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Bunuel
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6 - digits

9 x 10 x 10 x 10 x 10 x 10 <---- 9 options for digit 1 (since 0 cannot be the first) AND 10 options for each of the other (5) digits

7 - digits

9 x 10 x 10 x 10 x 10 x 10 x 10 <---- 9 options for digit 1 (since 0 cannot be the first) AND 10 options for each of the other (6) digits

(9 x 10^6) - (9 x 10^5)
= (9 x 10^5) (10^1 - 1)
= 9 x 9 x 10^5
= 81 x 10^5

Answer is D.
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Bunuel
A city's telephone numbers are made up of 6 digits. Knowing that the first digit can never be zero, if the telephone numbers change to 7 digits, how many more telephone numbers could be created?

(A) 81 x 10^3
(B) 90 x 10^3
(C) 81 x 10^4
(D) 81 x 10^5
(E) 90 x 10^5
The problem can be solved in two ways:
A. If only 6 digits, including a '0', are used
B. All digits are used.

B seems to be the case here, however.

A. 5 digits can take the 1st place and each of the rest 5 places can be taken in 6 ways.
Ways a telephone number can be formed = \(5*6^5\)

In case of a 7 digit number, total ways are \(5*6^6\)
Difference = \(5*6^6 - 5*6^5 = 5^2*6^5\)

B. 9 digits can take the 1st place and each of the rest 5 places can be taken in 10 ways.
Ways a telephone number can be formed = \(9*10^5\)

In case of a 7 digit number, total ways are \(9*10^6\)
Difference = \(9*10^6 - 9*10^5 = 9^2*10^5\)

Answer D.
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Hello from the GMAT Club BumpBot!

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