Challenge Math Challenge - March 2019

  • Thread starter fresh_42
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haha, so 06729 is not the solution for the leading 0, correct? No problem, same algorithm should find another solution starting with a digit. Lemme try it.
 
322
33
New attempt, same algorithm

Solution is n=15486, 2n=30972
Code:
0123456789 abcde fghik, 1 try a in 123456789, can't start with 0 per fresh_42
0..3456789 1bcde 2ghik, 2 bcde <= 4999, try b in 03
0..3456789 10cde 20hik, 3 rejected repeated 0
0..3456789 13cde 21hik, 3 rejected repeated 1
0.2.456789 1bcde 3ghik, 2 bcde >= 5000, try b in 56789
..2.4.6789 15cde 30hik, 3 cde <= 499 try c in 24
......6789 152de 304ik, 4 de <= 49 rejected no possible values for d
....4.6789 152de 305ik, 4 de >= 50 rejected repeated 5
..2...67.9 154de 308ik, 4 de <= 49 try d in 2
......67.9 1542e 3084k, 5 e <= 4 rejected repeated 4
......67.9 1542e 3085k, 5 e >= 5 rejected repeated 5
..2...678. 154de 309ik, 4 de >= 50 try d in 678
.......78. 1546e 3092k, 5 e <= 4 rejected no possible values for e
..2....78. 1546e 3093k, 5 e >= 5 rejected repeated 3
..2...6.8. 1547e 3094k, 5 e <= 4 rejected repeated 4
..2...6.8. 1547e 3095k, 5 e >= 5 rejected repeated 5
..2....7.. 1548e 3096k, 5 e <= 4 will try e in 2
.......7.. 15482 30964, 5 e <= 4 rejected repeated 4
.......... 15486 30972, stop
 

fresh_42

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2018 Award
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New attempt, same algorithm

Solution is n=15486, 2n=30972
Code:
0123456789 abcde fghik, 1 try a in 123456789, can't start with 0 per fresh_42
0..3456789 1bcde 2ghik, 2 bcde <= 4999, try b in 03
0..3456789 10cde 20hik, 3 rejected repeated 0
0..3456789 13cde 21hik, 3 rejected repeated 1
0.2.456789 1bcde 3ghik, 2 bcde >= 5000, try b in 56789
..2.4.6789 15cde 30hik, 3 cde <= 499 try c in 24
......6789 152de 304ik, 4 de <= 49 rejected no possible values for d
....4.6789 152de 305ik, 4 de >= 50 rejected repeated 5
..2...67.9 154de 308ik, 4 de <= 49 try d in 2
......67.9 1542e 3084k, 5 e <= 4 rejected repeated 4
......67.9 1542e 3085k, 5 e >= 5 rejected repeated 5
..2...678. 154de 309ik, 4 de >= 50 try d in 678
.......78. 1546e 3092k, 5 e <= 4 rejected no possible values for e
..2....78. 1546e 3093k, 5 e >= 5 rejected repeated 3
..2...6.8. 1547e 3094k, 5 e <= 4 rejected repeated 4
..2...6.8. 1547e 3095k, 5 e >= 5 rejected repeated 5
..2....7.. 1548e 3096k, 5 e <= 4 will try e in 2
.......7.. 15482 30964, 5 e <= 4 rejected repeated 4
.......... 15486 30972, stop
Too big. I think your algorithms doesn't consider the carries and rejects possible solutions.
Your first solution was close!
 
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Ops... big fingers got in the keyboard's way (error in the 4th line of previous post)... now I flunked high-school; trying again.

Btw, I wonder if there's a way to solve this through logic instead of searching for a solution; I tried a few times, but the solution through logic evaded me.

