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by talaangoshtari » Wed May 20, 2015 9:26 am
How many 5 digit numbers that are divisible by 9 can be formed using the digits 0, 1, 2, 4, 5, 6 if repeats are not allowed?

A.66
B.120
C.360
D.488
E.720
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by GMATGuruNY » Wed May 20, 2015 11:36 am
talaangoshtari wrote:How many 5 digit numbers that are divisible by 9 can be formed using the digits 0, 1, 2, 4, 5, 6 if repeats are not allowed?

A.66
B.120
C.360
D.488
E.720
For an integer to be a multiple of 9, the SUM OF ITS DIGITS must be a multiple of 9.
From {0, 1, 2, 4, 5, 6}, only one 5-digit combination will yield a sum that is a multiple of 9:
1+2+4+5+6 = 18.
Thus, the 5-digit integer must be composed of the digits 1, 2, 4, 5 and 6.
The number of ways to arrange these 5 digits = 5! = 120.

The correct answer is B.
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