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<p^4<999999. Therefore 900<p^2<1000. Therefore p is 31.
31^4=961^4=810000 + 108000 + 1800 + 3600 + 120 + 1 = 923521. The bottom
number in the 3rd row ends with the 3rd digit of either 130321 or
923521, and has 2 factors - thereby being prime. No prime ends with 0,
so the bottom row is 923521.

531441
....13
....90
....83
....42
923521

The bottom number on the 4th column has 3 factors, and is therefore the
square of a prime. It ends in 5.

531441
....13
....90
....83
...242
923521

The right hand number on the 4th row is of the form ?83 and has 4
factors. By checking the primes from 2 to 31, we find the prime
factorizations of the numbers ?83 to be 

183=3*61
283=283
383=383
483=3*7*23
583=11*53
683=683
783=3*3*3*29
883=883
983=983

Therefore the right hand number on the 4th row is 183 or 583.
Therefore the top number on the 4th column is an odd number of the form
4??? with 20 factors.

20=20=10*2=5*4=5*2*2. Therefore a number with 20 factors can be of the
forms p^19,pq^9,p^4q^3 and p^4qr. Suppose p,q,r are odd primes.
p^19>=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<p^2<529999
Let p=(700+a)
 30009<1400a+a^2<39999
Then certainly a<39999/1400<29. If a is as low as 21, then
1400a+a^2=21*1421=28420+1421=29841. Therefore a=22,23,24,25,26,27,28.
The only option giving p prime is p=727. Therefore the number is 
727^2=490049+28280+400+9800=528529

531441
211413
8.2590
5..583
2..242
923521

The left hand number in the 4th row is a square of a prime and is a
3-digit number starting with 5. Therefore it is 529.

531441
211413
8.2590
529583
2..242
923521

The bottom number in the 3rd column is a prime of the form 9?3. The
factor of 3 gets rid of possible digits 0,3,6,9. Possible factor of 7
gets rid of 7. Possible factor of 11 gets rid of 1. Possible factor of
11 gets rid of 2. Possible factor of 23 gets rid of 4. Therefore it is
either 5 or 8. 5242=2*2621 and 8242=2*4121=2*13*317. Thus the missing
digit is 5.

531441
211413
8.2590
529583
2.5242
923521


The left hand number in the 3rd row, 8?25, has 12 factors. 8?25 divided
by 25 is an integer between 321 and 357. Therefore we only need to check
primes up to 17 for factors to give:

8025=3*5*5*107 - 12
8125=5*5*5*5*13 - 10
8225=5*5*7*47 - 12
8325=3*3*5*5*37 - 18
8425=5*5*337 - 6
8525=5*5*11*31 - 12
8625=3*5*5*5*23 - 16
8725=5*5*349 - 6
8825=5*5*353 - 6
8925=3*5*5*7*17 - 24

Therefore the missing digit(being the start of another number) is either
2 or 5.

The left hand number in the 5th row, 2? has 6 factors.

20=2*2*5 - 6 factors
21=3*7 - 4 factors
22=2*11 - 4 factors
23=23 - 2 factors
24=2*2*2*3 - 8 factors
25=5*5 - 3 factors
26=2*13 - 4 factors
27=3*3*3 - 4 factors
28=2*2*7 - 6 factors
29=29 - 2 factors

Therefore the missing digit is either 0 or 8.

Therefore the bottom number in the 2nd column is one of 2202,2282,5202
or 5282. Their factorizations are 

2202=2*3*367
2282=2*7*163
5202=2*3*3*17*17
5282=2*19*139

The only one with 18 factors is 5202, completing the grid.

531441
211413
852590
529583
205242
923521
-------------------------------------------------------------------
Hi Ed,

I believe I found the solution to the crossnumber puzzle:

5 3 1 4 4 1
2 1 1 4 1 3
8 5 2 5 9 0
5 2 9 5 8 3
2 0 5 2 4 2
9 2 3 5 2 1

It looked a bit daunting at first, but a bit of code and the unique solution
for the top row made it much more manageable.  Thanks for the continued
puzzles and interesting links.

Mike Shafer

-------------------------------------------------------------------
Berend Jan van der Zwaag 

hi ed,
nice puzzle you found!
my solution:

531441
211413
852590
529583
205242
923521

cheers,
  berend
-------------------------------------
This is my solution to the crossnumber puzzle:

531441
211413
852590
529583
205242
923521

Best regards,
José H. Nieto
--------------------------------------
That was a fun puzzle, it took me a long time.  I had to use a table of primes and a graphing calculator extensively.
 
Here's the result:
 
531441
211413
852590
529583
205242
923521
 
The key was the first row, which has to be p^12.  Only 3^12 has 6 digits.
 
