Today's Puzzle

Yeah, correct answer. Next one please!

Will that logic work for all 3x3 prime puzzles?

If a Eagles fart will blow a Blue Jay's feather 200 feet how far does a rat turd have to fall to break a out house shingle?
I never could figure this one out.

drpepper said:

Will that logic work for all 3x3 prime puzzles?

If sum of any 9 primes can be divided by 3 so it should work with any 3 * 3 table using primes only. Sum of the primes (all 9) divided by 3 gives you a sum for each row and column, then continue using the same logics.

For someone who wants to test, here the primes up to 5,003:

2 3 5 7 11 13 17 19 23 29
31 37 41 43 47 53 59 61 67 71
73 79 83 89 97 101 103 107 109 113
127 131 137 139 149 151 157 163 167 173
179 181 191 193 197 199 211 223 227 229
233 239 241 251 257 263 269 271 277 281
283 293 307 311 313 317 331 337 347 349
353 359 367 373 379 383 389 397 401 409
419 421 431 433 439 443 449 457 461 463
467 479 487 491 499 503 509 521 523 541
547 557 563 569 571 577 587 593 599 601
607 613 617 619 631 641 643 647 653 659
661 673 677 683 691 701 709 719 727 733
739 743 751 757 761 769 773 787 797 809
811 821 823 827 829 839 853 857 859 863
877 881 883 887 907 911 919 929 937 941
947 953 967 971 977 983 991 997 1009 1013
1019 1021 1031 1033 1039 1049 1051 1061 1063 1069
1087 1091 1093 1097 1103 1109 1117 1123 1129 1151
1153 1163 1171 1181 1187 1193 1201 1213 1217 1223
1229 1231 1237 1249 1259 1277 1279 1283 1289 1291
1297 1301 1303 1307 1319 1321 1327 1361 1367 1373
1381 1399 1409 1423 1427 1429 1433 1439 1447 1451
1453 1459 1471 1481 1483 1487 1489 1493 1499 1511
1523 1531 1543 1549 1553 1559 1567 1571 1579 1583
1597 1601 1607 1609 1613 1619 1621 1627 1637 1657
1663 1667 1669 1693 1697 1699 1709 1721 1723 1733
1741 1747 1753 1759 1777 1783 1787 1789 1801 1811
1823 1831 1847 1861 1867 1871 1873 1877 1879 1889
1901 1907 1913 1931 1933 1949 1951 1973 1979 1987
1993 1997 1999 2003 2011 2017 2027 2029 2039 2053
2063 2069 2081 2083 2087 2089 2099 2111 2113 2129
2131 2137 2141 2143 2153 2161 2179 2203 2207 2213
2221 2237 2239 2243 2251 2267 2269 2273 2281 2287
2293 2297 2309 2311 2333 2339 2341 2347 2351 2357
2371 2377 2381 2383 2389 2393 2399 2411 2417 2423
2437 2441 2447 2459 2467 2473 2477 2503 2521 2531
2539 2543 2549 2551 2557 2579 2591 2593 2609 2617
2621 2633 2647 2657 2659 2663 2671 2677 2683 2687
2689 2693 2699 2707 2711 2713 2719 2729 2731 2741
2749 2753 2767 2777 2789 2791 2797 2801 2803 2819
2833 2837 2843 2851 2857 2861 2879 2887 2897 2903
2909 2917 2927 2939 2953 2957 2963 2969 2971 2999
3001 3011 3019 3023 3037 3041 3049 3061 3067 3079
3083 3089 3109 3119 3121 3137 3163 3167 3169 3181
3187 3191 3203 3209 3217 3221 3229 3251 3253 3257
3259 3271 3299 3301 3307 3313 3319 3323 3329 3331
3343 3347 3359 3361 3371 3373 3389 3391 3407 3413
3433 3449 3457 3461 3463 3467 3469 3491 3499 3511
3517 3527 3529 3533 3539 3541 3547 3557 3559 3571
3581 3583 3593 3607 3613 3617 3623 3631 3637 3643
3659 3671 3673 3677 3691 3697 3701 3709 3719 3727
3733 3739 3761 3767 3769 3779 3793 3797 3803 3821
3823 3833 3847 3851 3853 3863 3877 3881 3889 3907
3911 3917 3919 3923 3929 3931 3943 3947 3967 3989
4001 4003 4007 4013 4019 4021 4027 4049 4051 4057
4073 4079 4091 4093 4099 4111 4127 4129 4133 4139
4153 4157 4159 4177 4201 4211 4217 4219 4229 4231
4241 4243 4253 4259 4261 4271 4273 4283 4289 4297
4327 4337 4339 4349 4357 4363 4373 4391 4397 4409
4421 4423 4441 4447 4451 4457 4463 4481 4483 4493
4507 4513 4517 4519 4523 4547 4549 4561 4567 4583
4591 4597 4603 4621 4637 4639 4643 4649 4651 4657
4663 4673 4679 4691 4703 4721 4723 4729 4733 4751
4759 4783 4787 4789 4793 4799 4801 4813 4817 4831
4861 4871 4877 4889 4903 4909 4919 4931 4933 4937
4943 4951 4957 4967 4969 4973 4987 4993 4999 5003

