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# M09 Q16

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Intern
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19 Nov 2008, 07:07
2
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If $$X$$ and $$Y$$ are positive integers, is $$\frac{10^X + Y}{3}$$ an integer?

1. $$X \gt 5$$
2. $$Y = 2$$

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

soln:
------
S1 is not sufficient. Whatever is, we can vary to make either divisible or not divisible by 3.

S2 is sufficient. The key thing to notice is that if S2 holds, then the sum of digits of is divisible by 3, as is the expression itself.

---
how is it B ? From q stem (10X + Y)/3 ==> (10X+2)/3 ==> can be 12/3, 22/3,32/3.....
taking values for X as 1,2,3....

am I missing anything ??
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19 Nov 2008, 07:56
2
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How cant it be B?

Clear E
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19 Nov 2008, 08:06
2
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Very Much SUre .. E ..

There cant be two answers when you give different values. Jus try plugging in values.
You can make the answer integer and as well as fraction
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09 Dec 2008, 10:43
1
KUDOS
Thanks guys. +1 to everyone. We'll edit the question.
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18 Dec 2008, 18:18
In my M09 test, I see that part in the question as "10 to the power of X", not "10X". I guess mine was edited already.
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19 Dec 2008, 12:47
If it's in the power of X then both can give the answer alone as 10, 100, 1000, 10000 or whatever + 2 gives an integer.
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19 Dec 2008, 13:13
If it's in the power of X then both can give the answer alone as 10, 100, 1000, 10000 or whatever + 2 gives an integer.

For S1, Y is unknown, so it could be 10^6 + 2, which is divisible or 10^6 +3, which is not. Because there is no Y value for S1, it is not sufficient.
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19 Dec 2008, 14:10
Yeah, I mixed it up. Gee mistakes like that and I'm done!
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30 Aug 2010, 05:26
I am not getting it. Whats the official answer then? B?
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30 Aug 2010, 06:50
mbaobsessed wrote:
If $$X$$ and $$Y$$ are positive integers, is $$\frac{10^X + Y}{3}$$ an integer?

1. $$X \gt 5$$
2. $$Y = 2$$

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

soln:
------
S1 is not sufficient. Whatever is, we can vary to make either divisible or not divisible by 3.

S2 is sufficient. The key thing to notice is that if S2 holds, then the sum of digits of is divisible by 3, as is the expression itself.

---
how is it B ? From q stem (10X + Y)/3 ==> (10X+2)/3 ==> can be 12/3, 22/3,32/3.....
taking values for X as 1,2,3....

am I missing anything ??

The Q can be easily understood by splitting the numerator into two parts.

Part-1: (10^X)/3
Irrespective of the value of X (since its given that both X & Y are +ve integers we can neglect the negative values), the fraction will always yield a remainder of 1.

Part-2: (Y/3)
Based on the different values of Y, we can arrive at different values for the Remainder.
If S2 is taken into consideration, i.e. Y=2, the above fraction will yield a remainder of 2.

By adding the remainders from Part-1 and Part-2, we get a remainder of 3 which in other words means that there is no remainder. So the entire statement is divisible by 3 and will yield an integer as the answer.

The answer would be Statement 2 alone is sufficient to answer the Q.
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30 Aug 2010, 07:03
I am not getting it. Whats the official answer then? B?

B, exactly: Y=2 is sufficient.

to determine if (10^x+y)/3 is an integer, one just has to add all the digits of the numerator. if the numerator is a multiple of 3, then it can be divided by 3 with no remainder.

since 10^x is 10, 100, 1000, ... 10000000.... like this, 1 (on the left) adding y (y=2) equals 3 (1+2=3), (10^x+y)/3 is an integer.
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30 Aug 2010, 07:21
Hi guys,
I have a different approach, using rule of divisibility:
Question is: {(10^X)+y}/3 an integer.

For this expression to be an integer {(10^X)+y} needs to be divisible by 3 i.e sum of all digits of the expression should be divisible by 3.

X>5 => {(10^X)+y} = 1000000+y or = 10000000+Y and so on ...... the sum depends on value of y. If it is 2 , 5 or 8 then the expression {(10^X)+y} is divisible by 3 and the given expression is an integer but if it is not any of those three digits then the expression is not integer. Hence A is not sufficient.

y= 2 - sufficient, since whatever the value of x is the total of all the digits of 10^X is 1, and (10^X)+y will therefore always have digits whose sum is 3. Hence {(10^X)+y} is always divisible by 3 and {(10^X)+y}/3 will always give an integer value.

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30 Aug 2010, 08:50
Ans is clearly B
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30 Aug 2010, 09:39
1
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Ans B

Important thing to not here is the hidden constraint positive numbers, if it was not the case then both statements would have been required.

Using hidden constraint X is positive we know that 10^x will be at least 10 , further remainder on division by 3, will always be 1.

Applying divisibility rule for 3 + remainder sum , we know that numerator will be divisible by 3 if sum of remainders is multiple of 3, since we know the remainder of 10^x (ie. 1) , we shall focus on the remainder of y when divided by 3.

Now we have enough information to crack the question , lets go to the statements

A. X >5 , since x is positive therefore it doesn't add any new information.
we already know the remainder on division <however we shall stay alert if x wasn't given to be positive , then this statement would have provided valuable information to solve the question>.

B. Y =2 , remainder on division by 3 = 2.
By remainder some theory , Combined remainder of 10^x+y = 1+2=3
Since the remainder sum is multiple of 3 , 10^X +Y is divisible by 3

Ans B sufficient alone
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30 Aug 2010, 10:08
mbaobsessed wrote:
If $$X$$ and $$Y$$ are positive integers, is $$\frac{10^X + Y}{3}$$ an integer?

1. $$X \gt 5$$
2. $$Y = 2$$

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

IMO B

A.
$$\frac{(X^6 + 5)}{3}$$ integer
$$\frac{(X^6 + 4)}{3}$$ not integer

B.
$$\frac{(X^n + 2)}{3}$$ integer
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30 Aug 2010, 20:58
Its B...

Statement 1: with out the value of Y, it is not possible to say whether the resultant is an integer or not
Statemnt 2: given the Y value, the resultant is an integer for any value of X
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30 Aug 2010, 21:24
yes it is perfectly B.Divisibility by three is independent of x.and it will have its sum of digits as 1 and if y=2SUM OF DIGITS IS 3.hence completely divisible.
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01 Sep 2010, 03:09
BTW, what does IMO mean ?
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04 Sep 2011, 23:19
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Easy One ..B because whatever power 10 ..it does not matter if Y=2, as sum of digits is divisible by 3.

e.g. 10+2 = 12- divisible , 102- divisible, 1002-divisible, 10002-divisible
Re: M09 Q16   [#permalink] 04 Sep 2011, 23:19
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# M09 Q16

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