Maths: Decimal Digits Conundrum

**surkdidat** · 27-08-2006, 08:30 PM

Arrange the numbers 1 2 3 4 5 6 7 8 & 9 into four groups so that each number is a square number.

250 points for the correct answer!!

Can you do it in 3 groups?

500 points for the correct answer!!

And how about rearranging all nine numbers into one nine-digit long number so you end up with a square number?

LOWEST CORRECT ANSWER wins 1000 tokens

**Benny** · 27-08-2006, 09:33 PM

9 81 324 576

361 529 784

139854276

**surkdidat** · 28-08-2006, 10:33 AM

One is correct Benny, answer one.

Unfortunately 2 and 3 are wrong as I simply asked CAN you do it, to which the answers are Yes!!!!

1750 points on their way, I take it you knew that one before, and did not sit there trying to work it out!!!!

**Benny** · 28-08-2006, 07:42 PM

No, I'd never seen that one before. I worked it out as follows:

Parts 1 and 2
Compute all the squares from 1 upwards, until they become 4 digits long (I assumed that they were 3 digits and less).
Remove all those that have repeated digits like 100.
This is your "start list"
Pick the first number in the start list and then remove all other numbers with common digits from the list.
Pick the second number in the remaining list and then remove all other numbers with common digits from the list.
Etc.
Whenever you get to a "dead end" situation where the task becomes imposible you backtrack your last choice. If you cannot backtrack (because there are no more options left) you know the first number you picked was wrong so remove it from the start list and begin picking numbers again.

It only took me about 5-10 minutes to get them both after generating the start list.

Part 3
It wasn't too hard to do this with the help of a little computer program and some basic maths knowledge to help reduce the search time.

The smallest possible 9-digit number is 123456789 so calculate it's square root and begin with the next whole number x.

Compute x * x and check to see if it has common digits. If so or if it doesn't contain all the digits, then increase x by 1 and start again.

Luckily, the answer was relatively low in the search range (i.e. starts with a 1 rather than a 9!), so it only took about 700 tries.

**surkdidat** · 28-08-2006, 08:23 PM

Fair play, good explanation in how to solve it!

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27-08-2006, 09:33 PM	#2 (permalink)
Benny Super Duper Poster Join Date: Mar 2004 Location: UK Posts: 604 Credits: 1,203 Thanks: 0 Thanked 0 Times in 0 Posts	9 81 324 576 361 529 784 139854276

28-08-2006, 07:42 PM	#4 (permalink)
Benny Super Duper Poster Join Date: Mar 2004 Location: UK Posts: 604 Credits: 1,203 Thanks: 0 Thanked 0 Times in 0 Posts	No, I'd never seen that one before. I worked it out as follows: Parts 1 and 2 Compute all the squares from 1 upwards, until they become 4 digits long (I assumed that they were 3 digits and less). Remove all those that have repeated digits like 100. This is your "start list" Pick the first number in the start list and then remove all other numbers with common digits from the list. Pick the second number in the remaining list and then remove all other numbers with common digits from the list. Etc. Whenever you get to a "dead end" situation where the task becomes imposible you backtrack your last choice. If you cannot backtrack (because there are no more options left) you know the first number you picked was wrong so remove it from the start list and begin picking numbers again. It only took me about 5-10 minutes to get them both after generating the start list. Part 3 It wasn't too hard to do this with the help of a little computer program and some basic maths knowledge to help reduce the search time. The smallest possible 9-digit number is 123456789 so calculate it's square root and begin with the next whole number x. Compute x * x and check to see if it has common digits. If so or if it doesn't contain all the digits, then increase x by 1 and start again. Luckily, the answer was relatively low in the search range (i.e. starts with a 1 rather than a 9!), so it only took about 700 tries.

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