I'm thinking of a number (Robber's thread)

Question

Cop's thread here

In this cops-and-robbers challenge cops will think of a positive integer. They will then write a program or function that outputs one value when provided the number as input and another value for all other positive integer inputs. Cops will then reveal the program in an answer keeping the number a secret. Robbers can crack an answer by finding the number.

Your job as robbers is to find that number. Once found you can make an answer here detailing the number and how you found it. You should link the original answer being cracked in your answer.

Your score will be the number of cracks with more being better.

Answer 1

Java, by okx

Note: the original challenge to which this post answers has been deleted, it's however still accessible with the link to anyone having the privilege to see deleted answers. The explanations below include what the challenge was about.

Number: 18

The code is effectively looking for primes with the form 11²ⁿ-2.

By this Wikipedia list, we know that the 20th largest known prime number has 2,900,832 digits. No prime with the form 11²ⁿ-2 are in that list, so that digit number will be our limit.

All numbers of the form 11²ⁿ-2 that have at most that many digits have this constraint: n < 22. Factordb tells us that 11²²²-2 has 4,367,918 digits. This is proof enough.

The code by okx limits us to n above 6.

So, let's do the list:

11²ⁿ-2 is factorizable for n = 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 19, 20 and 21 (factors proofs are on factordb.com and wolfram alpha; factordb has some proofs that wolfram alpha hasn't and vice-versa).

The only value n for which we can't prove the factorization is 18 (factordb, WolframAlpha).

So (a) if there is a prime number of the form 11²ⁿ-2 with n > 6, then (b) 6 < n < 22, and (c) I proved each number with such n is composite except for 18, so (d) the prime we're looking for has n = 18.

Note that I don't exclude that 11²¹⁸-2 is composite. I haven't proved that that number is a prime: I only proved that if the OP thought about that number, it can only be 18. If that number is indeed a prime, congrats to okx for finding it :-)

However

There are no proof that 11²¹⁸-2 is the only number, though. It's the only number in the range of numbers for which we can currently prove the primality.

So I don't exclude the fact that one day someone might prove that 11²⁵¹-2 or 11²¹²⁴-2 is prime.

I therefore believe that the original entry by okx is invalid because there are no such proof that for n between 19 and the (2³¹-1)^231-1 (the theoretical limit of Java's BigInteger), there are no other primes.

Answer 2

Perl 6, by Ramillies

11

The function series($x) is simply the first 108 terms of the Taylor series for e^x evaluated at i*(x % 8*π). Since e^2πi = 1, this reduces to simply e^ix: the % 8*π serves only to bound the error of the truncated series.

The ∫ function integrates the provided Callable in the interval [0,2π] using the trapezoid method. The integrand in question is (e^ix)^11/(e^inx), or simply e^(11-n)ix. This integrates to 0 in that interval for all n != 11. At n = 11, the integrand reduces to simply 1. Hence the answer is 11, and the desired output is 6.28.

While 11 is the only answer in theory, the sampling in the integral is imperfect. Consequently, any integer of the form 87931k+11 should work.

Answer 3

Python 3, Erik Brody Dreyer

Number: 105263157894736842

Try it online!

---

How?

Everyone's favorite tool: OEIS!

1. Go to oeis.org
2. Search "last digit first doubles number"
3. Find Entry A087502 entitled

   Smallest positive integer which when written in base n is doubled when the last digit is put first.

4. Find the base 10 entry (105263157894736842)
5. Confirm it works

Answer 4

Tampio, fergusq

Number: 27

The program prints nolla for 27 and "1" negatiivisena for all other inputs. It times out on TIO, unfortunately, but works offline.

Whew, this was a fun one to work through. Wiktionary and this Finnish dictionary I found online actually helped a surprising amount, as a lot of Tampio syntax relies on inflection.

I first tried translating the program into Haskell-like pseudocode, but the "case system" of Tempio and how function calls worked kept tripping me up. Eventually, I ended up running all of the code from the definition of sata downward, which gave me surprisingly easily that sata == 55, and funktio wasn't difficult to reverse engineer from there as it was simple arithmetic that could be deduced from trial and error.

Answer 5

JavaScript, by Eli R

The number is 7464822360768511

I first disabled the anti-debugging function (by overwriting Function.prototype.constructor to a function with evals any code except debugger) then debugged the guess function by stepping into it.

From what I understood the guess function initializes a random number generator on a fixed seed then generate a random number and compare it to the input. As the input isn't used in the random number generation it can be ignored allowing to skip directly to the result.

Answer 6

Jelly, Erik the Outgolfer's answer

Number: 134

The following program:

5ȷ2_c⁼“ḍtṚøWoḂRf¦ẓ)ṿẒƓSÑÞ=v7&ðþạẆ®GȯżʠṬƑḋɓḋ⁼Ụ9ḌṢE¹’

can be cracked using the number 134. This outputs 0 for all numbers except for 134, which gives 1.

