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Given that \(n = 10^a + 10^b + 10^c\), where a, b, and c are

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by swerve » Fri Aug 16, 2019 2:53 pm

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Source: Magoosh

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

The OA is C
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Source: — Problem Solving |

by GMATGuruNY » Sat Aug 17, 2019 7:06 am
swerve wrote:Source: Magoosh

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
Since 10� = 1,000,000,000 -- and the value of n must be LESS than 1,000,000,000 -- the 3 distinct positive exponents must all be less than 9.

Number of options for the first exponent = 8. (Any exponent between 1 and 8, inclusive.)
Number of options for the second exponent = 7. (Any exponent between 1 and 8, inclusive, other than the exponent already used.)
Number of options for the third exponent = 6. (Any exponent between 1 and 8, inclusive, other than two exponents already used.)
To combine these options, we multiply:
8*7*6

Since the order of the 3 exponents does not matter -- 10¹ + 10² + 10³ will yield the same value for n as 10³ + 10² + 10¹ -- we must divide by the number of ways the 3 exponents can be arranged (3!):
(8*7*6)/(3*2*1) = 56.

The correct answer is C.
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