DS:
Number XYZ is the product of positive integer n and 9. Is X+Y+Z=9?
(1) x+y+z<15
(2) x+y+z>8
I have absolutely no idea how to solve this. Can someone please help me?
tricky DS
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This question could be worded better.allenh wrote:DS:
Number XYZ is the product of positive integer n and 9. Is X+Y+Z=9?
(1) x+y+z<15
(2) x+y+z>8
I believe that XYZ is supposed to represent some 3-digit number, where X is the hundreds digit, Y represents the tens digit and Z represents the units digit.
If this is the case, then the solution is based on the fact that, if a number is divisible by 9, the SUM of its digits will be divisible by 9
For example, we know that 13005 is divisible by 9, because 1+3+0+0+5 = 9, and 9 is divisible by 9.
Likewise, 88551 is divisible by 9, because 8+8+5+5+1 = 27, and 27 is divisible by 9.
The question tells us that XYZ = 9n, which means XYZ is divisible by 9.
So, the sum of its digits is divisible by 9. Also, the GREATEST sum of the 3 digits is 27
So, x + y + x, must equal 9 or 18 or 27
Now onto the question.............
Target question: Is x + y + z = 9?
Statement 1: x + y + z < 15
We already concluded that x + y + z must equal 9, 18 or 27
If x + y + z < 15, then x + y + z must equal 9
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: x + y + z > 8
We already concluded that x + y + z must equal 9, 18 or 27
If x + y + z > 8, then x + y + z COULD equal 9 or 18 or 27
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT
Answer = A
Cheers,
Brent
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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 equations ensures a solution.
Number XYZ is the product of positive integer n and 9. Is X+Y+Z=9?
(1) x+y+z<15
(2) x+y+z>8
the more accurate question would be,
If 3-digit positve integer xyz where x is the hundreds digit of n, y is the tens digit, and z is the units digit of n is divisible by 9, is x+y+z=9?
1) x+y+x<15
2) x+y+z>8
==> the remainder from dividing a positive integer n to 3 or 9 would be the same as dividing the addition of all 3 digit numbers by 3 or 9. Transforming the original condition and the question by variable appraoch method, xyz is divisible by 9 는 x+y+z is divisible by 9 means x+y+z=9,18,27. (Each digit can't be higher than 9, thus 27 is the maximum value)
1) x+y+z=9 sufificent because the answer is unique
2) x+y+z=9,18,27 NOT sufficient because the answer is not unique.
Therefore the answer is A.
If you know our own innovative logics to find the answer, you don't need to actually solve the problem.
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Number XYZ is the product of positive integer n and 9. Is X+Y+Z=9?
(1) x+y+z<15
(2) x+y+z>8
the more accurate question would be,
If 3-digit positve integer xyz where x is the hundreds digit of n, y is the tens digit, and z is the units digit of n is divisible by 9, is x+y+z=9?
1) x+y+x<15
2) x+y+z>8
==> the remainder from dividing a positive integer n to 3 or 9 would be the same as dividing the addition of all 3 digit numbers by 3 or 9. Transforming the original condition and the question by variable appraoch method, xyz is divisible by 9 는 x+y+z is divisible by 9 means x+y+z=9,18,27. (Each digit can't be higher than 9, thus 27 is the maximum value)
1) x+y+z=9 sufificent because the answer is unique
2) x+y+z=9,18,27 NOT sufficient because the answer is not unique.
Therefore the answer is A.
If you know our own innovative logics to find the answer, you don't need to actually solve the problem.
www.mathrevolution.com
- The one-and-only World's First Variable Approach for DS and IVY Approach for PS that allow anyone to easily solve GMAT math questions.
- The easy-to-use solutions. Math skills are totally irrelevant. Forget conventional ways of solving math questions.
- The most effective time management for GMAT math to date allowing you to solve 37 questions with 10 minutes to spare
- Hitting a score of 45 is very easy and points and 49-51 is also doable.
- Unlimited Access to over 120 free video lessons at https://www.mathrevolution.com/gmat/lesson
- Our advertising video at https://www.youtube.com/watch?v=R_Fki3_2vO8