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