"How many 5 digit numbers can be formed using the digits 0,2,3,4 and 5, when repitition is allowed such that the number formed is divisible by 2 or 5 or both? Can you describe the method used?
by 
Trixie
ANSWERS: 3
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Any number ending in 0, 2, 4, or 5 is divisible by either 2 or 5. So, with five digits, you can pick any of the 5 for the second through 4th digit, and only have 4 options for the first and last digit (you can't start with a zero, and you can't end on a 3) - so: 4*5*5*5*4 = 2000.
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I am sure it is 4*5*5*5*4=2000, but I have seen it in ONE of the GRE samples and it says 1500. Comments?
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Five digit number (cannot start with zero) so our first choices for the first digit are: 2,3,4,and 5 = 4 choice*something Second digit choices: 0,2,3,4,and 5 or 4*5*something Third digit: 0,2,3,4,and 5 = 4*5*5*something Fourth digit: 0,2,3,4,and 5 = 4*5*5*5*something Fifth digit: (remember that 2,3,& 4 are not divisible by 5 and 3&5 are not divisible by 2) the only number that works for both is 0 which means we only have one choice for our fifth digit = 4*5*5*5*1 which equal ***500*** The numbers in our made equation equal the number of choices we have to fill that spot or "digit". Because a five digit number can not start with "0" we only have four choices (0,2,3,4,and 5). Thus we have 4*something and so on and so on. Hope that is clear enough to help.
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