If you moved the 3 at the end of a certain large number to the beginning, it would be the same as multiplying it by 3. What is the number?
First, the first digit must be 1. Any number starting with 3, divided by 3, will yield a number that starts with a 1.
Therefore, a good place to start would be to write out the multiplication, using ... to indicate the unknown digits in both the product and the multiplicand.
1...3
<U>× 3</U>
31...
So, the last digit of the product must be 9 (3×3):
1...93
<U>× 3</U>
<U> 9</U>
31...9
The next to last digit must be 7 (3×9 is 27):
1...793
<U>× 3</U>
9
<U> 27 </U>
31...79
The third-to-last digit is 3 (the 2 carried over from 27, plus the last digit of 7×3=21):
1...3793
<U>× 3</U>
9
27
<U> 21 </U>
31...379
Continuing thus:
1...13793
<U>× 3</U>
9
27
21
<U> 9 </U>
31...1379
1...413793
<U>× 3</U>
9
27
21
9
<U> 3 </U>
31...41379
1...413793
<U>× 3</U>
9
27
21
9
3 </U>
31...41379
1...413793
<U>× 3</U>
9
27
21
9
3 </U>
31...41379
1...2413793
<U>× 3</U>
9
27
21
9
3
<U> 12 </U>
31...241379
1...72413793
<U>× 3</U>
9
27
21
9
3
12
<U> 6 </U>
31...7241379
1...172413793
<U>× 3</U>
9
27
21
9
3
12
6
<U> 21 </U>
31...17241379
1...5172413793
<U>× 3</U>
9
27
21
9
3
12
6
21
<U> 3 </U>
31...517241379
1...55172413793
<U>× 3</U>
9
27
21
9
3
12
6
21
3
<U> 15 </U>
31...5517241379
1...655172413793
<U>× 3</U>
9
27
21
9
3
12
6
21
3
15
<U> 15 </U>
31...65517241379
1...9655172413793
<U>× 3</U>
9
27
21
9
3
12
6
21
3
15
15
<U> 18 </U>
31...965517241379
1...89655172413793
<U>× 3</U>
9
27
21
9
3
12
6
21
3
15
15
18
<U> 27 </U>
31...8965517241379
1...689655172413793
<U>× 3</U>
9
27
21
9
3
12
6
21
3
15
15
18
27
<U> 24 </U>
31...68965517241379
1...0689655172413793
<U>× 3</U>
9
27
21
9
3
12
6
21
3
15
15
18
27
24
<U> 18 </U>
31...068965517241379
1...20689655172413793
<U>× 3</U>
9
27
21
9
3
12
6
21
3
15
15
18
27
24
18
<U> 0 </U>
31...2068965517241379
1...620689655172413793
<U>× 3</U>
9
27
21
9
3
12
6
21
3
15
15
18
27
24
18
0
<U> 6 </U>
31...62068965517241379
1...8620689655172413793
<U>× 3</U>
9
27
21
9
3
12
6
21
3
15
15
18
27
24
18
0
6
<U> 18 </U>
31...862068965517241379
1...58620689655172413793
<U>× 3</U>
9
27
21
9
3
12
6
21
3
15
15
18
27
24
18
0
6
18
<U> 24 </U>
31...5862068965517241379
1...758620689655172413793
<U>× 3</U>
9
27
21
9
3
12
6
21
3
15
15
18
27
24
18
0
6
18
24
<U> 15 </U>
31...75862068965517241379
1...2758620689655172413793
<U>× 3</U>
9
27
21
9
3
12
6
21
3
15
15
18
27
24
18
0
6
18
24
15
<U> 21 </U>
31...275862068965517241379
1...82758620689655172413793
<U>× 3</U>
9
27
21
9
3
12
6
21
3
15
15
18
27
24
18
0
6
18
24
15
21
<U> 6 </U>
31...8275862068965517241379
1...482758620689655172413793
<U>× 3</U>
9
27
21
9
3
12
6
21
3
15
15
18
27
24
18
0
6
18
24
15
21
6
<U> 24 </U>
31...48275862068965517241379
1...4482758620689655172413793
<U>× 3</U>
9
27
21
9
3
12
6
21
3
15
15
18
27
24
18
0
6
18
24
15
21
6
24
<U> 12 </U>
31...448275862068965517241379
1...34482758620689655172413793
<U>× 3</U>
9
27
21
9
3
12
6
21
3
15
15
18
27
24
18
0
6
18
24
15
21
6
24
12
<U> 12 </U>
31...3448275862068965517241379
1...034482758620689655172413793
<U>× 3</U>
9
27
21
9
3
12
6
21
3
15
15
18
27
24
18
0
6
18
24
15
21
6
24
12
12
<U> 9 </U>
31...03448275862068965517241379
1...1034482758620689655172413793
<U>× 3</U>
9
27
21
9
3
12
6
21
3
15
15
18
27
24
18
0
6
18
24
15
21
6
24
12
12
9
<U> 0 </U>
31...103448275862068965517241379
Finally, we have a digit 1 without a carry (the other ones were actually 11). This 1 can be used as the first digit, giving the final answer:
1034482758620689655172413793
<U>× 3</U>
9
27
21
9
3
12
6
21
3
15
15
18
27
24
18
0
6
18
24
15
21
6
24
12
12
9
0
<U>3 </U>
3103448275862068965517241379
Of course, that is just the smallest answer. If we didn't make that 1 be the first digit, and kept going, the whole process would just repeat itself. The next smallest number, then, would be 10344827586206896551724137931034482758620689655172413793, and so on.
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Posted by DJ
on 2004-03-01 11:52:19 |
Longhand...
