Is this a three-digit number ending in one?

Question

Given a nonnegative integer, return whether it is a three digit number ending in one, in any consistent integer base. In other words, the number needs to be represented in base-N, N being an integer greater than zero.

Rules

• This is code-golf, so shortest answer wins.
• Since unary behaves weirdly, the behavior with input 3₁₀ is undefined.
• Standard loopholes are prohibited.

Examples

True:

5   
73  
101 
1073
17
22
36
55
99  

False:

8
18
23
27
98
90
88
72
68

A handful of large numbers:

46656 true
46657 true
46658 true
46659 true
46660 true
46661 false
46662 false
46663 true
46664 false
46665 true
46666 true
46667 false
46668 false
46669 false
46670 true
46671 true

Answer 1

Jelly, 7 bytes

bRṫ€3ċJ

Returns the number of bases (non-zero being truthy, zero being falsy) in which the input is a three digit number ending in one.

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How it works

bRṫ€3ċJ  Main link. Argument: n

 R       Range; yield [1, ..., n].
b        Base; convert n to bases 1, ..., n.
  ṫ€3    Tail each 3; remove the first two elements of each digit array.
      J  Indices of [n]; yield [1].
     ċ   Count the number of times [1] appears in the result to the left.

Answer 2

JavaScript (ES7), 43 40 39 bytes

f=(n,b)=>n<b*b?0:n%b==1&n<b**3|f(n,-~b)

Test cases

f=(n,b)=>n<b*b?0:n%b==1&n<b**3|f(n,-~b)

console.log('[Truthy]')
console.log(f(5))
console.log(f(73))
console.log(f(101))
console.log(f(1073))
console.log(f(17))
console.log(f(22))
console.log(f(36))
console.log(f(55))
console.log(f(99))
console.log(f(46656))
console.log(f(46657))
console.log(f(46658))
console.log(f(46659))
console.log(f(46660))
console.log(f(46663))
console.log(f(46665))
console.log(f(46666))
console.log(f(46670))
console.log(f(46671))

console.log('[Falsy]')
console.log(f(8))
console.log(f(18))
console.log(f(23))
console.log(f(27))
console.log(f(98))
console.log(f(90))
console.log(f(88))
console.log(f(72))
console.log(f(68))
console.log(f(46661))
console.log(f(46662))
console.log(f(46664))
console.log(f(46667))
console.log(f(46668))
console.log(f(46669))

Commented

f = (n,           // given n = input
        b) =>     // and using b = base, initially undefined
  n < b * b ?     // if n is less than b²:
    0             //   n has less than 3 digits in base b or above -> failure
  :               // else:
    n % b == 1 &  //   return a truthy value if n is congruent to 1 modulo b
    n < b**3 |    //   and n is less than b³ (i.e. has less than 4 digits in base b)
    f(n, -~b)     //   or the above conditions are true for some greater value of b

Answer 3

Python 3, 50 47 bytes

-2 bytes thanks to @LeakyNun
-1 byte thanks to @Dennis

lambda n:any(i>n/i/i>n%i==1for i in range(2,n))

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Answer 4

Haskell, 41 40 bytes

f n=or[mod n k==1|k<-[2..n],k^2<n,n<k^3]

Thanks to @Zgarb for golfing off 1 byte!

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Answer 5

Brachylog, 10 bytes

≥ℕ≜;?ḃ₍Ṫt1

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Answer 6

05AB1E, 11 8 bytes

Saved 3 bytes thanks to Adnan.

Lв3ù€θ1å

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Explanation

Lв            # convert input to bases [1 ... input]
  ʒg3Q}       # keep only elements of length 3
       €θ     # get the last item of each
         1å   # is there any 1?

Answer 7

Jelly, 12 bytes

bRµL€żṪ€3,1e

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Answer 8

Mathematica, 43 bytes

!FreeQ[IntegerDigits[#,2~Range~#],{_,_,1}]&

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or Try it online! (large numbers)

Martin Ender saved 3 bytes

Answer 9

Wolfram Language (Mathematica), 35 bytes

Or@@Array[x~Mod~#==1<x/#^2<#&,x=#]&

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Explicitly checks whether n % i = 1 and i² < n < i³ for any possible base i. For golfing purposes, the inequality is rearranged to 1 < n/i² < i, so that it can be chained onto the equality.

Answer 10

Clean, 58 56 bytes

-2 thanks to Dennis

import StdEnv
@n=or[n>m^2&&n<m^3&&n rem m==1\\m<-[2..n]]

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Answer 11

Husk, 10 bytes

€;1ṠMo↓2Bḣ

Try it online! Pretty close to the Jelly answer of Dennis.

