Write the digits of 2015 in a column.
Then make another column on the right, and add 1 to each digit.
Then make a third column, where you multiply the second column by 2.
And then in a fourth column, raise the third column to the power 3.

    2         3         6         216    
0 1 2 8
1 2 4 64
5 6 12 1728

See what a logical progression you have: one, two, three; and plus, times, exponent!
Now add up the final column. -- you get 2016!

There is only one other four-digit number that does this.
The product of its digits is 120.
What is the number?
Solution to the Problem:

The answer is 8135.

Since the product of the digits of the number we're looking for is 120, the only digits possible are 1, 2, 3, 4, 5, 6, and 8.   There has to be a 5, since it is a prime factor of 120 and no other single digit has 5 as a factor.   The three digits other than 5 must have a product of 24.   Therefore they have to be either 1, 4, and 6; or 1, 3, and 8; or 2, 2, and 6; or 2, 3, and 4.

The four possible sets listed above for the remaining three digits would give these sets of numbers in the fourth column, respectively: 64, 1000, and 2744; or 64, 512, and 5832; or 216, 216, and 2744; or 216, 512, and 1000.

The sums of each of these sets of three numbers plus 1728 are respectively 5536; 8136; 4904; and 3456.   We can simply subtract 1 from each of these and check whether the resulting number has digits that multiply to make 120: only 8135 works.   So:
    8        9         18         5832    
1 2 4 64
3 4 8 512
5 6 12 1728
The sum of the last column is 8136.
8135 is the only number with this property, whose digits have a product of 120.

Correctly solved by:

1. James Alarie Flint, Michigan
2. Mr. Ng Ser Hong Singapore, Singapore
3. Marco Morelli Fermo, Italy
4. Sreeroopa Sankararaman Singapore, Singapore
5. Chase Smith Mountain View High School,
Mountain View, Wyoming