On 3rd try, solution is n=13485 2n=26970; if this is still wrong I'm giving up high school and going back to primary school
Code:
0123456789 abcde fghik, 1 try a in 123456789, can't start with 0 per fresh_42
0..3456789 1bcde 2ghik, 2 bcde <= 4999, try b in 03
...3456789 10cde 20hik, 3 cde <= 499 rejected repeated 0
...3456789 10cde 21hik, 3 cde >= 500 rejected repeated 1
0...45.789 13cde 26hik, 3 cde <= 499 try c in 04 (this is the line that had a bug... 2*13 = 26, not 21, haha)
....45.789 130de 260ik, 3 de <= 499 rejected repeated 0
....45.789 130de 261ik, 3 de >= 500 rejected repeated 1
0....5.7.9 134de 268ik, 3 de <= 499 try d in 0
.....5.7.9 1340e 2680k, 4 e <= 4 rejected repeated 0
.....5.7.9 1340e 2681k, 4 e >= 5 rejected repeated 1
0....5.78. 134de 269ik, 3 de >= 500 try d in 578
.......78. 1345e 2690k, 4 e <= 4 rejected no values for e
0......78. 1345e 2691k, 4 e >= 5 rejected repeated 1
0....5..8. 1347e 2694k, 4 e <= 4 rejected repeated 4
0.......8. 1347e 2695k, 4 e >= 5 try e in 8
0......... 13478 26956, rejected repeated 6
0....5.7.. 1348e 2696k, 4 e <= 4 rejected repeated 6
0....5.... 1348e 2697k, 4 e >= 5 try e in 5
.......... 13485 26970, stop
 

fresh_42

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Insights Author
2018 Award
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7,094
On 3rd try, solution is n=13485 2n=26970; if this is still wrong I'm giving up high school and going back to primary school
Code:
0123456789 abcde fghik, 1 try a in 123456789, can't start with 0 per fresh_42
0..3456789 1bcde 2ghik, 2 bcde <= 4999, try b in 03
...3456789 10cde 20hik, 3 cde <= 499 rejected repeated 0
...3456789 10cde 21hik, 3 cde >= 500 rejected repeated 1
0...45.789 13cde 26hik, 3 cde <= 499 try c in 04 (this is the line that had a bug... 2*13 = 26, not 21, haha)
....45.789 130de 260ik, 3 de <= 499 rejected repeated 0
....45.789 130de 261ik, 3 de >= 500 rejected repeated 1
0....5.7.9 134de 268ik, 3 de <= 499 try d in 0
.....5.7.9 1340e 2680k, 4 e <= 4 rejected repeated 0
.....5.7.9 1340e 2681k, 4 e >= 5 rejected repeated 1
0....5.78. 134de 269ik, 3 de >= 500 try d in 578
.......78. 1345e 2690k, 4 e <= 4 rejected no values for e
0......78. 1345e 2691k, 4 e >= 5 rejected repeated 1
0....5..8. 1347e 2694k, 4 e <= 4 rejected repeated 4
0.......8. 1347e 2695k, 4 e >= 5 try e in 8
0......... 13478 26956, rejected repeated 6
0....5.7.. 1348e 2696k, 4 e <= 4 rejected repeated 6
0....5.... 1348e 2697k, 4 e >= 5 try e in 5
.......... 13485 26970, stop
This is correct, keep on learning! :smile:

The manual solution isn't really better than your algorithm. It goes along the following lines:
  • If the 0 is in n, then 1 and 5 must be in 2n.
  • If the 0 is in 2n, then 5 and 1 must be in n.
  • If the 9 is in n, 8 and 4 in 2n must be in it.
  • If the 9 in 2n is in, 4 and 8 must be in n.
etc.
 
