What is the probability that a randomly selected three-digit number has the property that one digit is equal to the product of the other two? Express your answer as a common fraction.

15 answers

the product number (the third digit) could be
1x1, 1x2, 1x3 , ... 1x9 ----> 9 of them
e.g. 111, 122, 133, ... 199
each of the last 8 can be arranged in 3!/2! or 3 ways
so far we have 1 + 24 or 25 such numbers
could be
2x2, 2x3, 2x4,
e.g. 224 236 248
arrange 224 in 3 ways
arrange 236 in 6 ways
arrange 248 in 6 ways for 15 more

could be
3x3 --- 339 --> arranged in 3 ways for 3 more

could have a 0 as one of the factors
e.g. 100, 200, 300, ... 900 --->9 more
(we can't have a zero at the front)
so we have a total of 25+15+3+9 = 49

hope I didn't miss any
so prob of your event = 49/999

yer both wrong, it's
72/135

its 13/225

The answer is 13/225

Reiny:
25+15+3+9 = 52, not 49

52/900 = 13/225
13/225 is correct

13/225

yep. It's 13/225.

there are only 889 3 digit numbers

there is 900 as of 999-100=899 899+1 = 900

13/225

Yes, that is correct.

you homework! how could u