For single-digit numbers, we have $x^3 \equiv x$ mod $10,$ which translates to $x \equiv 0, \pm 1$ mod $5.$ There are five numbers in the range $[1,9]$ satisfying that congruence.

For two-digit numbers, we have $x^3 \equiv x$ mod $100,$ which translates to $25|x(x-1)(x+1)$ and $4|x(x-1)(x+1).$ This means $x \equiv 0, \pm 1$ mod $25$ and $x \equiv 0, \pm 1$ mod $4.$ There are nine numbers mod $100$ that satisfy these congruences (by the Chinese Remainder Theorem ). They're all of the form $25k$ or $25k \pm 1$ for some nonnegative integer $k,$ so the only ones less than $10$ that work are $0$ and $1$ , so there are seven two-digit numbers.

For three-digit numbers, we have $x^3 \equiv x$ mod $1000,$ which translates to $125|x(x-1)(x+1)$ and $8|x(x-1)(x+1).$ This means $x \equiv 0, \pm 1$ mod $125$ and $x \equiv 0, 1, 3, 5, 7$ mod $8.$ There are fifteen numbers mod $1000$ satisfying these congruences (by the Chinese Remainder Theorem ). They're all of the form $125k$ or $125k \pm 1$ for some nonnegative integer $k,$ so the only ones less than $100$ that work are, again, $0$ and $1.$ That leaves thirteen three-digit numbers.

So the final answer is $13 + 7 + 5 = \fbox{25}.$

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k = 0 ;
for m = 1:999
    if mod(m^3-m,10^(floor(log(m)/log(10))+1))==0
        k = k+1 ;
    end
end
k 

$\boxed{k=25}$

By direct calculation, the single digits with this property are $0$ , 1 , 4 , 5 , 6 , and 9 (its cubes are 0, 1, 64, 125, 216, and 729 respectively).

For two digit numbers, if we will let $a$ be the first digit and $b$ be the last digit, then $(10a + b)^3 \equiv 10a + b \mod{100}$ would describe this property, and this simplifies to $(3b^2 - 1)a \equiv \frac{b - b^3}{10} \mod{10}$ . Two digit numbers with this property must end in the single digits with this property, so we can use this formula by setting $b$ equal to the single digits listed above and solving for $a$ . Therefore:

If $b = 0$ , then $9a \equiv 0 \mod {10}$ for $0 \leq a \leq 9$ , so $a = 0$ , for 00 (a repeat, and not positive).

If $b = 1$ , then $2a \equiv 0 \mod{10}$ for $0 \leq a \leq 9$ , so $a = 0$ or $a = 5$ , for 01 (a repeat) and 51 .

If $b = 4$ , then $7a \equiv 4 \mod{10}$ for $0 \leq a \leq 9$ , so $a = 2$ , for 24 .

If $b = 5$ , then $4a \equiv 8 \mod{10}$ for $0 \leq a \leq 9$ , so $a = 2$ or $a = 7$ , for 25 and 75 .

If $b = 6$ , then $7a \equiv 9 \mod{10}$ for $0 \leq a \leq 9$ , so $a = 7$ , for 76 .

If $b = 9$ , then $2a \equiv 8 \mod{10}$ for $0 \leq a \leq 9$ , so $a = 4$ or $a = 9$ , for 49 and 99 .

For three digit numbers, if we let $a$ be the first digit and $b$ be the value of the last two digits, then $(100a + b)^3 \equiv 100a + b \mod{1000}$ would describe this property, and this simplifies to $(3b^2 - 1)a \equiv \frac{b - b^3}{100} \mod{10}$ . Three digit numbers with this property must end in the two digit numbers with this property, so we can use this formula by setting $b$ equal to the two digit numbers listed above and solving for $a$ . Therefore:

If $b = 00$ , then $9a \equiv 0 \mod{10}$ for $0 \leq a \leq 9$ , so $a = 0$ , for 000 (a repeat, and not positive).

