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.