Puzzles

Between 100 and 109 (inclusive), there are 5 integers whose digits add up to an even number, and 5 whose digits add up to an odd number. Between 110 and 119 (inclusive), there are 5 integers whose digits add up to an even number, and 5 whose digits add up to an odd number... In general, between \(10n\) and \(10n+9\) (inclusive), there are 5 integers whose digits add up to an even number, and 5 whose digits add up to an odd number.

The integers from 100 to 999 (inclusive) can be split into 45 sets of integers from \(10n\) to \(10n+9\) (and the digits of 1000 don't add to an even number), so there are 450 integers between 100 and 1000 whose digits add up to an even number.

195.

These numbers all have 3 digits are there are 5 choices for each digit (5, 6, 7, 8, or 9). The number of numbers is therefore 5×5×5=125.

There are two 1-digit numbers whose digits are all either 1 or 2 and who do not contain two 2s in a row (1, and 2).

There are three 2-digit numbers whose digits are all either 1 or 2 and who do not contain two 2s in a row (11, 12, and 21).

There are five 3-digit numbers whose digits are all either 1 or 2 and who do not contain two 2s in a row (111, 112, 121, 211, and 212).

There are eight 4-digit numbers whose digits are all either 1 or 2 and who do not contain two 2s in a row (1111, 1112, 1121, 1211, 1212, 2111, 2112, and 2121).

(It looks like these are the Fibonacci numbers.)

All these numbers either end in a 1, or end in 12. Therefore the \((n+1)\)-digit number can be made by appending a 1 to the end of the \(n\)-digit numbers or appending a 12 to the end of the \((n-1)\)-digit numbers: so the next term is always the sum of the previous two terms.

Continuing the pattern, we see that there are 987 14-digit numbers are there whose digits are all either 1 or 2 and who do not contain two 2s in a row.

There is 1 number whose digits are all non-zero and whose digits add up to 1 (1).

There are 2 numbers whose digits are all non-zero and whose digits add up to 2 (2, and 11).

There are 4 numbers whose digits are all non-zero and whose digits add up to 3 (3, 12, 21, and 111).

There are 8 numbers whose digits are all non-zero and whose digits add up to 4 (4, 13, 31, 112, 121, 211, and 1111).

The amount of numbers is doubling each time. You can justify this by noticing that every number whose digits are all non-zero and whose digits add to \(n+1\) can be made from a number adding to \(n\) by either adding 1 to the final digit or appending a 1 onto the end of the number.

Therefore, there are 128 numbers whose digits are all non-zero and whose digits add up to 8.

There is 1 number whose digits are all non-zero and whose digits add up to 1. There are 2 numbers whose digits are all non-zero and whose digits add up to 2. There are 4 numbers whose digits are all non-zero and whose digits add up to 3. There are 8 numbers whose digits are all non-zero and whose digits add up to 4. There are 16 numbers whose digits are all non-zero and whose digits add up to 5. There are 32 numbers whose digits are all non-zero and whose digits add up to 6. There are 64 numbers whose digits are all non-zero and whose digits add up to 7. There are 128 numbers whose digits are all non-zero and whose digits add up to 8. There are 256 numbers whose digits are all non-zero and whose digits add up to 9.

Are there 512 numbers whose digits are all non-zero and whose digits add up to 10?

If you add these red numbers to the start of each row and column, then every number in the grid is the sum of the red numbers in its row and column

However you pick one number from each row and column, you will always end up with the total of all the red numbers. This is 369.

20 can be written as the product of one-digit numbers greater than 1 in the following ways:

Our 12-digit numbers must contain one of these plus lots of 1s.

In the first case, there are 12×11=132 different possible positions for the 4 and the 5.

In the second case, there are 12×11×10×½=660 different possible positions for the 2s and the 5.

In total, this makes 132+660=792 numbers.

252 can be written as the product of one-digit numbers in the following ways:

Therfore, the smallest number whose digital product is 252 is 479.