# algebra 1

posted by .

Given any random, 3-digit number, what is the
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 -

when 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 -

number 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 -

Take Reiny's analysis. He paid a bit more attention to the details than did I.

• algebra 1 -

• algebra 1 -

"what is the probability" ...
cannot have 96 as the answer.

• algebra 1 -

ok thanks

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