Digit Cubes

Number Theory Level 3

Find the greatest 3-digt number (the first is not 0) where $\overline { abc }$ is the number and:

${ a }^{ 3 }{ +b }^{ 3 }{ +c }^{ 3 }=100a+10b+c$

For example, $153$ satisfies the conditions since ${ 1 }^{ 3 }{ +5 }^{ 3 }{ +3 }^{ 3 }=153$ .

The answer is 407.

4 solutions

Mardokay Mosazghi
Jul 19, 2014

Those numbers are called Armstrong numbers, and the highest 3 digit Armstrong number is $407 =4^3+0^3+7^3$ , in base 10 the Armstrong numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, 407.

how to SOLVE it????????

Chandrachur Banerjee - 6 years, 10 months ago

Is there any alternative way to solve this one?

ALLU PHANINDRA - 6 years, 10 months ago

do u know how to convert them to other bases for e.g. converting 407 to base 3???? @Mardokay Mosazghi

Harshi Singh - 5 years, 10 months ago

@Harshi Singh Yes i do

Mardokay Mosazghi - 5 years, 9 months ago

oh is it by using computer programs if not so then please tell the method

Harshi Singh - 5 years, 9 months ago

The way that I solved it is as follows(I did not include 0, so I got 371), but start with a = 9, then 729 + b^3 + c^3 = 900 + 10b + c, or b^3 - 10b = 171 + c - c^3. I made a table for c = 1 to c = 9, and computed the right hand side. Then make a table for b = 1 to b = 9 and compute b^3 - 10b. The nice thing is that the b table will be the same regardless of a. You then check for a match betweenthe b table and the c table. If none, lower a and repeat. Note that for each a, you get a new table for c, but use the same one for b. The first match came when a = 3, c = 1, and b = 7. The match was 273. Unfortunately, I neglected 0, but the scheme would work.

Edwin Gray - 2 years, 3 months ago

Deep Shah
Mar 28, 2015

This are Armstrong Numbers

Here is C code

include<stdio.h>

int cubes (int n) { return n^{3}; }

void main()

{

int i,j,k;

int ans;

for(i=1;i<=9;i++)

    for(j=0;j<=9;j++)

        for(k=0;k<=9;k++)

            if((cubes(i)+cubes(j)+cubes(k))==(i*100+j*10+k))

                ans = (i*100+j*10+k);


printf("%d\n",ans);

}

Vishwesh Ramanathan
Aug 16, 2014

To solve it you may have to make a table 1^3=1. 1+0 2^3=8. 8+4 3^3=27. 7+6 4^3=64. 4+0 5^3=125 5+0 6^3=216. 6+0 7^3=343. 3+4 8^3=512. 2+6 9^3=729. 9+0 The second column gives us the overall picture that if you want that number as last digit what last digit should you left out after addition of first two digits. Now apply brute force. This method reduces no: of hit and trials by a millionth .

Josh Speckman
Jul 29, 2014

You can just use a program to brute-force it:

for i in range(1,10):
  for j in range(0,10):
    for k in range(0,10):
      if 100*i + 10*j + k == i**3 + j**3 + k**3:
        print str(i) + str(j) + str(k)

Digit Cubes

The answer is 407.

4 solutions

include<stdio.h>

0 pending reports