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x, a, z, and b are single digit positive integers. x = ¼ a.

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x, a, z, and b are single digit positive integers. x = ¼ a.  [#permalink]

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New post 12 Mar 2013, 08:36
4
8
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A
B
C
D
E

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x, a, z, and b are single digit positive integers. x = ¼ a. z = ¼ b. (10a + b) – (10x + z) could NOT equal

A. 33
B. 36
C. 44
D. 63
E. 66

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Re: x, a, z, and b are single digit positive integers. x = ¼ a.  [#permalink]

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New post 12 Mar 2013, 08:51
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emmak wrote:
x, a, z, and b are single digit positive integers. x = ¼ a. z = ¼ b. (10a + b) – (10x + z) could NOT equal
a) 33
b) 36
c) 44
d) 63
e) 66


Without doing any algebra, you can reason out that the first part of the subtraction will be exactly four times bigger than the second part of the subtraction. Since all of the numbers are integers, it means that a and be will be divisible by 4 (while x and z can be anything). Once you subtract 1/4 from a and b, you'll be left with 3/4 of what they were orignally, which must be divisible by 3. The answer will necessarily be divisible by 3, so it elmininates answer choices a, b, d and e. Answer choice C is the only impossible outcome.

This type of question can lead people down rabbit holes, trying to calculate every possible number, especially with a premise as tempting as "single digit positive integers". There can't be that many choices, right? Doing this the long way may work out well, but you can't beat conceptual understanding for giving you the right answer in the most efficient way.

Hope this helps!
-Ron
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Re: x, a, z, and b are single digit positive integers. x = ¼ a.  [#permalink]

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New post 12 Mar 2013, 09:16
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emmak wrote:
x, a, z, and b are single digit positive integers. x = ¼ a. z = ¼ b. (10a + b) – (10x + z) could NOT equal
a) 33
b) 36
c) 44
d) 63
e) 66


(10a+b)-(10x+z) = 10(a-x)+(b-z) = 10(4x-x)+(4z-z) = 3*10(x+z). Thus the sum has to be a multiple of 3.Only option C is not a multiple of 3.

C.
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Re: x, a, z, and b are single digit positive integers. x = ¼ a.  [#permalink]

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New post 12 Mar 2013, 09:47
1
emmak wrote:
x, a, z, and b are single digit positive integers. x = ¼ a. z = ¼ b. (10a + b) – (10x + z) could NOT equal
a) 33
b) 36
c) 44
d) 63
e) 66


From the given equations, a = 4x, b = 4z

(10a+b) - (10x+z) = 40x+4z - 10x-z = 30x +3z = 3(10x+z)

Therefore the resultant should be divisible by 3 which isn't possible only in the case of C.

Hence C is the answer
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Re: x, a, z, and b are single digit positive integers. x = ¼ a.  [#permalink]

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New post 12 Jan 2015, 11:26
1
another way, not as short as the methods mentioned above but also helpfull:

"x, a, z, and b are single digit positive integers. x = ¼ a. z = ¼ b. (10a + b) – (10x + z) could NOT equal"

since they are all single digit positive integers, x and z could be 1 or 2, a and b would become 4 or 8.

if you look at the answer choies you can see a pattern in the single digit's. if you evaluate the possibility for the single digit of (10a + b) – (10x + z), you get:

single digit: b-z
if z=1, b=4
if z=2, b=8.

so 4-1=3 or 8-2=6

I hope its understandable :)
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Re: x, a, z, and b are single digit positive integers. x = ¼ a.  [#permalink]

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New post 13 Jan 2015, 02:03
pranbo wrote:
emmak wrote:
x, a, z, and b are single digit positive integers. x = ¼ a. z = ¼ b. (10a + b) – (10x + z) could NOT equal
a) 33
b) 36
c) 44
d) 63
e) 66


From the given equations, a = 4x, b = 4z

(10a+b) - (10x+z) = 40x+4z - 10x-z = 30x +3z = 3(10x+z)

Therefore the resultant should be divisible by 3 which isn't possible only in the case of C.

Hence C is the answer


Did in the same way. Just like to add.... instead of 44, if the option had been 48 or some multiple of 3, then again all yielded values would had to be checked
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Re: x, a, z, and b are single digit positive integers. x = ¼ a.  [#permalink]

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New post 04 Feb 2016, 02:20
did by algebra use

a=4x, b=4z

10a+b-10x-z
10(a-x)+b-z, by substitution get 30x+3z => 3(10x+z). So, should be multiple of 3

Only C does not fit
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Re: x, a, z, and b are single digit positive integers. x = ¼ a.  [#permalink]

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New post 04 Feb 2016, 02:43
1
emmak wrote:
x, a, z, and b are single digit positive integers. x = ¼ a. z = ¼ b. (10a + b) – (10x + z) could NOT equal

A. 33
B. 36
C. 44
D. 63
E. 66



Hi,
almost everyone has done by considerimg non multiple of 3..
But if it doe snot strike in the stress of the exam, another method would be..

1) firstly we are not given that all are different integers..
2) since each is a positive single digit integer and x=a/4... a can take only 4 and 8 as value, and x can take 1 and 2..
3) same as 2) for b and z.

so in (10a + b) – (10x + z) ..

10a+b can be
44,48,84,88..

10x+z can be
11,12,21,22..

lets see teh choices
A. 33.. 44-11 possible
B. 36..48-12
C. 44.. no combination possible
D. 63..84-21
E. 66..88-22

ans C
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x, a, z, and b are single digit positive integers. x = ¼ a.  [#permalink]

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New post 08 May 2016, 07:20
Original Explanation:

Solution: C

Spot the pattern here.

All four variables must represent integers, so if x = ¼ a and z = ¼ b, then both a and b must be either 4 or 8.

Now try all four possibilities.

    If a = 4 and b = 4, then (10a+b) – (10x+z) = 44 – 11 = 33.
    If a = 4 and b = 8, then (10a+b) – (10x+z) = 48 – 12 = 36.
    If a = 8 and b = 4, then (10a+b) – (10x+z) = 84 – 21 = 63.
    If a = 8 and b = 8, then (10a+b) – (10x+z) = 88 – 22 = 66.

(C) is impossible.
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Re: x, a, z, and b are single digit positive integers. x = ¼ a.  [#permalink]

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Re: x, a, z, and b are single digit positive integers. x = ¼ a. &nbs [#permalink] 19 Aug 2017, 16:49
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