Question 58446: I'm having trouble with a word problem, not from a book and don't understand how to find an equation for this one:
Use the digits 0,1,2,3,4,5,6,7,8,9 and form a 10 digit number. It must be divisible by every number starting at 1 through 18, including 18. Each digit must be used only one time.
Found 3 solutions by venugopalramana, jenrobrody, joyofmath:
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! I'm having trouble with a word problem, not from a book and don't understand how to find an equation for this one:
Use the digits 0,1,2,3,4,5,6,7,8,9 and form a 10 digit number. It must be divisible by every number starting at 1 through 18, including 18. Each digit must be used only one time.
FIRST FIND LCM TO GET THE LEAST MULTIPLE BY NUMBERS 1 TO 18
2|1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18
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2| 1,3,2,5,3,7,4,9,5,11,6,13,7,15,8,17,9
---------------------------------------------------
2| 1,3,1,5,3,7,2,9,5,11,3,13,7,15,4,17,9
---------------------------------------------------
3| 1,3,1,5,3,7,1,9,5,11,3,13,7,15,2,17,9
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3| 1,1,1,5,1,7,1,3,5,11,1,13,7,5,2,17,3
---------------------------------------------------
5| 1,1,1,5,1,7,1,1,5,11,1,13,7,5,2,17,1
-------------------------------------------------------
| 1,1,1,1,1,7,1,1,1,11,1,13,7,1,2,17,1
SO LCM = =2*2*2*3*3*5*7*11*13*17 = 6126120
OBVIOUSLY THIS DOES NOT SATISFY OUR REQUIREMENT.
SO TRY MULTIPLYING THS NUMBER WITH 2,3,4..ETC..TILL YOU GET THE ANSWER
THIS IS THE METHOD I CAN SUGGEST TO YOU.
Answer by jenrobrody(19)  (Show Source):
You can put this solution on YOUR website! There is not really an easy equation for this one, however we use a trial-and-error approach:
The 10 digit number needs to be divisible by numbers 1-18:
All numbers are divisible by one.
Even numbers are divisible by two.
If the sum of all the digits are divisible by 3, then the number is.(If we use the digits 0,1,2,3,4,5,6,7,8,and 9 in any order, their sum is 0+1+2+3+...=45 divisible by three, so any number with those ten digits is divisible by three.)
If the last two digits of a number is divisible by 4, the number is.
If the number ends in 5 or 0, the number is divisible by 5.
If the number is divisible by 2 and 3, then the number is divisible by 6.
We are going to just have to check divisiblility by 7 the hard way.
If the last three digits of a number is divisible by 8, the number is.
If the sum of all the digits are divisible by 9, then the number is. (Again, the digits 0,1,2,3,4,5,6,7,8,and 9 in any order will give a ten digit number that is divisible by 9)
If the last digit ends in 0(even and divisble by 5), then the number is divisible by 10.
We are going to just have to check divisiblility by 11,13,and 17 the hard way.
If the number is divisible by 4 and 3, then the number is divisible by 12.
If the number is divisible by 7 and 2, then the number is divisible by 14.
If the number is divisible by 5 and 3, then the number is divisible by 15.
If the last 4 digits of the number are divisible by 16, then the number is.
If the number is divisible by 9 and 2, then the number is divisible by 18.
In summary:
To be divisible by numbers 1 through 18 a number needs:
1. to end in zero (divisible by 2,5,and 10)
2. to be divisible by 16. (divisible by 2,4,8,16)
3. to use all ten digits 0 through 9. (divisible by 3,9)
----the first three combine to give a number divisible by 1,2,3,4,5,6,8,9,10,12,15,16,18
4. to be divisible by 7,11,13,17
A number divisible by 7,11,13,16,and 17 has all of those factors, so we start with 7x11x13x16x17=272272.
This number does not have 10 digits, so we count by 272272s:
544544,816816,....., until we have a ten digit number that ends in 0....
im trying to find it for you...
Answer by joyofmath(189) (Show Source):
You can put this solution on YOUR website! This problem is tough. I used a computer to help!
The least common multiple of 1*2*3...*18 = 
So, the answer is a multiple of 12252240.
I wrote a computer program to find all multiples of 12252240 that had 10 unique digits. The numbers that came back were 2438195760 3785942160 4753869120
and 4876391520.
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