- 19 Sep '09 06:16 / 2 editsIt is possible to place positive integers into the twenty-one vacant squaress of the 5x5 square shown below so that the numbers in each row and column form an arithmetic sequence.

*.............. *.............. *.............. *.............. *

*.............. 74............ *.............. *.............. *

*.............. *.............. *.............. *.............. 186

*.............. *.............. 103.......... *.............. *

0.............. *.............. *.............. *.............. *

The periods are only there so it lines up nicely. I hope you guys get what I'm asking

replace each asterix with a number and explain how you did it. - 19 Sep '09 15:06

There are 3 numbers adjacent to 0. call them a,b,c:*Originally posted by clandarkfire***It is possible to place positive integers into the twenty-one vacant squaress of the 5x5 square shown below so that the numbers in each row and column form an arithmetic sequence.**

*.............. *.............. *.............. *.............. *

*.............. 74............ *.............. *.............. *

*.............. *.............. *...... ...[text shortened]... you guys get what I'm asking

replace each asterix with a number and explain how you did it.

b c

0 a

they determine all the other numbers, for example

... ... ... ... ...

... ... ... ... ...

2b 2c-a ... ... ...

b c 2c-b 3c-2b 4c-3b

0 a 2a 3a 4a

you get 3 equations with 3 unknowns. - 19 Sep '09 19:37

in other words, how did you use mod 12 arithmetics to solve this problem?*Originally posted by geepamoogle***052...082...112...142...172**

039...074...109...144...179

026...066...106...146...186

013...058...103...148...193

000...050...100...150...200

The key was mod 12 arithmetic as far as it would go, trying out relatively few possibilities.

(Trial and error probably would have worked too.) - 20 Sep '09 02:42Well, I started by finding the remainder of each of the given numbers when divided by 12, then deducing what the possibilities were for reamining slots.

A good starting place was that the numbers 3 up and 3 over had to be divisible by 3, and the ones 4 up and 4 over divisible by 4. I'd have to work it out again to remember where I went from there, but the general idea was to eliminate the remainders which were impossible.

Eventually, I was able to determine that the number above the zero was of the form 12x + 1.

In examining the problem later, I have found a more useful pattern which can be used alongside multiple equation algebra to answer this sort of question.

Be it noted here, that the above solution is the only solution possible, even if you allowed negative numbers, fractions, and real numbers.

So now, here is a slightly modified form.

????...????...????...????...????

????...074....????...????...????

????...????...????...????...186

????...????...103...????...????

040...????...????...????...???? - 01 Oct '09 22:32 / 1 edit

Since each sequence is sequential in nature, and since I only have to deal with at most a difference of 4 terms in any sequence, I only need a number divisible by 1, 2, 3 and 4. 12 is the LCM. Alternately, one could track modulo 3 and modulo 4, since it has the same overall result. Apologies if this approach doesn't make any sense. I explain it poorly, although it makes perfect sense to me.*Originally posted by TheMaster37***Why 12?**

However, examining the problem afterwards resulted in a discovery of a much more direct and efficient way of solving it.

Examine the differences in each sequence in the grid and you should notice something interesting which proves very useful when coupled with multiple equation solving techniques.