How many odd 4-digit positive integers that are multiples of 5 can be

Question

How many odd 4-digit positive integers that are multiples of 5 can be formed without using the digit 3?

A. 648 
B. 729
C. 900
D. 1296
E. 3240

Divyadisha · Answer

I would choose 'A' for it.

Since the number is odd number and multiple of 5, we can fix digit '5' as the 4th digit.
Remaining 1st digit can be any 8 numbers other than '0' and '3', 2nd digit can be and 9 numbers other than '3' and 3rd digit can be any '9' numbers other than '3', making possibilities to be- 8X9X9X1= 648

Thanks

lucalimoncelli · Answer

I choose A.

Possible n° of 1st digit: 8 (0 can't be the first number, or else it wouldn't have 4 digits. 3 is exlcuded)
Possible n° of 2nd digit: 9 (3 is excluded)
Possible n° of 3rd digit: 9 (3 is excluded)
Possible n° of 4th digit: 1 (a number is a multiple of 5 if it ends in 5 or 0, here we are asked for the odd numbers, hence the last digit can't be 0)

So, 8*9*9*1=648 (A)

I hope my reasoning is correct.

sowragu · Answer

Start with most restrictive digit.

So there is only one possibility for unit digit.

8*9*9*1

here by finding the unit digit itself we can chose the correct answer. No need to do complete multiplication.

IMO A

Ferrarif1286 · Answer

A.

__ __ __ 5

There are 8 numbers for thousands place (can't be 3 or 0), 9 for hundreds place (can't be 3), and 9 for tens place (can't be 3). Unit digit has to be 5 or 0 in order for the number to be multiple of 5 but since it is given that the 4-digit integer is odd, the only unit digit is 5. Therefore, 8x9x9x1=648.

ScottTargetTestPrep · Answer

We need to determine the number of odd 4-digit positive integers that are multiples of 5 and can be formed without using the digit 3. Thus, we know that the last digit is 5.

So, we have 8 options for the first digit (because the first digit can’t be 0 or 3), 9 for the second (can’t be 3), 9 for the third (can’t be 3), and one for the fourth (must be 5). Thus, we have 8 x 9 x 9 x 1 = 648 options.

Answer: A

exc4libur · Answer

4-digit: ABCD
D: odd integers multiple of 5 must end in {5} = 1 option
A: first digit cannot be 0 or 3 {1,2,4,5,6,7,8,9} = 8 options
B: second digit cannot be 3 {0,1,2,4,5,6,7,8,9} = 9 options
C: third digit cannot be 3 {0,1,2,4,5,6,7,8,9} = 9 options

total: 1*8*9*9=648

Answer (A)

TheNightKing · Answer

__ __ ___ ___

The last (unit) digit has to be four for the integer to be 5 because it has to be odd and multiple of 5. So that is only 1 way.

The left most digit (cannot be 0 and cannot be 3) so it can take 8 possible values.

The middle two digits can only not take 3 which means 9 possible values for both of them.

8*9*9*1=648

Option A

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How many odd 4-digit positive integers that are multiples of 5 can be

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