2003 => 127 + 172 + 271 + 712 + 721
Also the uniq solutiions..
5106 == 689 + 698 + 869 + 896 + 968 + 986
5328 == 789 + 798 + 879 + 897 + 978 + 987
Interesting::
2001,2003,2005,2011,2014,2020 etc.. all are numbers
which are sum of numbers with the same digits in a
uniq way.
I wonder if there's a series for this type of
numbers..
A N Kadakia
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Barcelona, Catalonia, 10 d'abril de 2003
127 + 172 + 271 + 712 + 721 = 2003
Jordi Domènech
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127+172+271+712+721=2003
Zbigniew Zarzycki
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Hello!
2003 = 127 + 172 + 271 + 712 + 721
This wasn't very easy..
Juha Saukkola
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Erich Friedman: It is easy to express 2004 as the sum of distinct positive numbers with the same digits: 2004 = 725 + 752 + 527, 2004 = 617 + 671 + 716, 2004 = 509 + 590 + 905. It is hard to write 2003 as the sum of distinct positive numbers with the same digits. The answer appears to be unique.
For the above question.. there is one more answer..
that is 914+941+149 = 2004
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A quick analysis shows that 2 and 7 numbers would be too few or too many,
respectively, that 3 or 6 wouldn't work because of the property of nines
and because 2003 isn't a multiple of 3. The combinations with 4 numbers can
be breezed through pretty quickly, which leaves 5 numbers. Since all are
the same mod 9, and 5 ÷ 5 = 1 (mod 9), each has to be equivalent to 1 (mod
9). The sum of the digits, then, is most likely 10 (19 seems too high, and
1 is impossibly low). 2003 can be thought of as the six permutations of the
three digits added together, minus one of them. In these six, the sum
should be 2220 (since the sum of the digits in each place value is 20).
2220 - 2003 = 217. The other five must be:
127 + 172 + 271 + 712 + 721 = 2003
-Jonah Ostroff
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127+172+271+712+721=2003
James Lewis Melby
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Great problem! My answer is: 127 + 172 + 271 + 712 + 721 = 2003.
Peter Exterkate
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271 + 721 + 712 + 127 + 172
- Matt Elder
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