Try it online!

---

How did I find it?

TBH, I have no idea how it works. I have added Ç€G at the end and mapped over [1...1000] to see where the 1 is. Then, I ran script to count the number of zeros before the 1.

Answer 7

Ly, LyricLy

Answer: 239

I already wrote two Ly programs, so this wasn't too difficult. I think I would have somehow used the result from the loop in case someone tries to remove it. BTW, the fragment nI means that it doesn't always terminate successfully.

Answer 8

Python 2, Sisyphus

print~~[all([c[1](c[0](l))==h and c[0](l)[p]==c[0](p^q) for c in [(str,len)] for o in [2] for h in [(o*o*o+o/o)**o] for p,q in [(60,59),(40,44),(19,20),(63,58),(61,53),(12,10),(43,42),(1,3),(35,33),(37,45),(17,18),(32,35),(20,16),(22,30),(45,43),(48,53),(58,59),(79,75),(68,77)]] + [{i+1 for i in f(r[5])}=={j(i) for j in [q[3]] for i in l} for q in [(range,zip,str,int)] for r in [[3,1,4,1,5,9]] for g in [q[1]] for s in [[p(l)[i:i+r[5]] for p in [q[2]] for i in [r[5]*u for f in [q[0]] for u in f(r[5])]]] for l in s + g(*s) + [[z for y in [s[aprint~~[all([c[1](c[0](l))==h and c[0](l)[p]==c[0](p^q) for c in [(str,len)] for o in [2] for h in [(o*o*o+o/o)**o] for p,q in [(60,59),(40,44),(19,20),(63,58),(61,53),(12,10),(43,42),(1,3),(35,33),(37,45),(17,18),(32,35),(20,16),(22,30),(45,43),(48,53),(58,59),(79,75),(68,77)]] + [{i+1 for i in f(r[5])}=={j(i) for j in [q[3]] for i in l} for q in [(range,zip,str,int)] for r in [[3,1,4,1,5,9]] for g in [q[1]] for s in [[p(l)[i:i+r[5]] for p in [q[2]] for i in [r[5]*u for f in [q[0]] for u in f(r[5])]]] for l in s + g(*s) + [[z for y in [s[i+a][j:j+r[0]] for g in [q[0]] for a in g(r[0])] for z in y] for k in [[w*r[0] for i in [q[0]] for w in i(r[0])]] for i in k for j in k] for f in [q[0]]]) for l in [int(raw_input())]][0]

Try it online!

Answer:

126437958895621473374985126457193862983246517612578394269314785548769231731852649

Explanation

Slightly less convoluted:

print~~[all(
	[len(str(l))==81 and str(l)[p]==str(p^q)
		for p,q in [(60,59),(40,44),(19,20),(63,58),(61,53),(12,10),(43,42),(1,3),(35,33),(37,45),(17,18),(32,35),(20,16),(22,30),(45,43),(48,53),(58,59),(79,75),(68,77)]
	]
	+
	[{i+1 for i in range(9)}=={int(i) for i in l}
		for s in [[str(l)[9*u:9*u+9] for u in range(9)]]
			for l in s + zip(*s) + [[z for y in [s[i+a][j:j+3] for a in range(3)] for z in y] for k in [[w*3 for w in range(3)]] for i in k for j in k]]) for l in [int(raw_input())]][0]

Try it online!

Firstly, this part:

[len(str(l))==81 and str(l)[p]==str(p^q)
        for p,q in [(60,59),(40,44),(19,20),(63,58),(61,53),(12,10),(43,42),(1,3),(35,33),(37,45),(17,18),(32,35),(20,16),(22,30),(45,43),(48,53),(58,59),(79,75),(68,77)]
]

checks if the sudoku is in the following form:

.2.|...|...
...|6..|..3
.74|.8.|...
-----------
...|..3|..2
.8.|.4.|.1.
6..|5..|...
-----------
...|.1.|78.
5..|..9|...
...|...|.4.

Then this part checks for validity:

[{i+1 for i in range(9)}=={int(i) for i in l}
        for s in [[str(l)[9*u:9*u+9] for u in range(9)]]
            for l in s + zip(*s) + [[z for y in [s[a][j:j+3] for a in range(3)] for z in y] for k in [[w*3 for w in range(3)]] for i in k for j in k]]

Answer 9

Pyth, Mr. Xcoder

hqQl+r@G7hZ@c." y|çEC#nZÙ¦Y;åê½9{ü/ãѪ#¤
ØìjX\"¦Hó¤Ê#§T£®úåâ«B'3£zÞz~Уë"\,a67Cr@G7hZ

Try it here. Number: 9

Answer 10

Swift 3, TheIOSCoder

Number: 387849.

