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Please Help Solving this Word Problem!!!!!

 
 
Reply Tue 11 Oct, 2011 09:58 am
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 squared. What are all the numbers that I could be thinking of?
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fresco
 
  1  
Reply Tue 11 Oct, 2011 11:54 am
@Lindastanich,
The single digit factors of 24 involving odd numbers are 1x3x8, 2x3x4, 1x6x4
Of these, 1,3,8 and 2,3,4 add to a multiple of 3
So the possible solutions are all odd combinations of either bunch,
i.e. those not ending in an even number,
e.g 831, 423, etc...
And 15 squared =225, so chose all combinations above that.
(Since this only excludes one combination, it seems rather a poor question!)



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raprap
 
  1  
Reply Tue 11 Oct, 2011 12:54 pm
let a,b & cbe the three digits

100*a+10*b+c=3(2n+1)

where n is an integer

3(2n+1)>225

and

a*b*c=24

look at possibilities for c

3*1=3
3*3=9
3*5=15
3*7=21
3*9=27

24/3=8
24/9 no
24/5 no
24/1=24
24/7=no
so c=3 or c=1
so c=1 or c=3

c=3 a*b=8 (2*4=8) number is 243 or 423

423/3=141 243/3=81

c=1 possibilities 381, 461, 641 & 831

which are integers when divided by 3?

381/3=127 461/3=153 2/3 641/3=213 2/3 831/3=277

Possible solutions 423, 243, 381, 831

Rap
fresco
 
  1  
Reply Tue 11 Oct, 2011 01:27 pm
@raprap,
...and 813 Wink
raprap
 
  1  
Reply Tue 11 Oct, 2011 04:41 pm
@fresco,
Yes all permutations except 183

Rap
Lindastanich
 
  1  
Reply Wed 12 Oct, 2011 12:39 pm
@raprap,
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 squared.
What are the five numbers I could be thinking of?
;
The three digits whose product is 24
2 * 2 * 6 = 24; all even


2 * 3 * 4 = 24
1 * 3 * 8 = 24
:
Odd multiple of three, last digit has to be odd
Greater than 225 (15^2)
:
243, 423, 381, 831, 813 would be the 5 numbers



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Lindastanich
 
  1  
Reply Wed 12 Oct, 2011 12:40 pm
@Lindastanich,
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 squared.
What are the five numbers I could be thinking of?
;
The three digits whose product is 24
2 * 2 * 6 = 24; all even


2 * 3 * 4 = 24
1 * 3 * 8 = 24
:
Odd multiple of three, last digit has to be odd
Greater than 225 (15^2)
:
243, 423, 381, 831, 813 would be the 5 numbers


CORRECT???????
fresco
 
  1  
Reply Wed 12 Oct, 2011 01:12 pm
@Lindastanich,
Your 5 numbers are correct but you've missed the methodological step that the sum of the digits must be a multiple of 3. (Otherwise you could have gone for combinations of 1,4,6. Rap's trial and error method is not as economical)
0 Replies
 
 

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