SOLUTION: I am thinking of a three digit number. It is an odd multiple of three, and the product of its digits is 24. It is larger than 15 to the 2nd power. What are all the numbers i could

Algebra ->  Customizable Word Problem Solvers  -> Numbers -> SOLUTION: I am thinking of a three digit number. It is an odd multiple of three, and the product of its digits is 24. It is larger than 15 to the 2nd power. What are all the numbers i could       Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 355346: I am thinking of a three digit number. It is an odd multiple of three, and the product of its digits is 24. It is larger than 15 to the 2nd power. What are all the numbers i could be thinking of?
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
The only way 3 digits can have product 24 is 

1x3x8 = 24
1x4x6 = 24
2x2x6 = 24
2x3x4 = 24

So the digits consist of 1,3,8 or 1,4,6, or 2,2,6, or 2,3,4 

To be divisible by 3 the sum of the digits must be divisible by 3.

1+3+8=12
1+4+6=11
2+2+6=10
2+3+4=9

Of those sums of digits, only 12 and 9 are divisible by 3. 
So we have ruled out all but integers whose digits consist of
1,3,8, and 2,3,4.

Since they must be odd they either must end in 1 or 3.

The only ones which can end in 1 are 381 and 831.

The others must end in 3. 

They must be greater than 152 which is 225. So the
first digit cannot be 1. So the only way its digits can consist of
1,3,8 and end in 3 is to be 813.

The rest must consist of the digits 2,3,4, and the only way they can 
end in 3 is to be 243 or 423.  

So there are exactly five such three-digit integers: 

381, 831, 813, 243, and 423.

Edwin