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What is the remainder when the positive integer x is divided [#permalink]
 08 Apr 2008, 13:35
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.
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is
sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
your reasoning pls
Last edited by notahug on 08 Apr 2008, 20:41, edited 1 time in total.
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Statement 1:
Examples, 5, 17, 29, 41, 53, 65, 77, 89, 101...
If divided by 8 they leave either 1 or 5 as remainders, so not sufficient.
Statement 2:
Examples, 11, 29, 47, 65, 83, 101...
If divided by 8 they leave different remainders, so not sufficient.
Combining Statement 1 and 2
Examples 29, 65, 101
Divided by 8 leaves remainder 1, and 5, so again insufficient.
Answer E.
Last edited by abhijit_sen on 08 Apr 2008, 21:27, edited 1 time in total.
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is there other way to approach this?
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i thought it was E ... 29/8 gives remainder 5, but 65/8 gives remainder 1.
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yeah..this should be E..
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notahug wrote: is there other way to approach this? I have the same problem as your. I do not master this skill of number plugging. I did tried but it did not work for me. The other way does not help me more! 1. 12k + 5 = (8k +8)+(4k-3) 4k-4 divided by 8 produce some remainders so not sufficient 2. 18n+11 = 16n + 8 + (2n+3) 2n+3 also produces some remainders when it divided by 8, not sufficient When combining, I stop thinking,  I apply the way Walker suggest but it seems difficult for me. Any other help?
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What is the best way to solve this, any more idea?
Walker?
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Re: Remainder -- Walker --Join us! [#permalink]
08 Apr 2008, 21:13
My approach for the problems with remainder is the same as abhijit_sen wrote: I simply write out a few integers: 12k+5: 5, 17, 29, 41, 53, 65 18m+11: 11, 29, 47, 6529 and 65 satisfies both conditions but have different remainders: 5 and 1 So, E I do not know faster way....
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Re: Remainder -- Walker --Join us! [#permalink]
08 Apr 2008, 21:31
walker wrote: My approach for the problems with remainder is the same as abhijit_sen wrote: I simply write out a few integers:
12k+5: 5, 17, 29, 41, 53, 65
18m+11: 11, 29, 47, 65
29 and 65 satisfies both conditions but have different remainders: 5 and 1
So, E
I do not know faster way.... Tks I think most of us consider "plugging" the best way.
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Re: Remainder -- Walker --Join us! [#permalink]
08 Apr 2008, 21:50
At the beginning I tried to use "mod" approach to solve such problems but found it too clumsy and sometimes this approach lead me to deadlock ... On the other hand, the plugging numbers seems to be not so nice but work well for almost all reminder problems.
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Re: Remainder -- Walker --Join us!
[#permalink]
08 Apr 2008, 21:50
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