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+1 vote

Suppose that $6$-digit numbers are formed using each of the digits $1, 2, 3, 7, 8, 9$ exactly once. The number of such $6$-digit numbers that are divisible by $6$ but not divisible by $9$ is equal to

- $120$
- $180$
- $240$
- $360$

+4 votes

Best answer

test of divisibility by 9 is: sum of digits is divisible by 9.

here sum of digits is 1 + 2+ 3+ 7+ 8+9 = 30

so all numbers formed by these digits exactly once are not divisible by 9.

But all numbers formed by these digits are divisible by 3.

so we have to see the number of numbers which are even.

there are totally 6! = 720 numbers possible with these digits.

Out of which 1/3rd are even(2,8 are even digits out of 6 available digits)

So the answer is 240.

C

here sum of digits is 1 + 2+ 3+ 7+ 8+9 = 30

so all numbers formed by these digits exactly once are not divisible by 9.

But all numbers formed by these digits are divisible by 3.

so we have to see the number of numbers which are even.

there are totally 6! = 720 numbers possible with these digits.

Out of which 1/3rd are even(2,8 are even digits out of 6 available digits)

So the answer is 240.

C

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