Answer 12

APL (Dyalog Unicode), 21 20 14 bytes^SBCS

-5 thanks to @ngn.

Purely arithmetic solution (doesn't actually do any base conversions) and thus very fast.

3∊⊢(|×∘⌈⍟)⍨1↓⍳

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⊢(…)⍨1↓⍳ on one dropped from the ɩndices 1…argument and the argument, apply:

 | the division remainders

 ×∘⌈ times the rounded-up

 ⍟ log_N Argument

3∊ is three a member of that?

Answer 13

Pyth, 10 bytes

}]1>R2jLQS

Verify all the test cases.

Answer 14

Husk, 15 bytes

V§&o=1→o=3LṠMBḣ

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Explanation

V§&(=1→)(=3L)ṠMBḣ  -- implicit input, for example: 5
             ṠMB   -- map "convert 5 to base" over..
                ḣ  --   range [1..5]
                   -- [[1,1,1,1,1],[1,0,1],[1,2],[1,1],[1,0]]
V                  -- does any of the elements satisfy the following
 §&( 1 )( 2 )      --   apply functions 1,2 and join with & (logical and)
         =3L       --     is length equals to 3?
    =1→            --     is last digit 1?

Answer 15

PHP, 48+1 bytes

while(++$b**2<$n=$argn)$n%$b-1|$n>$b**3||die(1);

exits with 0 for falsy (or input 3), 1 for truthy.
Run as pipe with -nR or try it online.

Answer 16

C, 60 bytes

A function that returns non-zero if the argument can be represented as a three-digit number ending in 1:

i,j;f(n){for(j=0,i=sqrt(n);i>cbrt(n);)j+=n%i--==1;return j;}

Note: this works with GCC, where the functions are built-in. For other compilers, you probably need to ensure that the argument and return types are known:

#include<math.h>

Explanation

The lowest base in which n is represented in 3 digits is ⌊∛n⌋, and the lowest base in which n is represented in 2 digits is ⌊√n⌋, so we simply test whether the number is congruent to 1 modulo any bases in the 3-digit range. We return the count of the number of bases satisfying the condition, giving a non-zero (truthy) or zero (falsy) value as appropriate.

Test program

Pass any number of inputs as positional parameters:

#include<stdio.h>
int main(int c,char**v)
{
    while(*++v)
        printf("%s => %d\n", *v, f(atoi(*v)));
}

Answer 17

APL (Dyalog Unicode), 19 bytes^SBCS

Dennis' method.

(⊂,1)∊2↓¨⊢⊥⍣¯1¨⍨1↓⍳

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(⊂,1)∊ Is [1] a member of

2↓¨ two elements dropped from each of

⊢⊥⍣¯1¨⍨ the argument represented in each of the bases

1↓⍳ one dropped from the ɩndices 1 through the argument?

Answer 18

Julia, 31 bytes

n->any(i>n/i^2>n%i==1for i=2:n)

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Answer 19

Pyt, 35 33 bytes

←ĐĐ3=?∧∧:ŕĐ2⇹Ř⇹Ľ⅟⌊⁺3=⇹Đ2⇹Ř%*ž1⇹∈;

Explanation:

←ĐĐ                                             Push input onto stack 3 times
   3=?  :                       ;               If input equals 3, execute code after the question mark;otherwise, execute code after the colon. In either case, afterwards, execute the code after the semicolon
      ∧∧                                        Get 'True'
        :                                       Input not equal to 3
         ŕ                                      Remove 'False'
          Đ2⇹Ř                                  Push [2,3,...,n]
              ⇹Ľ⅟⌊⁺                             Push [floor(log_2(n))+1,floor(log_3(n))+1,...,floor(log_n(n))+1]
                   3=                           Is each element equal to 3
                     ⇹                          Swap the top two elements on the stack (putting n back on top)
                      Đ2⇹Ř                      Push [2,3,...,n]
                          %                     Push [n%2,n%3,...,n%n]
                           *                    Multiply [n%x] by the other array (i.e. is floor(log_x(n))+1=3?)
                            ž                   Remove all zeroes from the array
                             1⇹∈                Is 1 in the array?
                                ;               End conditional
                                                Implicit print

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Answer 20

><>, 42 bytes

1<v?)}:{*::+
n~\1-:::**{:}):?!n~:{:}$%1=:?

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Returns 10 for truthy, 00 for falsey.