Last edited:
32,707
8,570
Minimal python implementation:
Code:
for i in range(10000,100000):
  if ''.join(sorted(str(i)+str(2*i)))=="0123456789":
    print i
    break
 

Bosko

Gold Member
16
3
3.) Show that $$\int_{0}^{1} dx \int_{x}^{\sqrt{x}} f(x,y)\,dy = \int_{0}^{1} dy \int_{y^2}^{y} f(x,y)\,dx\,.$$
The integration area for f(x,y) function is ... https://www.wolframalpha.com/input/?i=plot+y=sqrt+x,y=x,++x=0,1

Screenshot 2019-03-07 at 16.08.43.png

1) If we calculate first by x from 0 to 1 and then by y from x to x1/2 we get
$$\int_{0}^{1} dx \int_{x}^{\sqrt{x}} f(x,y)\,dy $$

2) If we calculate first by y from 0 to 1 and then by x from y2 to y we get
$$\int_{0}^{1} dy \int_{y^2}^{y} f(x,y)\,dx\,.$$

1) and 2) are equal because it is the same function f(x,y) on the same area but calculated on two different ways
$$\int_{}^{} \int_{Area}^{} f(x,y)\,dx dy\,.$$
You could do it with the upload button at bottom right of the edit field, but I would appreciate if you wouldn't. Here's a guideline to write formulas on PF:
https://www.physicsforums.com/help/latexhelp/
Thanks
 

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Last edited:
322
33
Trying 5c

(100000A + 10000B + 1000C + 100D + 10E + F).3 = (100000B + 10000C + 1000D + 100E + 10F + A)
(100000A + 10000B + 1000C + 100D + 10E + F).2 = (100000C + 10000D + 1000E + 100F + 10A + B)
(100000A + 10000B + 1000C + 100D + 10E + F).6 = (100000D + 10000E + 1000F + 100A + 10B + C)
(100000A + 10000B + 1000C + 100D + 10E + F).4 = (100000E + 10000F + 1000A + 100B + 10C + D)
(100000A + 10000B + 1000C + 100D + 10E + F).5 = (100000F + 10000A + 1000B + 100C + 10D + E)

(300000A + 30000B + 3000C + 300D + 30E + 3F) = (100000B + 10000C + 1000D + 100E + 10F + A)
(200000A + 20000B + 2000C + 200D + 20E + 2F) = (100000C + 10000D + 1000E + 100F + 10A + B)
(600000A + 60000B + 6000C + 600D + 60E + 6F) = (100000D + 10000E + 1000F + 100A + 10B + C)
(400000A + 40000B + 4000C + 400D + 40E + 4F) = (100000E + 10000F + 1000A + 100B + 10C + D)
(500000A + 50000B + 5000C + 500D + 50E + 5F) = (100000F + 10000A + 1000B + 100C + 10D + E)

as A,B,C,D,E,F in 0..9, then

3F mod 10 = A
2F mod 10 = B
6F mod 10 = C
4F mod 10 = D
5F mod 10 = E

(30E+3F) mod 100 = (10F + A)
= (30(5F mod 10)+3F) mod 100 = 10F + (3F mod 10)

so can just try all numbers from F=1 to F=9

(30.(5.1 mod 10) + 3.1) mod 100 = (30.5+3) mod 100 = 53 <> 10.1 + (3.1 mod 10) = 13
(30.(5.2 mod 10) + 3.2) mod 100 = (30.0+6) mod 100 = 6 <> 10.2 + (3.2 mod 10) = 26
(30.(5.3 mod 10) + 3.3) mod 100 = (30.5+9) mod 100 = 59 <> 10.3 + (3.3 mod 10) = 39
(30.(5.4 mod 10) + 3.4) mod 100 = (30.0+12) mod 100 = 12 <> 10.4 + (3.4 mod 10) = 42
(30.(5.5 mod 10) + 3.5) mod 100 = (30.5+15) mod 100 = 65 <> 10.5 + (3.5 mod 10) = 55
(30.(5.6 mod 10) + 3.6) mod 100 = (30.0+18) mod 100 = 18 <> 10.6 + (3.6 mod 10) = 68
(30.(5.7 mod 10) + 3.7) mod 100 = (30.5+21) mod 100 = 71 = 10.7 + (3.7 mod 10) = 71
stop

so F=7, therefore
A = 3.7 mod 10 = 1
B = 2.7 mod 10 = 4
C = 6.7 mod 10 = 2
D = 4.7 mod 10 = 8
E = 5.7 mod 10 = 5

142857.1 = 142857
142857.3 = 428571
142857.2 = 285714
142857.6 = 857142
142857.4 = 571428
142857.5 = 714285

What a beautiful problem!!!! I'd never imagine a 6-digit number multiplied like that would just twist around itself. Really amazing! Well done, fresh_42, you really have an impressive mind!!!

