Question 368581: what is the largest multiple of 12 that can be written using each 0, 1,2,3,4,5,6,7,8,9 exactly once?
Found 2 solutions by jim_thompson5910, Edwin McCravy:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Since the digits add up to 45, which is divisible by 3, this means that if you construct a 9 digit number using all the digits, then that number will be divisible by 3.
So we only need to make sure that the number is also divisible by 4. So we only need to worry about the last two digits. So the last two digits could be 04, 08, 12, 16, 20, etc
It turns out that because 1 and 2 are small and close together, this minimizes the change if we were to swap them.
Naturally, the number 987654321 is the largest possible number using all the given digits. However, it is not divisible by 4. If we switch 1 and 2, we get
987654312
which is now divisible by 4
So because 987654312 is both divisible by 3 and 4, this means that 987654312 is divisible by 12. Also, this is the largest multiple of 12.
Answer by Edwin McCravy(20065)  (Show Source):
You can put this solution on YOUR website!
The other tutor's answer is incorrect because it does not contain a 0.
Here's my reasoning:
The largest possible integer using all ten digits once each is 9876543210
But it's not divisible by 12 because dividing that by 12 gives 823045267.5
However that tells us that the whole part of that, 823045267, times 12, is
the largest multiple of 12 less than 9876543210
823045267×12 = 9876543204
Unfortunately that does not use all 10 ten digits once. However the first
8 digits are all different.
So the problem is at least with the last two digits.
However neither 9876543210 nor 9876543201 is a multiple of 12
So the problem is at least with the last three digits
Let's check 9876543000 to see if it's divisible by 12
Yes it is because 9876543000/12 = 823045250
Now if we add any multiple of 12 to 9876543000, we will get
a multiple of 12. Can we think of a multiple of 12 that
uses the remaining digits 0, 1, and 2. Indeed we can! For
120 is a multiple of 12 and it uses those three digits. No
other arrangement of those three digits is a multiple of 12.
So we add 120 to 9876543000 and get 9876543120.
That's the largest integer that uses all the digits and is also
a multiple of 12.
Answer: 9876543120
Edwin
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