Question

How many 5 digit numbers divisible by 9 can be formed from the digits 0,1,2,4,6,8 & 9 such that each digit occurs at most once in any number?

Please also provide explanation.
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Answer 1

Answer:

61290,62190,86490,84690,69840

Step-by-step explanation:

9*7760=69840

9*9410=84690

9*9610=86490

9*6910=62190

9*6810=61290

Answer 2

Answer and explanation:

The given digits are 0, 1, 2, 4, 6, 8 and 9.

Number of digits = 7

But we have to form 5 digit number.

So, according to combinations, we have

Number of ways in which 5 digit number can be formed = ${}_7 C_5$

                                                                                       = (7×6)/(2×1)

                                                                                      = 42/2

                                                                                      = 21 ways

Out of these, we will select those numbers which are divisible by 9.

According to divisibility rule for 9, if a number is divisible by 9 then the sum of its digits should also be divisible by 9.

Now, 0 + 1 + 2 + 4 + 6 + 8 + 9 = 30

But 30 is not divisible by 9. So, we will choose only those 5 digits whose sum is 27.

We will exclude any two digits whose sum is 3.

∴ We will exclude 2 and 1 from the digits. Now we have only 5 digits left i.e., 0, 4, 6, 8, 9

⇒ 0, 4, 6, 8, 9 (1 and 2 excluded)

Sum = 0 + 4 + 6 + 8 + 9 = 27 which is divisible by 9.

These can be arranged in 5! - 4! ways (cause 0 in the beginning) = 4!(5-1)

                                                                         = 4×3×2×1×4

                                                                        = 96 ways.

Hence, the number of 5 digit numbers divisible by 9 is 96.

96 five-digit numbers can be formed.

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