drpepper said:

solution:

Code:

 103    7   109
  79   73    67
  37  139    43

How did I arrive at the solution?
I don't know if this works in every case, but I used some logic as follows.

• given: 3,37,43,67,73,79,103,109,139
• the sum of the numbers is 657
• for the sum of each row or column to be equal, the sum has to be 657/3=219
• because there are 9 squares, I decided to divide the total by 9 and see what comes up: 657/9=73
• looking at the sequence, I found 73 to be the median --> why not try that in the center square?
• so 219-73=146 --> is there a pattern?
• the sums of the outer pair, second from outer pair, etc. all are 146
• this provides the following pairs: 7,139; 37,109; 43,103; 67,79
• because a center out perspective is working so far, let's examine using the two center pairs above (37,109; 43,103) for the diagonals
• placing the pairs in diagonal opposite corners provides the same sums as calculatd earlier for the rows and columns
• remaining squares are just a matter of simple math to provide a column or row sum of 219
• magically, all of these solutions used one of the remaining numbers given for the puzzle

I was posting my solution while you were providing the hint. I did not peak.

Nice one! I couldn't for the life of me get the diagonals sorted, I had the same logic about the lines totalling 219 but at that point, I found all of the sets of numbers that totalled 219 and worked them into a grid. I got the rows and columns sorted, it was just those bloody diagonals that threw me


109 103 7
43 37 139
67 79 73


Was as far as I got before I had to go. Now it seems obvious to get that 7 out of the corner if it's in the same row/column as the 109 and 103!

As far as I know. A sum of any three primes (excluding 2) can be divided by 3 so it should work with any 3 * 3 table using primes only.

Not quite :)

5+7+11=23
13+17+19=49

tom982 said:

As far as I know. A sum of any three primes (excluding 2) can be divided by 3 so it should work with any 3 * 3 table using primes only.

Not quite :)


5+7+11=2313+17+19=49

There goes my theory, also shows how to do testing wrong. I took randomly 5 sets of 9 primes and when the sum was always dividable with 3, I generalized and decided that it is a fact.

Really bad mistake from someone who has often posted "Do not post your thoughts as facts" .

Edited my post accordingly.

Easy mistake to make :) I've also found one that includes 2:


2+11+17=30

Confusing, this interest me. I had a typo in my original post, instead of

A sum of any three primes (excluding 2) can be divided by 3 so it should work with any 3 * 3 table using primes only.

... I meant

A sum of any nine primes (excluding 2) can be divided by 3 so it should work with any 3 * 3 table using primes only.

I think I am still partially right. Testing it all the time as we speak, all groups of 9 primes I test will give a sum that is dividable with 3. If so, the puzzle works with any 9 primes excluding 2, making it possible to place them so that all rows, column and both diagonal lines give the same sum as long as the sum of these 9 primes divided with 9 equals the 5th (middle) prime.

Confusing, really confusing...

3+5+7+11+13+17+19+23+41=139 :)