This returns 222 for the number above, and 212 for others.

Test Here.

---

How?

The relation is quite obvious: .pi*123456.0 multiplies 3.1415... by 123456.0 and converts it to an Integer. This means it is equal to 387848. Then, it checks if the parameter equals the number above incremented by 1 (n==1+...). The value is 387848 + 1, which is 387849...

Answer 11

Java, Sleafar

23

The program traces n iterations of a fractal, and checks whether a specific point is removed in the nth iteration. In each iteration, each segment is replaced by eight smaller segments in the following pattern:

  +
  |
+-+
|
+-+-+
    |
  +-+
  |
  +

Since no points are ever added to the line y=.5 after the first iteration, I only had to handle the set of points included along this line. This set is similar to the construction of the Cantor set, but with the middle half removed instead of the middle third.

The program I used to crack it:

import java.math.BigDecimal;

public class Main {

        public static int f(BigDecimal x) {
                BigDecimal[] result = x.divideAndRemainder(BigDecimal.ONE);
                if (result[0].compareTo(BigDecimal.ZERO) == 0 || result[0].compareTo(BigDecimal.valueOf(3)) == 0) {
                        return 1+f(result[1].multiply(BigDecimal.valueOf(4)));
                }
                return 0;
        }

	public static void main(String[] args) {
		BigDecimal x = new BigDecimal(args[0]);
		System.out.println(f(x));
	}
}

Try it online!

Answer 12

Haskell, flawr

I think the number is 12018.

I found it by defining in PARI/GP f(x,k)=1/x^2-1/x^(2-1/k) and g(k)=sumpos(x=1,f(x,k))-sumpos(x=12345679,f(x,k)) and some binary searching.

I don't like that one could trivially transform the program into one that returns the number and uses the same time and memory as the original program needs for any input.

Answer 13

Brain-Flak, Nitrodon

The number is:

1574

Try it online!

Here's the program I use to crack it

n!(0:_)=n
n!a=(n+1)!zipWith(+)a(tail a++[0])
main=print$0![45646,253224,392678,8676,-24]

Try it online!

Answer 14

Java, user902383

The secret number is 3141592

Explanation:

After expanding all the \u00xx escapes, you get:

public class Mango {
static void convert(String s){for(char c : s.toCharArray()){ System.out.print("\\u00"+Integer.toHexString(c));}}
public static void main(String[] args) {int x  = Integer.parseInt(args[0]);
double a= x/8.-392699;double b = Math.log10((int) (x/Math.PI+1))-6;
System.out.println((a/b==a/b?"Fail":"OK" ));
}}

(the convert method isn't used).

This computes some floating point math on the result, and then requires that a/b isn't equal to itself. The number for which that is true is nan, which is produced by 0/0. It turns out that if a is 0, then b is log10(0/pi + 1) which is also 0, so a has to be 0 and x has to be 392699*8.

Answer 15

Python 3, user71546

def check(x):
    if x < 0 or x >= 5754820589765829850934909 or pow(x, 18446744073709551616, 5754820589765829850934909) != 2093489574700401569580277 or x % 4 != 1:
        return "No way ;-("
    return "Cool B-)"

print(check(141421356237))

Try it online!

The number is 141421356237.

The other three numbers which satisfy the first two conditions are 1459265341309891512760823, 5754820589765688429578672, and 4295555248455938338174086.

How I solved it

The gist of this challenge is to find the number that meets the following, given the prime p = 5754820589765829850934909 and a value a = 2093489574700401569580277:

• 0 <= x < p
• x ** (2 ** 64) % p == a
• x % 4 == 1

The second one can be rephrased to "find the modular square root of modular square root of ... (64 times) of a modulo p."

Since the given prime is 5 modulo 8, we can use the following formula from the above link (though I couldn't find the reference that supports it):

\$ v = (2a)^{(p-5)/8} \mod p \$

\$ i = 2av^2 \mod p \$

\$ r = av(i-1) \mod p \$

\$ r' = p - r \$

Then we have two output values r and r' for an input value a. But not every number has such a square root, so we have to check if r*r % p == a is actually met.

So I wrote a quick J script to find the 64th modular square roots.

p =: 5754820589765829850934909x
a =: 2093489574700401569580277x
powmod =: p&|@^

f =: monad define
for. i.64 do.
  v =: p | (2*a) powmod (p-5)%8
  i =: p | 2*a*v*v
  x =: p | a*v*i-1    NB. Apply the above formula
  a =: (a=p|x*x) # x  NB. Filter by actually being the modular square root
  a =: a, p-a         NB. Concat r with r'
  a
end.
)

res =: f ''

Try it online!

res has the four values shown above; the last filtering by modulo 4 gives the answer.