## The Properties of Numbers II by Piccadilly

### Puzzle solution process

This is one possible solution path. We’ll use square brackets to distinguish between the grid entry  and the number 1, etc.

 is both a square and a cube, so it’s a power of 6. The only 5-digit ones are 56 = 15625 and 66 = 46656, so  starts with 6 and  starts with 5 or 6. But  is prime (therefore odd) and the sum of its digits is even, so it must be one of {53, 59} and  = 15625 .

Now  starts with 6 and is prime, so the middle digit of  is one of {1, 3, 7, 9}. The reverse of  is prime and > , so  starts with {1, 3, 7, 9} and its last digit is ≥ 6. It’s also a multiple of . The three-digit multiples of 53 whose first two digits are in {1, 3, 7, 9} are {318, 371, 795}; but 813 isn’t a prime (its digit sum is 12, so it’s a multiple of 3) and 173 and 597 are both < . For 59, the candidate multiples are {118, 177}; but 771 isn’t a prime (its digit sum is 15, so it’s a multiple of 3) so  = 118 and  = 59 .

 has three digits and  =  +  has four, so  starts with 5 or higher. If  is in the range 590 to 599 then  is in the range 1180 to 1198; if  starts with 69 then  starts with 13, etc; generally, if the first digit of  is n then the first two digits of  are 2n + 1, ie, 1 followed by an odd digit.

As  ends in 9 and  ends in 1,  = × must end in 9, so the palindrome  is _9_9_, with the middle digit odd. Also, it’s a multiple of , so  = 59n for some integer n. If  ends (and starts) with 1, n must end in 9 and be between 19191/59 ≈ 325.3 and 19991/59 ≈ 338.8; the only suitable number in that range is 329, but 59×329 = 19411 isn’t a palindrome. If  ends with 2, n ends in 8 and is between 29192/59 ≈ 494.8 and 29992/59 ≈ 508.3, ie one of {498, 508}, but neither 59×498 = 29382 nor 59×508 = 29972 is a palindrome. Similarly, if  ends in 3, n is in {667, 677}, neither giving a palindrome, and so on. The only palindromes that result are {59295, 89798}, but the middle digit must be odd, so  = 89798 ,  is 17__ and  is 89_.

For  we have 6_1. The middle digit can’t be any of {2, 5, 8} because then the digit sum would be a multiple of 3 so  wouldn’t be prime. We can rule out 601 because that would make  end in 0 and therefore  would end in 0 and  would start with 0. Also, 671 may be recognised as a multiple of 11 (as the middle digit is the sum of the other two). For  in {611, 631, 641, 661, 691} the corresponding values for  = × are {36049, 37229, 37819, 38999, 40769}, of which the last is invalid because it would make  start with 0.  is a four-digit square starting with 5, so it’s one of {5041, 5184, 5329, 5476, 5625, 5776, 5929}. The maximum for  =  +  +  +  is then 5929 + 661 + 661 + 59 = 7310, so  starts with 6 or 7, ruling out 38999 for , so  is now one of {611, 631, 641}. If we have a prepared list of primes we could look up these values to see which are prime, or we could check whether they’re divisible by any primes less than √641 ≈ 25.3 (ie, {7, 11, 13, 17, 19, 23}); alternatively, we can continue as follows.

The last digits of  +  +  add up to 11, so  ends with the last digit of  + 1, ie one of {0, 2, 5, 6, 7} is the second digit of .  is a factor of , both with four digits, so the first digit of  is ≤ 4; but  is prime, so  starts with one of {1, 3}. The only squares starting with those combinations of first two digits are {1024, 1089, 1225, 1296, 1521, 1600, 1681, 1764, 3025, 3249, 3600, 3721}; but  is also square, so ’s third digit must be one of {0, 1, 4, 5, 6, 9}, narrowing it to {1296, 1600, 1764, 3249, 3600}. So  ends in one of {2, 6, 7} and  ends in one of {1, 5, 6} and is one of {5041, 5476, 5625, 5776}.

If  is 641 then  is 37819, and any of {5041, 5476, 5625} for  makes  start with 6, clashing with  in the grid; but if  is 5776 then  is 7117, a palindrome, which isn’t allowed; so  isn’t 641.

If  is 611 then  is 36049; if  is 5776 then  is 7057, clashing with ; if  is 5041 then  is 6322 and  =  +  is 12644, but that would make  end in 4, not a prime. If  is 5476 then  is 6757 and  is a four-digit cube with 5 as the second digit, but there’s no such cube. So  would be 5625, with  being 6906,  being one of {1000, 8000} (not 4096, which is a square) and  being one of {1600, 3600}. That gives _0_00 for , the maximum being 90900 with a digit sum of 18; but the digit sum has to be greater than that of  = 891, which is also 18. Therefore  isn’t 611, so  = 631 , which makes  = 37229 . The grid now has  = 893 , so  =  +  gives  = 1786 .

