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Given that n = 10^a + 10^b + 10^c, where a, b, and c are distinct

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Given that n = 10^a + 10^b + 10^c, where a, b, and c are distinct  [#permalink]

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New post Updated on: 28 Nov 2017, 20:33
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Given that \(n = 10^a + 10^b + 10^c\), where a, b, and c are distinct positive integers, how many different positive values of n result if n is less than 1 billion (1,000,000,000) ?

A. 28
B. 36
C. 56
D. 84
E. 120

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Originally posted by Gnpth on 28 Nov 2017, 18:47.
Last edited by Bunuel on 28 Nov 2017, 20:33, edited 2 times in total.
Renamed the topic.
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Given that n = 10^a + 10^b + 10^c, where a, b, and c are distinct  [#permalink]

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New post 28 Nov 2017, 20:31
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Gnpth wrote:
Given that \(n = 10^a + 10^b + 10^c\), where a, b, and c are distinct positive integers, how many different positive values of n result if n is less than 1 billion (1,000,000,000) ?

A. 28
B. 36
C. 56
D. 84
E. 120



Hi,

The max value of a, b or c can be 8 to Keep the value \(< 10^9\)..
Any of the three a,b, or c being 9 means \(10^9+..+..>10^9\)

Also a,b,c are distinct, so the answer is nothing but choosing 3 out of 8..
here arrangement is not important, so work on COMBINATION
\(8C3=\frac{8!}{5!3!}=\frac{8*7*6}{3*2}=56\)
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Re: Given that n = 10^a + 10^b + 10^c, where a, b, and c are distinct  [#permalink]

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New post 24 Jan 2019, 13:15
1 billion = \(10^9\)

so the max. number could be \(10^8 + 10^7 + 10^6\) (a,b and c are distinct)

order doesn't matter so you could rearange the three numbers in 3! ways

full calculation\(\frac{8*7*6}{3!}\)\(= 56\)
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Re: Given that n = 10^a + 10^b + 10^c, where a, b, and c are distinct   [#permalink] 24 Jan 2019, 13:15
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