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Friday puzzler -- make-up for last week

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Friday puzzler -- make-up for last week

#1

Friday puzzler -- make-up for last week

Alex Y

Sorry for going AWOL last week. Lee was all over that last one, so let's try a little harder one.

You need to find three 5-digit perfect squares such that:

1) only five digits are used among them (but with repeats and missing digits permitted in any one number)

2) each of the five digits is the count of the number of times another digit appear in the three numbers. None of the digits refers to its own count. I.e., if there are 3s and 7s in the answer, it is possible that "7" appears three times, but "3" does not appear three times.

3) if I told you which digit appeared only once, you would be able to deduce the answer.

But I won't tell you which digit appears only one time, and you can still deduce the answer!

Re: Friday puzzler -- make-up for last week

#2

Hints

Alex Y

A spreadsheet will let you quickly generate all 216 five-digit perfect squares.

From the first two properties given in the puzzle, you can derive another property that lets you quickly whittle the list of candidates down to nine. (Or if you find that property first, you could use the spreadsheet to calculate only 72 candidates, and quickly visually inspect them to find the nine possibilities.

Re: Friday puzzler -- make-up for last week

#3

Will start the discussion

Dan Donaldson

We know that the answers cannot contain a 0 because of the counts.

Re: Friday puzzler -- make-up for last week

#4

Good start

Alex Y

And while that seems obvious once you state it, it totally eluded me for the longest time as I solved this!

Re: Friday puzzler -- make-up for last week

#5

Re: Good start

Larry Barrett

And I believe the group of three perfect squares has to have at least one 1, because of clue 3, and it has to have more than one 1 because of clue 2. Furthermore, the last digit in each of the three 5-digit perfect squares must be a number from this set: {0,1,4,9,6,5} since every square ends in one of these (and Dan eliminated 0).

Re: Friday puzzler -- make-up for last week

#6

Another criteria

Dan Donaldson

We know 0 cannot be one of the digits, but we also know that the 5 digits must be 1,2,3,4,5 because each one represents the number of another and the sum of 1-5 is 15, which is the total number of digits in the three answers.

Re: Friday puzzler -- make-up for last week

#7

The 9

Dan Donaldson

Here are the 9 taking out all of the 0's and numbers greater than 5

35344

34225

12544

52441

33124

13225

21316

12321

14161

Re: Friday puzzler -- make-up for last week

#8

Re: The 9

Larry Barrett

Double check your list, but what you have includes a set of three that satisfy the criteria.

Re: Friday puzzler -- make-up for last week

#9

Re: The 9

Dan Donaldson

OOPS, that is what I get for doing stuff like this at 3 in the morning. ;)

Re: Friday puzzler -- make-up for last week

#10

Getting there!

Alex Y

No 0's and Sum of digits = 15 -> digits are 1,2,3,4,5 is the corollary of the first two conditions that let you narrow the candidates to 9.

Re: Friday puzzler -- make-up for last week

#11

Re: Getting there!

Larry Barrett

My answer has 3 1s, 5 2s, 4 3s, 2 4s, and 1 5. I did not check to see if there are other combinations.

Re: Friday puzzler -- make-up for last week

#12



Alex Y

Kudos to Dan and Larry for the cooperative effort to get this one!

The unique solution is indeed 12321, 33124, 34225.

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