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If x, y, and z are digits is (x + y + z) a multiple of 9?

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If x, y, and z are digits is (x + y + z) a multiple of 9? [#permalink]

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New post 17 Mar 2011, 17:04
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If x, y, and z are digits is (x + y + z) a multiple of 9?

(1) The two-digit number yz is a multiple of 9
(2) If xyz is a three-digit number, 3(xyz - 1) + 3 is a multiple of 9
[Reveal] Spoiler: OA
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Re: Is X + Y + Z divisible by 9? [#permalink]

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New post 17 Mar 2011, 18:37
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Well what you have to do is simplify the expression in (2) into 3(xyz) since the +3 and -3 cancel out.

Next step is you break this down into 300(x)+3(yz), so that you can relate this to (1).

If we take from (1) that (yz) is a multiple, then 3(yz) will be one too. For (2) to be true then, 300(x) has to be a multiple and so x will have to be either 3, 6, or 9.

Also (y+z) must equal 9 (18-> 1+8, 27-> 2+7, etc).

Since x can be either 3,6,or 9 obviously we dont know if the total will be 12,15, or 18, so we dont know if it is a multiple of 9.
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Re: Is X + Y + Z divisible by 9? [#permalink]

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New post 17 Mar 2011, 19:32
Thanks a lot, bostonrb.

I see that all 2-digit multiples of 9 has 2 digits that add up to 9. 0,9, 1,8, 27, etc.
Thanks for helping me with this question.
+1

bostonrb wrote:
Well what you have to do is simplify the expression in (2) into 3(xyz) since the +3 and -3 cancel out.

Next step is you break this down into 300(x)+3(yz), so that you can relate this to (1).

If we take from (1) that (yz) is a multiple, then 3(yz) will be one too. For (2) to be true then, 300(x) has to be a multiple and so x will have to be either 3, 6, or 9.

Also (y+z) must equal 9 (18-> 1+8, 27-> 2+7, etc).

Since x can be either 3,6,or 9 obviously we dont know if the total will be 12,15, or 18, so we dont know if it is a multiple of 9.
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Re: Is X + Y + Z divisible by 9? [#permalink]

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New post 17 Mar 2011, 20:08
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From (1), if yz is a multiple of 9, then y+z is also a multiple of 9, but no information about x, so insufficient.

From (2), 3(xyz-1)+3 = 9k (where k is an integer) => 3xyz -3 + 3 = 9k => xyz = 3k

so x+y+z is a multiple of 3, but it may or may not be a multiple of 9, so insufficient.

Now From (1) and (2), xyz may or may not be a multiple of 9

e.g x = 3, y = 6, z = 3, so xyz = 3*6*3, but 363 is not a multiple of 9, but 63 is.

or x = 9, y = 6, z = 3, so xyz = 9*6*3, which is a multilple of 9, and 63 is as well.

Answer - E
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Re: If x, y, and z are integers is (x+y+z) a multiple of 9? [#permalink]

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New post 30 Sep 2013, 11:47
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josemarioamaya wrote:
If x, y, and z are digits is (x+y+z) a multiple of 9?

1) the two digit number yz is a multiple of 9
2) If xyz is a three digit number, 3(xyz-1)+3 is a multiple of 9

I'm happy to help with this. :-)

First of all, if the sum of the digits of a number equals a multiple of 9, then the number itself is a multiple of 9. For example, 657 has a sum of digits 6 + 5 + 7 = 18, which is a multiple of 9, so this automatically means 657 is a multiple of 9. See
http://magoosh.com/gmat/2012/gmat-divis ... shortcuts/

Statement #1: the two digit number yz is a multiple of 9
This gives us zero information about x, so we cannot determine an answer. The three digit number xyz could be 645 or 945 --- the latter is divisible by 9, but the former isn't. This statement, alone and by itself, is insufficient.

Statement #2: If xyz is a three digit number, 3(xyz-1)+3 is a multiple of 9
First of all, 3(xyz-1)+3 = 3*xyz - 3 + 3 = 3*xyz. If 3*xyz is a multiple of 9, then all this means is that xyz must be a multiple of 3. Thus, either 645 or 945 would work. This statement, alone and by itself, is insufficient.

Combined statements:
Notice that the two numbers 645 and 945 each satisfy both statements, but the former is not a multiple of 9, and the latter is. Thus, even with both statements, we can still come up with different numbers that yield different answer to the prompt question. Thus, even together, the combined statements are insufficient.

OA = (E)

Does all this make sense?
Mike :-)
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Re: If x, y, and z are integers is (x+y+z) a multiple of 9? [#permalink]

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New post 30 Sep 2013, 12:58
Thanks mike, you make it look easy!!!
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Re: If x, y, and z are digits is (x + y + z) a multiple of 9? [#permalink]

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New post 05 Jan 2014, 14:01
Yalephd wrote:
If x, y, and z are digits is (x + y + z) a multiple of 9?

(1) The two-digit number yz is a multiple of 9
(2) If xyz is a three-digit number, 3(xyz - 1) + 3 is a multiple of 9


Here's what I did. Now we have that x,y,z are digits and we need to know if x+y+z is a multiple of 9.

First statement yz is a multiple of 9. Therefore the sum of y+z is in fact a multiple of 9, but we know nothing about 'x' so this is clearly insufficient.

Second statement, we have that 3(xyz-1) +3 is a multiple of 9. Now, 3's cancel out and we are left with 3xyz is a multiple of 9 which is the same as xyz is a multiple of 3, or that x+y+z is a multiple of 3. Now, x+y+z could be a mulitple of 9 too or not. Insufficient.

Finally, with both statements together so we have that y+z is a multiple of 9, let's assume for instance that y+z = 1+8= yz=18. Now, we also know that xyz is a multiple of 3, if we say that x=3 then 318 divisible by 3 but not divisible by 9 hence insufficient. If we say that x=6, same here 618 divisible by 3 but not divisible by 9. However, if we try 918, we have that xyz divisible by 3 and 9, hence we also have a YES answer.

Therefore, answer is E

Hope this helps
Cheers!
J :)
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Re: If x, y, and z are digits is (x + y + z) a multiple of 9? [#permalink]

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Re: If x, y, and z are digits is (x + y + z) a multiple of 9? [#permalink]

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New post 21 Aug 2016, 04:17
Taking x,y,z as digits
statement 1=> 000=> yes
100=> no
Not suff
Statement 2=> x=000=> yes
001=> no
Not suff
Combining them => 000=> yes
100=> No
Push that E
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Re: If x, y, and z are digits is (x + y + z) a multiple of 9?   [#permalink] 21 Aug 2016, 04:17
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