If $b = 01$ , then $2a \equiv 0 \mod{10}$ for $0 \leq a \leq 9$ , so $a = 0$ or $a = 5$ , for 001 (a repeat) and 501 .

If $b = 24$ , then $7a \equiv 2 \mod{10}$ for $0 \leq a \leq 9$ , so $a = 6$ , for 624 .

If $b = 25$ , then $4a \equiv 4 \mod{10}$ for $0 \leq a \leq 9$ , so $a = 1$ or $a = 6$ , for 125 and 625 .

If $b = 49$ , then $2a \equiv 4 \mod{10}$ for $0 \leq a \leq 9$ , so $a = 2$ or $a = 7$ , for 249 and 749 .

If $b = 51$ , then $2a \equiv 4 \mod{10}$ for $0 \leq a \leq 9$ , so $a = 2$ or $a = 7$ , for 251 and 751 .

If $b = 75$ , then $4a \equiv 2 \mod{10}$ for $0 \leq a \leq 9$ , so $a = 3$ or $a = 8$ , for 375 and 875 .

If $b = 76$ , then $7a \equiv 1 \mod{10}$ for $0 \leq a \leq 9$ , so $a = 3$ , for 376 .

If $b = 99$ , then $2a \equiv 8 \mod{10}$ for $0 \leq a \leq 9$ , so $a = 4$ or $a = 9$ , for 499 and 999 .

The positive integers less than 1000 that have all of its digits also appear at the end of its cube in the same order are 1, 4, 5, 6, 9, 24, 25, 49, 51, 75, 76, 99, 125, 249, 251, 375, 376, 499, 501, 624, 625, 749, 751, 875, and 999 for a total of $\boxed{25}$ different numbers.

A lazy solution in python 3:

count = 0
for i in range(1, 1000):
    cube = i ** 3
    if str(cube)[-len(str(i)):] == str(i):
        print("The cube of %s is %s" % (str(i), str(cube)))
        count += 1
print("The total number found is: " + str(count))

brute force with a calculator is not too hard:

1 digit: 1, 4, 5, 6, 9

extend each of the above to 2 digits: 51, 24, 25, 75, 46, 49, 99

extend all of the above to 3 digits

501, 251, 751, 624, 125, 625, 375, 875, 376, 249, 749, 499, 999

(I almost missed 501 by forgetting to expand 1 to 3 digits. I caught it by noticing they come in pairs that add to 10, 100, and 1000 with the sole exception of 5.)

It is better to read the Number theory answer. However, when considering this problem is just in a small scale as 1000, the cube will only reach 10^9. Hence, it can be solved using brute force by a simple program.

Here's python implementation of the brute-force solution, you can paste this solution to the web environment here on brilliant :

import math
sum = 0
for num in range(1, 1000):
    cube = pow(num, 3)
    num_length = int(math.log10(num)) + 1
    last_digits_of_cube = cube % pow(10, num_length)
    if last_digits_of_cube == num:
        sum += 1
        # print(str(num) + " has the cube of " + str(cube)) # Show each answer; can be commented.
print(sum)

Mathematica

Select[Range@1000, FromDigits[IntegerDigits[#^3, 10, IntegerLength@#]] == # &]

{1,4,5,6,9,24,25,49,51,75,76,99,125,249,251,375,376,499,501,624,625,749,751,875,999}

include<stdio.h>

include<conio.h>

main() { long int n,v=0,k=0,c=0; int q,i=10,p=0; for(n=1;n<1000;n++) { k=n; v=n n n; c=v-n; if (n<10) q=c%10; else { if (n<100 && n>9) q=c%100; else q=c%1000; } if (q==0) p++; } printf("\n %d",p); getch(); }

A simple solution in R:

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n = 0
for (i in 1:999) {
        if (i^3 %% 10^nchar(i) == i) {
                n = n + 1
        }
}
print(n)

https://oeis.org/A033819

Lazy but adaptable Excel Solution

Column 1 , 1-999.

