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Suppose x is a positive even number, all of whose digits are either 3

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Suppose x is a positive even number, all of whose digits are either 3  [#permalink]

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New post 30 Jan 2015, 07:21
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Suppose x is a positive even number, all of whose digits are either 3 or 4. If x is also divisible by both 3 and 4, how many digits are in the smallest possible value of x?

A. Three
B. Four
C. Five
D. Six
E. Seven

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Re: Suppose x is a positive even number, all of whose digits are either 3  [#permalink]

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New post 30 Jan 2015, 09:52
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Hello Bunuel, I gave it a try:

CORRECT ANSWER A. Three digits

Since the number is positive and even and contains only 3 and 4 as digits, then its numbers digit is 4 and is fixed.

As there are not many posibilites for x, I found the most simple method to just start plug in numbers starting from the first and least value of x, which is answer key A. three

If x would have 3 digits, then we have 4 posibilites (since 4 is always the number digit for an even number):

334: is divisible by 4 but not by 3
344: is divisible by 4 but not by 3
434: is not divisible by 4 nor by 3
444: is divisible by 4 AND by 3 - then CORRECT ANSWER A. Three digits

Bunuel wrote:
Suppose x is a positive even number, all of whose digits are either 3 or 4. If x is also divisible by both 3 and 4, how many digits are in the smallest possible value of x?

A. Three
B. Four
C. Five
D. Six
E. Seven


Kudos for a correct solution.

The OA will be revealed on Sunday

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Re: Suppose x is a positive even number, all of whose digits are either 3  [#permalink]

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New post 30 Jan 2015, 09:59
ans A...
as per given terms of reference , to be even last digit has to be 4..
to be div by 4 , last two digits have to div by 4 so it becomes 44..
finally to be div by 3, sum shud be div by 3.. so it becomes 444...
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Re: Suppose x is a positive even number, all of whose digits are either 3  [#permalink]

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New post 31 Jan 2015, 02:33
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last two digits of a multiple of 4 must be divisible by 4. thus the unit and ten digit must be 44. Moreover the sum of all digits must be a multiple of 3. 444 fits perfectly

Answer A
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Re: Suppose x is a positive even number, all of whose digits are either 3  [#permalink]

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New post 02 Feb 2015, 04:23
Bunuel wrote:
Suppose x is a positive even number, all of whose digits are either 3 or 4. If x is also divisible by both 3 and 4, how many digits are in the smallest possible value of x?

A. Three
B. Four
C. Five
D. Six
E. Seven


Kudos for a correct solution.

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VERITAS PREP OFFICIAL SOLUTION:

Since x is even, its units digit must be 4. Since x divides by 4, its last two digits must divide by 4: 34 doesn't work, so the number must end in 44. Since x divides by 4, the sum of its digits must divide by 3. 344 doesn't work, but 444 does ... so the LEAST possible value of x is 444, which has three digits.
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Re: Suppose x is a positive even number, all of whose digits are either 3  [#permalink]

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New post 10 Feb 2015, 14:11
Since it is an even number it has to end in 44..Any such number will be divisible by 4. Now it is divisible by both 3 and 4 and the lcm of 3 and 4 is 12..so the number's digits must add up to 12 or a multiple or it. In our case we need the smallest such number so 12=4+4+4.444 is divisible by 3 as well.

Ans A.


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Re: Suppose x is a positive even number, all of whose digits are either 3  [#permalink]

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New post 10 Feb 2015, 19:37
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Test to check divisibility by 4 >> Last 2 digits MUST be divisible by 4 >> xxxxxxx04, xxxxxxxx36, xxxxxxx44

Test to check divisibility by 3 >> Addition of all digits should be divisible by 3 >> 444

Answer = A
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Re: Suppose x is a positive even number, all of whose digits are either 3  [#permalink]

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New post 09 Oct 2015, 22:17
Bunuel wrote:
Suppose x is a positive even number, all of whose digits are either 3 or 4. If x is also divisible by both 3 and 4, how many digits are in the smallest possible value of x?

A. Three
B. Four
C. Five
D. Six
E. Seven


Kudos for a correct solution.

The OA will be revealed on Sunday


Since x is even so Unit digit of x must be 4 as it can't be 3

Since x is Divisible by 4 so Last two digits of x must be divisible by 4 as Last two digits can't be 34 so last two digits of x must be 44

Since x must be divisible by 3 so sum of the digits of x must be divisible by 3 so x may be 444 whose sum of digits is 12 i.e. Divisible by 3

i.e. Smallest x = 444

Answer: option A
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Re: Suppose x is a positive even number, all of whose digits are either 3  [#permalink]

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New post 11 Oct 2015, 13:05
Hi All,

This question is packed with Number Property and logic shortcuts that you can take advantage of. I'm going to give you some hints so that you can try this question again:

1) Each of the digits of X are either a 3 or a 4.
2) Notice how the prompt is NOT asking for the actual value of X; it's asking for the number of digits in the shortest possible answer.
3) To be divisible by 3, the sum of the digits of X must be divisible by 3.
4) To be divisible by 4, the number that is created by the last two digits of X must be divisible by 4.

Since the question asks for the X that has least digits, you can take advantage of the answers. We'll start with Answer A - can you find a 3-digit number that fits all of the above facts? If you can, then you're done. If you can't, then you should look for a 4-digit number that fits all of the above facts.

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Re: Suppose x is a positive even number, all of whose digits are either 3  [#permalink]

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New post 14 Oct 2015, 02:03
x is even, so 4 in the end, but only 4 is not divisible by 3. Cannot be 34 (not divisible by 3 and 4) or 44 (not divisible by 3). Cannot be 344 (not divisible by 3) but can be 444 (divisible by 3 and 4)

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Re: Suppose x is a positive even number, all of whose digits are either 3  [#permalink]

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New post 30 Oct 2015, 21:31
Bunuel,

I arrived at 444 fairly quickly, but other similar problems' solutions have said something to the effect of "because '3' is noted in the question, there must be at least one '3' in the answer." As a result, I quickly added on a 3,444 to the number, yielding 4 digits and an answer of (B). Any guidance as to what the real GMAT will have in store for us? Obviously wording of the question will play an important role, but if that's the case, perhaps this one should be worded a little more clearly.
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Re: Suppose x is a positive even number, all of whose digits are either 3  [#permalink]

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New post 31 Oct 2015, 14:39
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Hi iwantstanford,

Since you've made a reference to "similar questions' solutions", can you list any of those questions here?

Ultimately each Quant question is a set of 'rules/restrictions/facts' that you have to work with. This prompt tells us that each of the digits in a number is either 3 or 4 - but that does NOT mean that at least one 3 and at least one 4 will appear in the correct answer (because the prompt did NOT provide THAT 'restriction'). IF that extra restriction were in place, then the answer 3444 would have been the smallest number that would fit all of the rules.

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Re: Suppose x is a positive even number, all of whose digits are either 3  [#permalink]

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New post 13 Sep 2017, 08:10
Let us disect the problem and proceed step by step.

Even number -> units digit is 4
Divisible by 4 -> tens digit is also 4, because 34 is not divisible by 4
Divisible by 3-> Hundreds digit should also be 4 as 344 is not divisible by 3.

So the number is 444, a three digit number.
Re: Suppose x is a positive even number, all of whose digits are either 3 &nbs [#permalink] 13 Sep 2017, 08:10
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