This seems like a hard puzzle to create, but I wouldn't mind seeing others like it.
 
-Jeremy Galvagni
---------------------------------------------------------
Tricky little puzzle, but I enjoyed it The answer:

531441
211413
852590
529583
205242
923521

The key to the puzzle is a singular math fact: If the number n can be
written as (p1^e1) * (p2^e2) * (p3^e3)..., where p1, p2, p3,... are
distinct prime numbers, and e1, e2, e3,... represent whole-number
exponents, then n has exactly (e1+1) * (e2+1) * (e3+3)... factors,
including 1 and itself. This means, for example, that the 6-digit number
in the first row must be the 14th power of a prime number, and that any
number with an odd number of factors must be a perfect square.
-----------------------------------------------------------
David Corbett

13, First row: twelfth power of a prime. Only
candidate: 3^12=531441.
5, Sixth column: Fourth power of a prime. Only
candidate: 19^4=130321.
22, Fifth column: product of prime and tenth power.
Tenth power must be 2^10. 41, 43 and 47 all give
answers that fit in: only 41 gives a 12 factor number
on third row. Answer: 41*2^10=41984.
4, Third row (right): 90.
10, Third column (top):product of prime and fourth
power. Possible answers: 112, 162, 176.
14, Second row: product of prime and sixth power,
which must be 3^6. Prime ends in 9: only 29 gives the
right digit for the 3rd column. Answer: 29*3^6=21141.
10, Third column (top): 112.
3, First column: square starting with 52. Answer:
727^2=528529.
3, Fourth row (left): square: 23^2=529.
5, Sixth row: fourth power. Answer: 31^4=923521.
3, Fourth column (bottom): square: 5^2=25.
20, Fourth column (top): product of two primes and
fourth power, or cube and fourth power. Starts with
44. 2^4 gives no solution to either. 3^4 gives only
solution as 5*11*3^4=4455.
4, Fourth row (right): 583.
4, Fifth row (right): product of two primes, one of
which is 2.
2, Fourth column (bottom): prime. Only digit which
satisfies both conditions is 5. Answer: 953.
4, Fifth row (right): 5242.
12, Third row (left): product of two primes and square
(5^2), or square and cube. Latter gives no solution.
Former gives two possibilities: 7*47*5^2=8225, or
11*31*5^2=8525.
6, Fifth row (left): product of square and prime.
Possible answers: 20 or 28.
18, Second column (bottom): single factor of 2 (from
above): product of 2, and two squares. Of the four
possible answers, only 2*(3*17)^2=5202 is correct.

So the final square is:

531441
211413
852590
529583
205242
923521
--------------------------------------------------
Hi Ed.

Presented on mathpuzzle.com:

"I found a nice crossnumber puzzle in the August 1974 issue of Games and
Puzzles.  Your only clue for the numbers going across and down is the 
number of divisors that number has."

Answer:

531441
211413
852590
529583
205242
923521

Thanks, and keep 'em comin'.
Gary Mulkey
----------------------------------------------------
531441
211413
852590
529583
205242
923521

Sadly, I took some of the fun out of the puzzle by using short computer
programs to find some of the clues :)

--Nathan
Nathan Stohler
-----------------------------------------------------
531441
211413
852590
529583
205242
923521

A nice puzzle. Who would have guessed there is only one 6-digit number with exactly 13 divisors.
I'll try to make some similar puzzles with smaller numbers.
Regards from Slovenia,
Bostjan Kuzman
--------------------------------------------------------

531441
211413
852590
529583
205242
923521

Prof. Dr. rer. nat. Heinrich Hemme
Deutschland

-------------------------------------------------------
Andrej_Jakobčič
Hi Ed!

Very nice puzzle, here is solution.
I wrote short pascal program for counting divisors and solve step by step
with computer and by hand (mind?).

Best regards
Andrej

5 3 1 4 4 1
2 1 1 4 1 3
8 5 2 5 9 0
5 2 9 5 8 3
2 0 8 2 4 2
9 2 3 5 2 1
-------------------------------------------------------

Ed,
 
I believe that I have a solution to the crossnumber puzzle, from  the August 1974 issue of Games and Puzzles, that you have on your web site (http://www.mathpuzzle.com).
 
The following is in  Courier font.
 
            3      2     10    20     22      5
   +---+---+---+---+---+---+
13 | 5 | 3 | 1 | 4 | 4 | 1 | 
   +---+---+---+---+---+---+
14 | 2 | 1 | 1 | 4 | 1 + 3 |
   +---+++++---+---+---+---+
12 | 8 | 5 | 2 | 5 + 9 | 0 | 12
   +---+---+++++---+---+---+
 3 | 5 | 2 | 9 + 5 | 8 | 3 |  4
   +---+---+---+++++---+---+
 6 | 2 | 0 + 5 | 2 | 4 | 2 |  4
   +---+---+---+---+++++---+
 5 | 9 | 2 | 3 | 5 | 2 | 1 |
   +---+---+---+---+---+---+
         18   2   3  
  
I have also attached a gif (crossnumber.solution.gif) of the solution.
 
