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January 30, 2015

January 30, 2015

Posted by **naomi** on Friday, January 18, 2013 at 11:17am.

probability that a number will be:

a) a multiple of 5

b) divisible by 2

c) a square number

i know the answer but i don't know how to solve it

- algebra 1 -
**Steve**, Friday, January 18, 2013 at 12:25pmwhen you divide a number by 5, there are only 5 possible remainders: 0,1,2,3,4. So, since multiples of 5 have zero remainder, there's a 1/5 chance that a random number will be a multiple of 5.

In fact, there's a 1/n chance that a number will be a multiple of n.

For the squares, you need to figure out how many perfect squares there are between 100 and 999. If there are, say, 35, then there's a 1/35 chance that a random 3-digit number will be a square.

- algebra 1 -
**Reiny**, Friday, January 18, 2013 at 12:32pmnumber of 3-digit numbers :

999-99 = 900

multiples of 5:

5, 10, 15, ... , 995 ---> 199 of them

prob(multiple of 5) = 199/900

divisible by 2 --- > must be even

numbers are :

100, 102, 104, ... 998 ----> 445 of them

prob(even) = 445/900 = 89/180

numbers which when squared produce a 3 digit number

10^2 = 100

11^2 = 121

..

31^2 = 961

32^2 = 1024 ... too large

so there are 22 of these 3-digit squares

prob(square) = 22/900

= 11/450

I hope you know how I am getting these counts.

e.g. how many even numbers from 100 to 998

consider them to be an arithmetic sequence

where a = 100 , d = 2 and t(n) = 998

a + (n-1)d = term(n)

100 + (n-1)(2) = 998

2(n-1) = 898

n-1 = 449

n = 449+1 = 450

- algebra 1 - go with Reiny -
**Steve**, Friday, January 18, 2013 at 12:51pmTake Reiny's analysis. He paid a bit more attention to the details than did I.

- algebra 1 -
**da**, Friday, January 18, 2013 at 4:28pmbut the answer is 96

- algebra 1 -
**Steve**, Friday, January 18, 2013 at 4:57pm"what is the probability" ...

cannot have 96 as the answer.

- algebra 1 -
**da**, Saturday, January 19, 2013 at 2:04pmok thanks

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