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10^x + 10^y + 10^z = n, where x, y, and z are positive integers. Which

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10^x + 10^y + 10^z = n, where x, y, and z are positive integers. Which  [#permalink]

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New post 28 Oct 2018, 22:56
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D
E

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Question Stats:

28% (01:33) correct 73% (01:49) wrong based on 80 sessions

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\(10^x + 10^y + 10^z = n\), where x, y, and z are positive integers. Which of the following could be the total number of zeroes, to the left of the decimal point, contained in n?

I. x + y
II. y – z
III. z

A. III only
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III

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Re: 10^x + 10^y + 10^z = n, where x, y, and z are positive integers. Which  [#permalink]

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New post 29 Oct 2018, 00:02
Consider x as 1, y as 3, z as 2. Then n = 1110. the total number of zeros to the left of the decimal point is 1.

Statement 2 is possible as y-z = 3-2 = 1.

Since x,y and z are positive integers. min value of x+y is 2. But if x is 1 and y is 1 and z is 2. then n = 120. only 1 zero. cannot be possible.
Statement 1 is not possible.

If x is 3, y is 2 and z is 1. then n will contain 1 zero. Hence statement 3 is possible.

Only 2 and 3 are possible.

D is the answer.
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10^x + 10^y + 10^z = n, where x, y, and z are positive integers. Which  [#permalink]

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New post 30 Oct 2018, 12:14
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Afc0892 wrote:
Since x,y and z are positive integers. min value of x+y is 2. But if x is 1 and y is 1 and z is 2. then n = 120. only 1 zero. cannot be possible.
Statement 1 is not possible.



Statement 1 is also possible. Consider x = 1, y = 1, z = 3.

Then you have 10 + 10 + 1000 = 1020
1020 has 2 zeros to the left of the decimal point, i.e. in the units place and the hundred's place and hence Statement 1 is also possible.

I think the catch is "total" number of zeros. The question does not ask number of zeros until you reach a non-zero number starting from the units digit and going left.

Statement 2 is possible. Consider, x = 1, y = 3, z = 2, hence n = 1110, y-z = 1 is satisfied.

Statement 3 is possible. Consider, x = 3, y = 2, z = 1, hence n = 1110, z = 1 is satisfied.

Hence all three statements are possible.

Option E
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10^x + 10^y + 10^z = n, where x, y, and z are positive integers. Which &nbs [#permalink] 30 Oct 2018, 12:14
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