Column 2 , (1-999)^3

Column 3 , Used =RIGHT(value,#chars) to cut off the last one, two and three values according to number of corresponding decimals in the first column.

Then =LEN function is used to determine how many characters are cut off of the #Rightmost values of Column 2 for Column 3.

Column 4 , Checks if matching, 1 for true 0 for false.

Used =SUM(Column4) to count matches.

• Answer = 25.

I wrote an AWK script, which produced the number 25.

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# to get the answer to a puzzle at Brilliant
#
#  the cube 24 problem: that all the digits of a number appear at
#  the end of its cube, i.e.:
#    24^3 => 13824
#  -- for all positive integers less than 1000.
#

BEGIN {
  count = 0;
  for ( t=1 ; t < 1000; t++ ) {
    tstr = t "";
    t3 = t*t*t;
    t3str = t3 "";
    found = index( t3 "", t "" );
    if ( found > 0 ) {
      if ( 0 == found ) print " found == " found " at " tstr;
      # print t "" " in " t3 "" " at " found;
      if ( ( length(t3 "") + 1 - length(t "") ) == found ) {
        count++;
        print count ": " t "" " in " t3 "" " at " found ;
      }
      if ( 0 == found ) print " found == " found " at " t " (after)";
    }
  }
  print count " numbers";
  exit(0);
  }

END {
  }

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25

1: 1 in 1 at 1
2: 4 in 64 at 2
3: 5 in 125 at 3
4: 6 in 216 at 3
5: 9 in 729 at 3
6: 24 in 13824 at 4
7: 25 in 15625 at 4
8: 49 in 117649 at 5
9: 51 in 132651 at 5
10: 75 in 421875 at 5
11: 76 in 438976 at 5
12: 99 in 970299 at 5
13: 125 in 1953125 at 5
14: 249 in 15438249 at 6
15: 251 in 15813251 at 6
16: 375 in 52734375 at 6
17: 376 in 53157376 at 6
18: 499 in 124251499 at 7
19: 501 in 125751501 at 7
20: 624 in 242970624 at 7
21: 625 in 244140625 at 7
22: 749 in 420189749 at 7
23: 751 in 423564751 at 7
24: 875 in 669921875 at 7
25: 999 in 997002999 at 7

25 numbers

Coding via Dev C++

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#include<stdio.h>
#include<conio.h>
#include<iostream>
#include<math.h>
unsigned int nm=1,dg,mod,power10,nm3,total=1;
main()
{
while(nm<1000)
    {
    using namespace std;
    dg=ceil(log10(nm));

    if(dg==floor(log10(nm)))
        dg=dg+1;

    power10=ceil(pow(10,dg));
    nm3=ceil(pow(nm,3));
    mod=nm3%power10;
    if(mod-nm==0)
    {
        printf("%d satisfies the criteria.[%d]\n",nm,total);
        total=total+1;
    }
    nm=nm+1;
    }
}

A one-line python solution. Pretty simple for an advanced problem to be honest.

I just used the string constructor str() , in order to easily check if the final digits of the cube are equal to the initial integer.

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n=0
for i in range(1,1000):
    if str(i) == str(i**3)[-len(str(i)):]:
        n+=1
print('Total:',n) 

Answer: $\boxed{25}$

Very lazy solution, which yields the result, 25 :

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print(sum([str(i**3).endswith(str(i)) for i in range(1, 1000)]))

Try it here!

This makes use of:

• The str.endswith() method .
• The fact that booleans are actually subclasses of integers in Python, so you can safely sum them.

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v=[]
for n in 1...1000
  v<<n if n==(n**3)%(10**(Math.log10(n).to_i+1))
end
puts "#{v} length: #{v.length}" 

$[1, 4, 5, 6, 9, 24, 25, 49, 51, 75, 76, 99, 125, 249, 251, 375, 376, 499, 501, 624, 625, 749, 751, 875, 999] length: \boxed{25}$