### Brain Teasers

# Make 2003

It is easy to express 2004 as the sum of distinct positive numbers comprising the same digits:

2004 = 725 + 752 + 527 or

2004 = 617 + 671 + 716 or

2004 = 509 + 590 + 905.

It is harder to write 2003 as the sum of distinct positive numbers with the same digits.

What are they?

2004 = 725 + 752 + 527 or

2004 = 617 + 671 + 716 or

2004 = 509 + 590 + 905.

It is harder to write 2003 as the sum of distinct positive numbers with the same digits.

What are they?

### Hint

You will need five numbers of three digits each.### Answer

127 + 172 + 271 + 712 + 721 = 2003Hide Hint Show Hint Hide Answer Show Answer

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## Comments

Great puzzle! I am going to think about this. Why is 2003 different to 2004 in this respect? Anyone got any ideas?

I would think that any even numbered year would be easier than the year 2003, but that's just my opinion. Gr8 teaser!

It has little to do with odd or even, and a lot to do with "casting out nines".

If you divide a the sum of the digits of a number by nine and keep only the remainder, you are "casting out nines".

If you do any arithmetic operation on numbers and the same operation on their "cast out nines" remainder, you will get a result and the "cast out nines" remainder of that result.

For example, cast out nines from 617 or any number with the same digits, you get 6+1+7-9=5. Add three of those numbers, for example 617+671+716 and their cast out nines remainders 5+5+5

you get 2004 and 15. 2+4=1+6.

In the case of 2003, 2+3=5, so any numbers that add up to 2003 must have cast out nines remainders that add up to 5 plus any number of nines, for example 14, 23, 32, 41, 50.

But in this puzzler the numbers we add up have the same digits, so they have the same cast out nines remainder. 23 and 41 are prime numbers, so are out of the question, 14=2x7, but there are only 6 different three-digit numbers with the same digits, and two three-digit numbers or two four-digit numbers with the same digits cannot add up to 2003.

So 5 is our best bet.

We need 5 three-digit numbers with the same three digits whose cast out nines remainder is 1 each, i.e. the three digits have to add up to 10 or 19, but 19 would require numbers that are too big.

If we add up all 6 of the numbers with those 3 digits, they will appear twice each in the ones, the tens, and the hundreds places. So the sum of all six will be 222 times the sum of the digits, which we have already decided is 10. So all 6 numbers add up to 2220. But we must leave one out to get 2003. Which one? 2220-2003=217 of course.

The other five add up to 2003:

127+172+271+712+721.

This explanation is a lot longer than actually doing the operations.

If you divide a the sum of the digits of a number by nine and keep only the remainder, you are "casting out nines".

If you do any arithmetic operation on numbers and the same operation on their "cast out nines" remainder, you will get a result and the "cast out nines" remainder of that result.

For example, cast out nines from 617 or any number with the same digits, you get 6+1+7-9=5. Add three of those numbers, for example 617+671+716 and their cast out nines remainders 5+5+5

you get 2004 and 15. 2+4=1+6.

In the case of 2003, 2+3=5, so any numbers that add up to 2003 must have cast out nines remainders that add up to 5 plus any number of nines, for example 14, 23, 32, 41, 50.

But in this puzzler the numbers we add up have the same digits, so they have the same cast out nines remainder. 23 and 41 are prime numbers, so are out of the question, 14=2x7, but there are only 6 different three-digit numbers with the same digits, and two three-digit numbers or two four-digit numbers with the same digits cannot add up to 2003.

So 5 is our best bet.

We need 5 three-digit numbers with the same three digits whose cast out nines remainder is 1 each, i.e. the three digits have to add up to 10 or 19, but 19 would require numbers that are too big.

If we add up all 6 of the numbers with those 3 digits, they will appear twice each in the ones, the tens, and the hundreds places. So the sum of all six will be 222 times the sum of the digits, which we have already decided is 10. So all 6 numbers add up to 2220. But we must leave one out to get 2003. Which one? 2220-2003=217 of course.

The other five add up to 2003:

127+172+271+712+721.

This explanation is a lot longer than actually doing the operations.

The scip model that I used to solve this teaser can be found at https://pastebin.com/biGiCSPd

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