Luke Pebody The top row has 13 factors and must therefore be some 6-digit 12th power of a prime. 2^12=4096 5^12=(5^3)^4>(100)^4=10^8 Therefore the top row must be 3^12. 3^12=(3^6)^2=729^2 729x729=490000+28000+12600+400+360+81=531441 531441 ...... ...... ...... ...... ...... The right column has 5 factors and must be the 4th power of some prime. It is between 100000 and 200000. Therefore the square of the prime is between 300 and 500. Therefore the prime is between 18 and 22 inclusive. Therefore the prime is 19. 19^4=361^2 361*361=90000 + 3600 + 1 + 36000 + 600 + 120 =93601+36720=130321 531441 .....3 .....0 .....3 .....2 .....1 The right hand number on the 3rd row is a number of the form ?0 with 12 factors. 10 has 4 factors, 20 has 6, 30 has 8, 40 has 8, 50 has 6, 60 has 12, 70 has 8, 80 has 10, 90 has 12. Therefore it is 70 or 90. The number in the 5th column has 22 factors. It cannot be a number of the form p^21, as 2^21>2^20=(2^10)^2>(1000)^2>10^6. Therefore it must be a number of the form p*q^10. Now 3^10=(243)^2>40000. Therefore if you multiply it by any prime, you get a number more than 80000, which cannot be of the required form 4????. Therefore we have a number of the form p*2^10=p*1024. 37*1024=37888 is too small and 50*1024 is too big. Therefore p=41,43 or 47. 41*1024=41984 43*1024=44032 47*1024=48128 The middle number has to be 7 or 9 to make the right hand number on the 3rd row 70 or 90. Therefore 41984 is the required number. 531441 ....13 ....90 ....83 ....42 .....1 The left hand side is the square of a prime, which prime is a lot greater than 5. Thus it must end with 1 or 9. The bottom row is the 4th power of a prime, ending with 1 and starting with 1 (in which case it is 130321 from above) or 9. Suppose it starts with 9. 900000=3^19>3^18=(3^3)^6>10^6>5000. pq^9>=5*3^9=5*(27)^3>5*10^3=5000. p^4q^3>=3^4*5^3>80*100=8000. Therefore the odd number 4??? with 20 factors is of the form p^4qr. If p>=5 then p^4qr>=5^4*3*7>=5^4*20=100*5^3=12500>5000. Therefore p=3. Then qr is between 4000/81>49 and 5000/81<62 and is the product of two distinct odd primes, neither equal to 3. Therefore it must be 55. Then the number is 81*55=4455. 531441 ...413 ...590 ...583 ...242 923521 The number in the 2nd row has 14 factors and is of the form ???41. Note that it is not of the form p^13, as it would have to be at least as large as 3^13>3^12=(3^6)^2=729^2>400^2=160000. Therefore it is of the form p^6q, where p and q are primes not dividing into 10. If p>=7, p^6q>=3*7^6>=3*40^3=192000. Therefore p=3. Then 3^6q = 41 (mod 100). Dividing by 9 repeatedly, we see that 3^4q = 441/9 = 49 (mod 100), 3^2q = 549/9 = 61 (mod 100), q = 261/9 = 29 (mod 100). 129 is not prime. 229*729>200*700=140000 is too large. Therefore q=29, and the number is 29*729=14580+6561=21141. This gives the top number in the 2nd column as 31, which fits. 531441 211413 ...590 ...583 ...242 923521 The top number in the 3rd column is 11? and has 10 factors. Checking for primes up to 7, we get the following factorizations 110 2*5*11 - 8 factors 111 3*37 - 4 factors 112 2*2*2*2*7 - 10 factors 113 113 - 2 factors 114 2*3*19 - 8 factors 115 5*23 - 4 factors 116 2*2*29 - 6 factors 117 3*3*13 - 6 factors 118 2*59 - 4 factors 119 7*17 - 4 factors 531441 211413 ..2590 ...583 ...242 923521 The left hand column is the square of a prime and is of the form 52???9. 520009