Posted by **Anonymous** on Friday, July 27, 2012 at 5:27pm.

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.

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**Reiny**, Friday, July 27, 2012 at 9:29pm
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

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**Anonymous**, Saturday, November 24, 2012 at 9:22am
yer both wrong, it's

72/135

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**a**, Saturday, January 26, 2013 at 1:06am
its 13/225

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**yes**, Monday, July 29, 2013 at 6:45am
The answer is 13/225

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**Correct**, Sunday, August 25, 2013 at 1:49am
Reiny:

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

52/900 = 13/225

13/225 is correct

- math -
**yea**, Sunday, October 19, 2014 at 11:38am
13/225

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