2. N+5 is divisible by 9

D

Statement 1: W+Y+Z+X= 13
If the sum of an integer's digits is a multiple of 9, then the INTEGER ITSELF is a multiple of 9.
Here, the sum of the digits is 13 -- 4 MORE than a multiple of 9.
The implication is that integer N = (multiple of 9) + 4.
(For a proof, see below.)
Thus, when N is divided by 9, the REMAINDER is 4.
SUFFICIENT.

Statement 2: N+5 is divisible by 9
N+5 = 9, 18, 27, 36...
Subtracting 5 from each value in this list, we get:
N = 4, 13, 22, 31...
In each case, when N is divided by 9, the remainder is 4:
4/9 = 0 R4.
13/9 = 1 R4.
22/9 = 2 R4.
SUFFICIENT.

The correct answer is D.

Proof for statement 1:

N = 1000W + 100Y + 10X + Z

= 999W + 99Y + 9X + (W+Y+X+Z)

= 9(111W + 11Y + X) + 13

= (multiple of 9) + (9 + 4)

= (multiple of 9 + 9) + 4

= (multiple of 9) + 4.

Thus, when N is divided by 9, the remainder is 4.

GMATGuruNY wrote:

avada wrote:If W Y X Z are the digits of the 4 digit number N,a positive integer, what is the remainder when N is divided by 9?
1. W+Y+Z+X= 13
2. N+5 is divisible by 9

D

If the sum of an integer's digits is a multiple of 9, then the INTEGER ITSELF is a multiple of 9.
Here, the sum of the digits is 13 -- 4 MORE than a multiple of 9.
The implication is that integer N = (multiple of 9) + 4.
(For a proof, see below.)
Thus, when N is divided by 9, the REMAINDER is 4.

So for a takeaway we can say in DS questions if they ask about remainders and they give us the sum like this we should know it's sufficient?

EX:
If xyz is a positive three digit integer. what is the remainder of xyz when divided by 3?
1) x+y+z = 14. So, here we don't need to do any math, we just know the rule stated above.

Let me know if I'm on the right (or wrong) track.
Thanks

alex.gellatly wrote:

GMATGuruNY wrote:

avada wrote:If W Y X Z are the digits of the 4 digit number N,a positive integer, what is the remainder when N is divided by 9?
1. W+Y+Z+X= 13
2. N+5 is divisible by 9

D

If the sum of an integer's digits is a multiple of 9, then the INTEGER ITSELF is a multiple of 9.
Here, the sum of the digits is 13 -- 4 MORE than a multiple of 9.
The implication is that integer N = (multiple of 9) + 4.
(For a proof, see below.)
Thus, when N is divided by 9, the REMAINDER is 4.

So for a takeaway we can say in DS questions if they ask about remainders and they give us the sum like this we should know it's sufficient?

EX:
If xyz is a positive three digit integer. what is the remainder of xyz when divided by 3?
1) x+y+z = 14. So, here we don't need to do any math, we just know the rule stated above.

Let me know if I'm on the right (or wrong) track.
Thanks

Yes, we can use the same reasoning for a multiple of 3.
If the sum of an integer's digits is a multiple of 3, then the integer itself is a multiple of 3.
Here, the sum of the digits is 14 -- 2 more than a multiple of 3.
Thus, when the integer is divided by 3, the remainder will be 2.

Proof:

100X + 10Y + Z

= 99X + 9Y + (X+Y+Z)

= 9(11X + Y) + 14

= (multiple of 3) + (12 + 2)

= (multiple of 3 + 12) + 2

= (multiple of 3 + multiple of 3) + 2

= (multiple of 3) + 2.

Thus, when the integer is divided by 3, the remainder will be 2.

GMATGuruNY wrote:

alex.gellatly wrote:

GMATGuruNY wrote:

avada wrote:If W Y X Z are the digits of the 4 digit number N,a positive integer, what is the remainder when N is divided by 9?
1. W+Y+Z+X= 13
2. N+5 is divisible by 9

D

If the sum of an integer's digits is a multiple of 9, then the INTEGER ITSELF is a multiple of 9.
Here, the sum of the digits is 13 -- 4 MORE than a multiple of 9.
The implication is that integer N = (multiple of 9) + 4.
(For a proof, see below.)
Thus, when N is divided by 9, the REMAINDER is 4.

So for a takeaway we can say in DS questions if they ask about remainders and they give us the sum like this we should know it's sufficient?

EX:
If xyz is a positive three digit integer. what is the remainder of xyz when divided by 3?
1) x+y+z = 14. So, here we don't need to do any math, we just know the rule stated above.

Let me know if I'm on the right (or wrong) track.
Thanks

Yes, we can use the same reasoning for a multiple of 3.
If the sum of an integer's digits is a multiple of 3, then the integer itself is a multiple of 3.
Here, the sum of the digits is 14 -- 2 more than a multiple of 3.
Thus, when the integer is divided by 3, the remainder will be 2.

Proof:

100X + 10Y + Z

= 99X + 9Y + (X+Y+Z)

= 9(11X + Y) + 14

= (multiple of 3) + (12 + 2)

= (multiple of 3 + 12) + 2

= (multiple of 3 + multiple of 3) + 2

= (multiple of 3) + 2.

Thus, when the integer is divided by 3, the remainder will be 2.

Does this only work for multiples of 3 and 9? Or can we use this same logic for other integers?

alex.gellatly wrote:
Does this only work for multiples of 3 and 9? Or can we use this same logic for other integers?

Alex have a look at this site. Focus on "The Divisibility Rules" and you will understand the topic better.
https://www.mathsisfun.com/divisibility-rules.html