Question 776097
<pre>
Suppose the 5-digit number N is  

"ABCDE", that is 

N = 10000A+1000B+100C+10D+E = (9999+1)A+(999+1)B+(99+1)C+(9+1)D+E
= (9999A+999B+99C+9D)+A+B+C+D+E = 9(1111A+111B+11C+D)+A+B+C+D+E

Let S = A+B+C+D+E, then

N = 9K+S where K = 1111A+111B+11C+D
 
There are P(5,3) = 60 3-digit numbers. If we were to add all 
60 of these digits, the 5 digits of N will therefore occur 
12 times among: 

1. the units digits, which will give a sum of 12(A+B+C+D+E)
2. the tens digits, which will give a sum of 120(A+B+C+D+E)
3. the hundreds digits, which will give a sum of digits of 1200(A+B+C+D) 

So the sum of all 60 3-digit numbers will be

12(A+B+C+D+E) + 120(A+B+C+D+E) + 1200(A+B+C+D+E) = 1332(A+B+C+D+E) = 1332S

N = 9K+S = 1332S

So therefore the 5-digit number must be 1332S

N = 9K+S = 1332S
      9K = 1331S
       S = {{{9K/1331}}}

Since S is an integer, K is a multiple of 1331,

Let K = 1331M

Then S = 9M

and so S is a multiple of 9.

The smallest S could be is 1+2+3+4+5 = 15
The largest S could be is 9+8+7+6+5 = 35

The only multiples of 9 between those values are 18 and 27

If S = 18, M=2, K=1331(2) = 2662, and 

N = 9(2662)+18 = 23976 and 2+3+9+7+6 = 27  

If S = 27, M=3, K=1331(3) = 3993, and 

N = 9(3993)+27 = 35964 and 3+5+9+6+4 = 27 

So there are two possibilities for N, 23976 and 35964,
but in either case the sum of the digits is 27.

27 is 3<sup>3</sup> so the answer is (b) cube.

Edwin</pre>