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Five digit combination problem

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by Yamna95 » Sat Apr 21, 2018 3:03 am
From what I can understand, if they are asking for numbers divisible by 2 or 5 or both, we have to first find out the combination using numbers divisible by both 2 and 5 at the end: 4*5*5*4 = 2000
But since this includes repetition of numbers divisible by either 2 or 5, we have to subtract combination of numbers divisible by 2 and those divisible by 5 from 2000:
Divisible by 5: 4*5*5*2= 200
Divisible by 2: 4*5*5*3=300

That's 2000 - (300+200) = 1500 i.e. answer choice D
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by GMATGuruNY » Sat Apr 21, 2018 3:43 am
Naruto wrote:In how many ways can five digit numbers be formed using digits 0, 2, 3, 4, 5 , when repetition is allowed such that the number formed is divisible by 2 or 5 or both.

A 100
B 150
C 3125
D 1500
E 125
A five-digit integer formed from the five given digit options will be divisible by 2 and/or 5 if its units digit is 0, 2, 4 or 5.
Thus:
Number of options for the units place = 4. (0, 2, 4, or 5).
Number of options for the ten-thousands place = 4. (Any of the 5 digits except 0.)
Number of options for the thousands place = 5. (Any of the 5 digits.)
Number of options for the hundreds place = 5. (Any of the 5 digits.)
Number of options for the tens place = 5. (Any of the 5 digits.)
To combine the options above, we multiply:
4*4*5*5*5 = 2000.

The correct value is not among the answer choices.
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