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Difficulty:
75%
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Question Stats:
56% (02:14) correct
44%
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wrong
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What is the remainder when the positive integer x is divided by 8?
(1) When x is divided by 12, the remainder is 5.
(2) When x is divided by 18, the remainder is 11.
(1) When x is divided by 12, the remainder is 5.
(2) When x is divided by 18, the remainder is 11.
When a number must satisfy two different divisible and remainder conditions, you could use what is known as "the Chinese remainder theorem" that uses modular arithmetic. Does anyone know how to apply that theorem to solve this problem?
Or how would you guys solve this in 2 min?
(I picked numbers
)
Or how would you guys solve this in 2 min?
(I picked numbers
)
20 Apr 2009, 04:16
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tenaman10
In many remainders questions, it's enough just to find a couple of numbers that 'work' with the given information, and if you simply list the first few numbers that satisfy each statement, it's easy to judge if the statements together are sufficient:
1) x = 5, 17, 29, 41, 53, 65, 77, 89 ...
2) x = 11, 29, 47, 65, 83, 101 ...
Since 29 and 65 give different remainders when you divide by 8, the answer is E.
More abstractly, when combining two statements like the above, the pattern will be based on the LCM of the two divisors. Here, we can consider dividing x by 36 = LCM(12, 18).
Notice that, if Statement 1 is true, the remainder will be 5, 17 or 29 when x is divided by 36.
If Statement 2 is true, the remainder will be 11 or 29 when x is divided by 36.
If both Statements are true, the remainder therefore must be 29 when x is divided by 36.
That's still not sufficient, as above; x could be 29, or x could be 65.
albany09
Let's look at the question:
What is the remainder when the positive integer x is divided by 8?
This means: What is leftover when you make groups of 8?
Statement 1: When x is divided by 12, the remainder is 5.
When you make groups of 12, 5 balls are leftover. When you make groups of 8 instead, each of the groups of 12 balls leaves 4 balls. If no. of groups of 12 is even, you can combine 2 groups of 4 balls each to make more groups of 8. In that case, 5 balls will be still leftover. So a remainder of 5 is possible.
If no. of groups of 12 is odd, 4 balls will be leftover from one group of 12 and 5 balls will be still leftover. So a total of 9 balls will be leftover. We can make another group of 8 out of these 9 balls and 1 ball will be leftover. So a remainder of 1 is also possible.
Since remainder can be 5 or 1, this statement alone is not sufficient.
Statement 2: When x is divided by 18, the remainder is 11.
When you make groups of 18, 11 balls are leftover. When you make groups of 8 instead, each of the groups of 18 balls makes 2 groups of 8 balls each and leaves 2 balls.
Now there are 4 possibilities:
1. We are left with 2 balls + the original 11 remaining balls = 13 balls
When you make another group of 8 from 13, remainder will be 5
2. We are left with 2+2 balls + the original 11 remaining balls = 15 balls
When you make another group of 8 from 15, remainder will be 7
3. We are left with 2+2+2 balls + the original 11 remaining balls = 17 balls
When you make 2 groups of 8 from 17, remainder will be 1
4. We are left with no groups of 2 balls since they all make a complete group of 8. Only the original 11 balls are remaining. When you make a group of 8 from 11, remainder will be 3.
Since remainder can be 5, 7, 1 or 3, this statement alone is not sufficient.
Using both statements, remainder can be either 5 or 1 so they both together are not sufficient.
Answer (E)