I wonder what deep property lies buried in this number, so that it does that, other than it looks suspiciously like 1/7 = 0.142857...
 
322
33
Hmmm... maybe that comes down to this...? This is super!

1/7 = 0.abcdefabcdefabcdef..
1E6.1/7 = abcdef.abcdefabcdef..
1E6.1/7 - 1/7 = abcdef
(1E6-1)/7 = abcdef = 999999/7 = 142857 => 999999 = 142857 * 7

similarly

1/7-0.1 = 0.0bcdefabcdefabcdef..
1E7.(1/7-7/70) = bcdefa.bcdefabcdef...
1E6.(3/7) = bcdefa.bcdefabcdef...
1E6.(3/7) - (3/7) = bcdefa = (999999)*3/7 = 428571 => 999999 = (428571)*7/3 = 142857 * 7

therefore 428571 = 142857 * 3

I bet the other ones go through the pattern. What a wonderful trick!! The digits on this 1/7 dude must make a group of some kind, I guess!
 

fresh_42

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2018 Award
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I wonder what deep property lies buried in this number, so that it does that, other than it looks suspiciously like 1/7 = 0.142857...
It does not only look like it, it is the reason:
If ##\sigma## notes the cyclic shift by one digit (##\sigma(ABCDEF)=BCDEFA##) we get with ##x=ABCDEF##
$$
x \cdot 10^k \equiv \sigma^k(x) \operatorname{mod}7
$$
i.e. ##\sigma## acts like the multiplication by ##10## in ##\mathbb{Z}_7\,,## ##10:7=1.42857 \ldots \,.##
 
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DeathByKugelBlitz

Gold Member
22
11
5b could be done in Python?

First calculate 1000! then use python to count the zeroes?
 

DeathByKugelBlitz

Gold Member
22
11
5b:

import math
A = str(math.factorial(1000))
A.count('0')

472
 

DeathByKugelBlitz

Gold Member
22
11
I got this but i made mistake somewhere
import math
A = (str(math.factorial(1000)))
B = (str(int(math.factorial(1000)))[::-1])
C = len(A) - len(B)
print(C)
 

fresh_42

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2018 Award
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I got this but i made mistake somewhere
import math
A = (str(math.factorial(1000)))
B = (str(int(math.factorial(1000)))[::-1])
C = len(A) - len(B)
print(C)
It can be done in one line without coding, two, if an explanation line is added!
 

Ibix

Science Advisor
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I'm new to this, only know what we learnt in lecture :frown:
Your approach should work in principle. Expand your calculation of B over several steps and make sure each one is doing what you think it is - I can think of a couple of likely failure points.

As fresh says, there are more elegant solutions.
 

DeathByKugelBlitz

Gold Member
22
11
import math
def reverse_int(n):
return int(str(n)[::-1])
A = (math.factorial(1000))
B = (reverse_int(A))
C = int(B)
D = A.count(0)
E = C.count(0)
print(D - E)

NameError: name 'D' is not defined
 
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I tried the 1000! problem with logic, by counting the x10 and x100 numbers, then counting the x2 . x5 combinations that produce 1, 2, 3 and 4 zeroes... but I got the wrong total of zeroes compared to the real number from Wolfram-Alpha's ...

That puts me back in kindergarten, oh no.. :(
 

fresh_42

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2018 Award
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I tried the 1000! problem with logic, by counting the x10 and x100 numbers, then counting the x2 . x5 combinations that produce 1, 2, 3 and 4 zeroes... but I got the wrong total of zeroes compared to the real number from Wolfram-Alpha's ...

That puts me back in kindergarten, oh no.. :(
Why did you count the twos? Were you afraid there might not be enough of them?
 
