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If X, Y and Z are digits, is (X + Y + Z) a multiple of 9? [#permalink]
 17 Mar 2011, 17:04
Question Stats:
50% (01:56) correct
50% (01:30) wrong based on 6 sessions
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 don't understand the OE for this one. Please explain.
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Re: Is X + Y + Z divisible by 9? [#permalink]
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]
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: Is X + Y + Z divisible by 9? [#permalink]
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]
17 Mar 2011, 19:32
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