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

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How many odd 4-digit positive integers that are multiples of 5 can be [#permalink]

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New post 09 Mar 2016, 11:46
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A
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E

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Re: How many odd 4-digit positive integers that are multiples of 5 can be [#permalink]

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New post 09 Mar 2016, 19:56
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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

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Re: How many odd 4-digit positive integers that are multiples of 5 can be [#permalink]

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New post 12 Mar 2016, 18:40
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.

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Re: How many odd 4-digit positive integers that are multiples of 5 can be [#permalink]

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New post 13 Mar 2016, 05:39
Bunuel wrote:
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


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

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Re: How many odd 4-digit positive integers that are multiples of 5 can be [#permalink]

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New post 14 Sep 2016, 10:18
Bunuel wrote:
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


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.

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Re: How many odd 4-digit positive integers that are multiples of 5 can be [#permalink]

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New post 20 Nov 2017, 11:20
Bunuel wrote:
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


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
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Re: How many odd 4-digit positive integers that are multiples of 5 can be   [#permalink] 20 Nov 2017, 11:20
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How many odd 4-digit positive integers that are multiples of 5 can be

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