How many four-digit numbers formed only of odd digits are divisible by five?

Complete step-by-step answer:
The number should be formed with a number of digits is 4.
It should be noted that the four digits odd number that we have to form should be less than 10000 (the last 4 digit number) and greater than 1000 ( the last 3 digit number)
Now, the four digit number consists of one’s, ten’s, hundreds and thousands digits, so the possible one’s digit will be 1, 3, 5, 7 and 9 because we have to form the odd number, so 1, 3, 5, 7and 9 are odd numbers it means one of the total 5 numbers can be placed at one’s digit.
Then the thousand’s place can be one of the 1, 2, 3, 4, 5, 6, 7, 8, and 9, but one of them is already occupied by one’s digit, then only one of the 8 numbers is possible to place in the thousand's number.
And the hundred’s also the same, 8 possible numbers, and coming to the ten’s place, we will have only 7 possible numbers to place in it.
Then the number could be \[8 \times 8 \times 7 \times 5 = 2240\]
So the number of four-digit odd numbers can be formed is 2240. It means option (A) is correct.

Note: Here, we have to know the definition of the odd number and the probability. We have to come through how a number can be formed with the help of the probability, while concluding the digit in one’s place, we have to take only odd numbers.

Quite appropriately, the word hippopotomonstrosesquipedalianism is used to describe words that are enormously long. In how many ways can its letters be arranged?

In the $33$ letters of the word:

• there are 5 occurrences of o
• p, i and s both occur 4 times each, and
• t, m, n, e, and a all occur 2 times each

resulting in $\displaystyle{\frac{33!}{5! (4!)^3 2^5}}$ possible ways to arrange the letters.

For the curious, the above expression evaluates to $163576434454494269015808000000$.

R
> counts = c(1,4,4,5,2,2,2,4,1,2,1,1,1,2,1)
          #  h i p o t m n s r e q u d a l
> factorial(sum(counts))/prod(factorial(counts))

If repetition is allowed, then every number ending with either 0 or 5 is divisible by five.  There are 9000 values and 2 out of 10 (or 1 out of 5) of them are divisible by 5, so 9000/5 = 1800 are divisible by 5 -- if repetition is allowed.

However, if repetition of digits is not allowed, we have:

   4th digit must be either 0 or 5  --  2 choices

   1st digit may be (1,2,3,4,5,6,7,8,9) -- 9 choices

   2nd digit may be (0,1,2,3,4,5,6,7,8,9) -- 10 choices, but not same as 1st digit or 4th digit

   3rd digit may be (0,1,2,3,4,5,6,7,8,9) -- 10 choices, but not same as 1st or 2nd digit or 4th digit

So, since the 1st digit cannot be 0 but can be 5, there are:

  If the last digit is 0:

   9 * 8 * 7 * 1 = 504 if the 4th digit is 0, the 1st digit can be (1,2,3,4,5,6,7,8,9), the 2nd digit can be (1,2,3,4,5,6,7,8,9 but not same as 1st digit) and the 3rd digit can be (1,2,3,4,5,6,7,8,9 but not same as 1st digit or 2nd digit).

  If the last digit is 0:

   8 * 8 * 7 * 1 = 448 if the 4th digit is 5, the 1st digit can be (1,2,3,4,6,7,8,9), the 2nd digit can be (1,2,3,4,6,7,8,9 but not same as 1st digit) and the 3rd digit can be (1,2,3,4,6,7,8,9 but not same as 1st digit or 2nd digit).

That gives a total of 504+448 = 952 numbers that are divisible by 5 with no repeating digits.

Elyse R.

Good Day, sir, I think in the 2nd digit of the equation that equates to 448, we can use 0 already. That's why there are still 8 chances, and 7 chances on the 3rd digit. Is it correct?

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10/09/20

Rajeev S.

In the eqn =448, the first digit would belong to {1,2,3,...,9} - {5}, so now a number from this set and '5' would be taken away. for the second digit, the other 8 numbers would be left, and 7 for the 3rd digit

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12/23/20

David W.

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count = 0 FOR a = 1 TO 9 FOR b = 0 to 9 IF (a<>b) THEN FOR c = 0 TO 9 IF ( (a<>b) AND (a<>c) AND (b<>c) ) THEN FOR d = 0 TO 5 BY 5 IF ( (d<>a) AND (d<>b) AND (d<>c) ) THEN count = count + 1 ENDIF ENDFOR ENDIF ENDFOR ENDIF ENDFOR ENDFOR OUTPUT x&" done." END 952 done."

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12/23/20

Kenneth S. answered • 05/18/16

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