If  is any of {5041, 5476, 5625} then  starts with 6 and clashes with , so  = 5776 ,  = 7097 and  = 1764 . The second digit of  is 9, and the only matching cube is  = 4913 .  is now _2_16, for which the only matching square is  = 92416 .  =  + , so  = 14194 .

 is a four-digit multiple of  = 1764, starting with {1, 3, 7, 9} because  is prime, so it’s one of {3528, 7056}; but 7056 is a square, so  = 3528 .

 is a three-digit cube, one of {125, 216, 343, 512, 729}.  is square, so  can’t start with 2, 3, or 7, which leaves {125, 512}.  is 6_1 or 6_2 and its digits add up to 10, the same as for  = 118; it can’t be 631 because that’s already used at , so  = 622 and  = 125 .

 has two three-digit multiples,  and , so  ≤ 333.  is _41 and it’s < , ie one of {141, 241}. The first is disallowed because it’s a palindrome (and a multiple of 3), so  = 241 .  < , so it starts with 1 or 2.  is a palindromic prime with 7 in the middle; it can’t be 171 or 777 because they’re multiples of 3, so it’s one of {373, 979}. The possibilities for  are now {139, 199, 239, 299} and it’s a factor of  = 8__. For 199 and 239, there are no multiples in the 800s; the only suitable multiple of 299 is 897, but  =  + , so  can’t start with 9. That leaves  = 139 with its multiple  = 834 , making  = 373 . The prime  is now 36_; it can’t be 363 or 369 (multiples of 3), and 361 may be recognised as 192, so  = 367 , which makes  = 734 .

 ends in 9 and  ends in 1, so  =  −  ends in 8.  is now _78 and it has the same digit sum as  = 92416, so  = 778 , which makes  = 1409 . The palindrome  is now 71_17.  is a multiple of  = 241 matching _71_, so  = 7712 .

 =  +  + , so  = 15659 .  +  < , so  <  −  = 13873 and  ends with 1.  is __3 and its digit sum is greater than that of  = 834, so both unknown digits of  are ≥ 4.  <  < , so  is between 241 and 367. We can rule out {261, 291, 321, 351} (multiples of 3), {301, 311, 321, 331} ( has to start with 4 or greater), 341 (multiple of 11, the middle digit being the sum of the other two) and 361 (square), leaving {251, 271, 281}.

The square  ending in 6 is one of {196, 256, 576} (not 676 because that’s a palindrome). So  is a multiple of  starting with one of {5, 7, 9}. The matching multiples of 251 are {502, 753} but the first would make  start with 0 and the second would make it a palindrome; so  isn’t 251. For 271 the only matching multiple is 542 and for 281 there’s only 562, so  starts with 5 (and ends with 2) and  = 256 . If  is 271, the only other three-digit multiple for  is 813; for 281, there’s only 843; so  starts with 8 and ends with 3.  is now 32_ and has the same digit sum as ;  can’t be 843, which would require the last digit of  to be 10, so  = 813 ,  = 327 ,  = 271 and  = 542 , making  = 45 .

The square  is now _88_1, for which the only match is  = 48841 .  is a three-digit multiple of  = 59 ending in 1, which can only be  = 531 . The digit sum of  is twice that of , so  = 71217 .  is 3_31 and has the same digit sum as  = 271, so  = 3331 .

We know  < 13873 (above) and the grid has 1_54_. Its digit sum + the digit sum of  = the digit sum of  + the digit sum of , so ’s digits add up to 23 + 9 − 10 = 22, ie the two unknown digits add up to 12 and are both ≥ 3. The only possibility is  = 13549 .

The prime  is 7_3 and to have a digit sum greater than that of  = 834, the middle digit must be ≥ 6. We can eliminate 763 (700 + 63, both obvious multiples of 7) and 783 (multiple of 3), leaving {773, 793}. There’s now no alternative to testing these for primeness. Dividing by primes up to the square root (ie, {7, 11, 13, 17, 19, 23}), we can find 793 = 13×61, so 793 isn’t prime and  = 773 .

The prime  is _14_ and has the same digit sum as  = 734, so the two unknown digits add up to 9, allowing {2147, 6143, 8141}. The prime  starts with 2, the second digit is one of {1, 3, 7} and its digit sum is greater than that of  = 45, so the candidates are {217, 219, 237, 239, 271, 273, 277, 279}. We can eliminate {219, 237, 273, 279} (multiples of 3), 271 (same as ) and 217 (210 + 7, both obvious multiples of 7), leaving {239, 277} and eliminating 8141 for . There’s now no alternative to testing 2147 and 6143 for primeness. If we’re lucky enough to choose 2147 first, we don’t have to divide by all the primes up to its square root to find 19×113, so 2147 isn’t prime,  = 6143 and  = 239 .

The prime  is __21_ and its digit sum is (digit sum of ) − (digit sum of ) = 8, so the last digit can’t be 7 or 9. If the last digit is 1 we have __211 where at least one of the first two digits is > 1, which means the reverse of  is less than , not greater. Therefore the last digit is 3 and the candidates are {11213, 20213}. There’s now no alternative to testing these for primeness. Dividing by primes up to the square roots (ie, 7 to 103 for 11213, 7 to 139 for 20213), we can find 20213 = 17×1189, so  = 11213 and the grid is complete.