I wrote a C++ program to help me solve the puzzle.  The program has a function called numDivisors:
 
    ( f(n) =  numDivisors(n) ).
 
Given a number n the function returns the number of divisors of  n.  Using the function within a for loop,
I was able to create 18 tables for the 18 divisors shown in the puzzle.  For example: 
 
    For the two 12 divisors on row  3, I created a table of a all 4 digit numbers n (1000 <= n <= 9999) with
    12 divisors, and a table of all 2 digit numbers n (10 <= n <= 99) with 12 divisors.
 
The table of all 6 digit numbers n (100000 <= n <= 999999) that have 13 divisors contained only one number: 531441
 
Once the tables were created the solution was a manual effort.  It would have been nice to write a program to solve the puzzle, but - alas - that is beyond my expertise.
 
Regards,
John Callas

-----------------------------------------------------------


This was a nice puzzle! The answer is:

531441
211413
852590
529583
205242
923521

I'm a masochist, so I did it all by hand...no calculator, even! :)

-- 
Jarom Lechner
"Man is certainly stark mad; he cannot
make a flea, yet he makes gods by the
dozens." --Michel de Montaigne

------------------------------------------------------------
Here you go.

5 2 1 4 4 1
2 1 1 4 1|3
8 5 2 5|9 0
5 2 9|5 8 3
2 0|5 2 4 2
9 2 3 5 2 1

Feel free to delete the extensive solution notes.

First, I made the assumption that there were no lead zeros and all entries were unique.

Anything with 2 factors is prime.
Anything with 3 factors is the square of a prime number.
Anything with 5 factors is the fourth power of a prime number.
This showed me that the top row, with 13 factors, was the twelth power of a prime number.  The only six digit twelth power is 531441.  So much for the top row.

In the fourth column, the two digit number must be 25 or 49.

The six digit fourth powers with the fourth digit being 5 or 9 are...

104976
456976
923521

We can eliminate the first two because the first column is the square of a prime number and cannot end in 6.  Therefore the sixth row is 923521 and the two digit number in the fourth column is 25.

The sixth column is the fourth power of a prime number beginning and ending with 1.  The two possibilities are 130321 and 194481.

These two options allow the two digit number with twelve factors in row 3 to be 60, 84 or 90.

The only five digit numbers with 22 factors that start with a 4 are 41984, 44032 and 48128.  Only the first is compatible with the previous statement.  In turn, this means the 2 digit number with 12 factors is 90 and the sixth column is 130321

The first column is a six digit sqaure of a prime number, starting with 5, ending with 9 and containing no zeros.  There are only four of these.

522729
528529
537289
597529

In the fifth row, the only two digit numbers with 6 factors that start with 2 or 8 are 20 and 28.  This eliminates 537289 from contention.  

In the fourth row, since there are no three digit squares of prime numbers beginning with 7, 522729 is eliminated.  Either way, the last three numbers of the first colum are 529.  Since the only three digit square of a prime number starting with 5 is 529, those are also the first three numbers of the fourth row.

Thus the three digit prime number in the third column is 953 or 983.  Since 5242 (fifth row) has four factors and 8242 does not, it must be 953.

In the fourth row, the three digit number with four factors ends with 83.  Since a number with four factors is the result of multiplying two primes, the only possibilities are 183 or 583.

The list of possibles for the fourth column four digit number with 20 factors that starts with 4 only has one result that ends in a 1 or a 5 and that's 4455.  That finishes off the fourth row and the fourth column.

Returning to the first column, the number is either 528529 or 597529.  The prime number in the second column is 31 or 37.  The only possible option for the five digit number with 14 factors in the second row is therefore 21141.  (Meaning the first column must be 528529.)

The three digit number with 10 factors in the third colum is therefore 112.

The four digit number with 12 factors in row threww is 8225 or 8525.  From earlier, the two digit number in the fifth row is 20 or 28.  This leaves four possibilities for the four digit number with 18 factors in the second column.  Of 2202, 2282, 5202 and 5282 only 5202 has the right number of factors and therefore completes the puzzle.
Warren Phillips
-------------------------------------------------------
Jordi Domčnech

5 3 1 4 4 1
2 1 1 4 1 3
8 5 2 5 9 0
5 2 9 5 8 3
2 0 5 2 4 2
9 2 3 5 2 1

J. D. A.

From Barcelona, Catalonia
-------------------------------------------------------
John Boozer sent a GIF.