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If x is a positive integer, is the remainder 0 when (3x +

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If x is a positive integer, is the remainder 0 when (3x + [#permalink] New post 06 Jan 2009, 18:06
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If x is a positive integer, is the remainder 0 when (3x + 1)/10?
(1) x = 3n + 2, where n is a positive integer.
(2) x > 4
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Re: DS: Remainder [#permalink] New post 06 Jan 2009, 20:24
E.
1) 9n+7 is divisible by 10 when n, given as a positive integer, is any # that is ends with 7 (ex:n=7, 17,27,...). Otherwise it is not. INSUFF.
2) if x=6 NO, but x=13 then YES so INSUFF.
1+2) x=7 then Yes, but x=5 NO so INSUFF.
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Re: DS: Remainder [#permalink] New post 06 Jan 2009, 20:57
umb, for #1... how do you figure that 'n' has to be something that ends with '7'?
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Re: DS: Remainder [#permalink] New post 06 Jan 2009, 21:10
pmenon wrote:
If x is a positive integer, is the remainder 0 when (3x + 1)/10?
(1) x = 3n + 2, where n is a positive integer.
(2) x > 4



from 1

is 9n+7/10 = intiger.......insuff

from 2

x>4.....insuff

both

E
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Re: DS: Remainder [#permalink] New post 07 Jan 2009, 02:56
My choice is E.
1. x=3n+2, 3x+1 equivalent to 9n+7, n=1=> insuf
2. x>4, ex x=5 => 3x+1 is not divided by 10
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A difficult math, pls help me [#permalink] New post 03 Sep 2011, 04:36
If x is a positive integer, is the remainder 0 when (3x + 1)/10?

(1) x = 3n + 2, where n is a positive integer.
(2) x > 4

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.
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Re: DS: Remainder [#permalink] New post 03 Sep 2011, 07:06
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pmenon wrote:
If x is a positive integer, is the remainder 0 when (3x + 1)/10?
(1) x = 3n + 2, where n is a positive integer.
(2) x > 4


n=7; x=23; 3x + 1=70; is the remainder 0 when 70 is divided by 10? Yes.
n=5; x=17; 3x + 1=52; is the remainder 0 when 52 is divided by 10? No.
Not Sufficient.

Ans: "E"
****************************************

Arriving at "n=5" or any number that x may be above 4 that will lead to some remainder other than 0 is easy.

Question is: How do we take care of the condition to rule out whether the expression may be evenly divisible by 10. Here's what I've done:

We need to find out whether:
\frac{3x+1}{10}, upon division by 10, will lead to a 0 remainder.

Let's also consider the other condition from statement 1:
x = 3n + 2

Merging the above two,

\frac{3(3n + 2)+1}{10}=\frac{9n+7}{10}

What do we know now?
\frac{7}{10} always gives us a remainder of 7. So, that's constant.

If we somehow make \frac{9n}{10} to give us a remainder of 3, then earlier 7+3=10 will definitely give us a remainder of 0. Remainders of individual term is the remainder of the entire expression.

And how do we get a number like that?
What if the units digit of 9n=3; that'll solve our problem, will it not? All integers that end with 3 will give a remainder of 3 upon division by 10.

So, n's units digit must be 7, because 7*9=63; we get our 3 as the unit digit. In fact, any positive integer whose units digit is 7 will fit in well as n to make the entire expression give us 0 as remainder.

Put, n=7; find x; x>4. Conditions satisfy both statements and give us an affirmative answer.
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Re: DS: Remainder   [#permalink] 03 Sep 2011, 07:06
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