Last edited:
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33
Because 2.5 = 10, so each group of 1.2.3.4.5.6.7.8.9 = 2.5.(1.3.4.6.7.8.9) = 10.(36,288)

So numbers that end in 1,3,4,6,7,8,9 don't add any zeroes to the result ( ex x3.x6 = (10x+3).(10x+6) = 100x^2 + 10.(3+6).x + 6.3... no zeroes in any x6.x3 for all x in {0..99} ).

As any number ending in 2 will be followed by a number ending in 5 in the sequence {1..1000}, then for x in 0..99 we have x2.x5 = 100x^2 + 10.(2+5)x + 10 = 100x^2+10(7x+1), therefore x2.x5 add zeroes. So we get 1x3 zeroes from 1000, 9x2=18 zeroes from 100, 200, .. , 900, 90x1 zeroes from 10, 20, ..., 90, 110, 120, ..., 190, 210, ... , 290, .. , 990, = 111 zeroes, plus the zeroes from the multiplications of x2.x5.

The zeroes for x2.x5 will be 1 for each x in 0..99... which is 100 numbers (ex 22.25 = 550); then 1 more for each (10.x^2 mod 10 + 7x+1) that is multiple of 10; then 1 more for each (x^2 mod 10 + 7x+1 ) that is multiple of 100. Inspection shows that only x=7, 17, 27, ..., 97 produces that expression as multiple of 10 (ex 372.375 = 139,500), and only 87 produces an expression that is multiple of 100 (872.875 = 736,000). So the x2.x5 terms produce 100 + 10 + 1 = 111 zeroes.

This calculation yields 222 zeroes, but Wolfram-Alpha shows 249 zeroes, so my result is wrong... I'm missing something and I have no clue what... :-(

Back to kindergarten, fbs7! :H
 
Last edited:

fresh_42

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2018 Award
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Because 2.5 = 10, so each group of 1.2.3.4.5.6.7.8.9 = 2.5.(1.3.4.6.7.8.9) = 10.(36,288)

So numbers that end in 1,3,4,6,7,8,9 don't add any zeroes to the result ( ex x3.x6 = (10x+3).(10x+6) = 100x^2 + 10.(3+6).x + 6.3... no zeroes in any x6.x3 for all x in {0..99} ).

As any number ending in 2 will be followed by a number ending in 5 in the sequence {1..1000}, then for x in 0..99 we have x2.x5 = 100x^2 + 10.(2+5)x + 10 = 100x^2+10(7x+1), therefore x2.x5 add zeroes. So we get 1x3 zeroes from 1000, 9x2=18 zeroes from 100, 200, .. , 900, 90x1 zeroes from 10, 20, ..., 90, 110, 120, ..., 190, 210, ... , 290, .. , 990, = 111 zeroes, plus the zeroes from the multiplications of x2.x5.

The zeroes for x2.x5 will be 1 for each x in 0..99... which is 100 numbers (ex 22.25 = 550); then 1 more for each (10.x^2 mod 10 + 7x+1) that is multiple of 10; then 1 more for each (x^2 mod 10 + 7x+1 ) that is multiple of 100. Inspection shows that only x=7, 17, 27, ..., 97 produces that expression as multiple of 10 (ex 372.375 = 139,500), and only 87 produces an expression that is multiple of 100 (872.875 = 736,000). So the x2.x5 terms produce 100 + 10 + 1 = 111 zeroes.

This calculation yields 222 zeroes, but Wolfram-Alpha shows 249 zeroes, so my result is wrong... I'm missing something and I have no clue what... :-(

Back to kindergarten, fbs7! :H
I have serious trouble decoding that! I assume that "xN." doesn't mean the hexadecimal number N?
You shouldn't worry about the last digits, just write ##1000!## as a product of primes.
 

DeathByKugelBlitz

Gold Member
22
11
import math
def primeFactors(n):
while n % 2 == 0:
print 2,
n = n / 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
print i,
n = n / i
if n > 2:
print n
(primeFactors(math.factorial(1000)))

2 2 13 107 36791
 

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