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# Hardest Math Problems in ACT! Can you solve it?

+2
1049
3
+444

This is a very hard problem and I do not think ANYONE will solve it, but that's ok!

Mr.Owl  Oct 26, 2017
#1
+7339
+7

From  100 - 199, These are the numbers containing a zero:

100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 120, 130, 140, 150, 160, 170, 180, 190

There are  19  of them.

From 100 - 199, there are 19 .

From 200 - 299, there are 19 .

From 300 - 399, there are 19 .   etc.

From 100 - 999, there are  19 * 9  numbers containing zero.

And there are 900 numbers from 100 - 999, including 100 and 999.

So the probability  =  $$\frac{19*9}{900}\,=\,\frac{171}{900}$$       ???

hectictar  Oct 26, 2017
#3
+444
+3

Wow, we got some smart people on here! This math problem was not mine but instead from a math program called ACT.

Great job hectictar! You got it right!

Mr.Owl  Oct 26, 2017
#2
+2295
+3

This appears to be correct, so Mr.Owl has met his/her match.

TheXSquaredFactor  Oct 26, 2017