1. W+Y+Z+X= 13
N + 5 is divisible by 92. N+5 is divisible by 9
Statement 1: W+Y+Z+X= 13avada 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
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?GMATGuruNY wrote:If the sum of an integer's digits is a multiple of 9, then the INTEGER ITSELF is a multiple of 9.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
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.
Yes, we can use the same reasoning for a multiple of 3.alex.gellatly wrote: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?GMATGuruNY wrote:If the sum of an integer's digits is a multiple of 9, then the INTEGER ITSELF is a multiple of 9.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
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.
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
Does this only work for multiples of 3 and 9? Or can we use this same logic for other integers?GMATGuruNY wrote:Yes, we can use the same reasoning for a multiple of 3.alex.gellatly wrote: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?GMATGuruNY wrote:If the sum of an integer's digits is a multiple of 9, then the INTEGER ITSELF is a multiple of 9.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
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.
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
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.
Alex have a look at this site. Focus on "The Divisibility Rules" and you will understand the topic better.alex.gellatly wrote:
Does this only work for multiples of 3 and 9? Or can we use this same logic for other integers?
