# Puzzles

## 24 December

There are six 3-digit numbers with the property that the sum of their digits is equal to the product of their digits. Today's number is the largest of these numbers.

## 22 December

In bases 3 to 9, the number 112 is:
\(11011_3\),
\(1300_4\),
\(422_5\),
\(304_6\),
\(220_7\),
\(160_8\), and
\(134_9\).
In bases 3, 4, 6, 8 and 9, these representations contain no digit 2.

There are two 3-digit numbers that contain no 2 in their representations in all the bases between 3 and 9 (inclusive). Today's number is the smaller of these two numbers.

## 17 December

Eve picks a three digit number then reverses its digits to make a second number. The second number is larger than her original number.

Eve adds her two numbers together; the result is 584. What was Eve's original number?

#### Show answer

#### Hide answer

For the first and last digits of the result to be different, there must be a 1 carried from the middle column. Therefore the middle digit of Eve's number is 9.
Eve's number could be 292 or 193, but it cannot be the first as reversing this is not larger than her original number, so here number is **193**.

## 1 December

If you write out the numbers from 1 to 1000 (inclusive), how many times will you write the digit 1?

#### Show answer

#### Hide answer

100 numbers will have 1 in the units column. Another 100 with have a 1 in the tens column; and other 100 will have a 1 in the hundreds column.
Finally, 1000 has a 1 in the thousands column.

In total, this makes **301** copies of the digit 1.

## Not Roman numerals

The letters \(I\), \(V\) and \(X\) each represent a different digit from 1 to 9. If

$$VI\times X=VVV,$$

what are \(I\), \(V\) and \(X\)?

#### Show answer

#### Hide answer

For any digit \(V\), \(VVV\) is a multiple of 111, and \(111=37\times3\). 37 is prime, so \(VI\) must be a multiple of 37 (as \(X\) is less than 10 so cannot be a multiple of 37). Therefore \(VI\) is either 37 or 74.

If \(VI\) was 74, then \(VVV\) is 777. But then \(VI\times X\) is even and \(VVV\) is odd. This is impossible, so \(VI\) must the 37.

Now that we know that \(VI\) is 37, we see that

$$37\times X=333,$$

and so \(X=9\) and the final solution is

$$37\times9=333.$$

## 22 December

In base 2, 1/24 is
0.0000101010101010101010101010...

In base 3, 1/24 is
0.0010101010101010101010101010...

In base 4, 1/24 is
0.0022222222222222222222222222...

In base 5, 1/24 is
0.0101010101010101010101010101...

In base 6, 1/24 is
0.013.

Therefore base 6 is the lowest base in which 1/24 has a finite number of digits.

Today's number is the smallest base in which 1/10890 has a finite number of digits.

Note: 1/24 always represents 1 divided by twenty-four (ie the 24 is written in decimal).

#### Show answer

#### Hide answer

If 1/10890 has a finite number of digits in base \(b\), then \(b^n/10890\) is an integer for some integer \(n\). In order for this to be
possible \(b\) must be a mutiple of each prime factor of 10890: the smallest such \(b\) will be when the exponent of each prime factor is 1.

The prime factorisation of 10890 is 2×3^{3}×5×11^{2}. Therefore the smallest base in which 1/10890 has a
finite number of digits is 2×3×5×11=**330**.

## 15 December

Today's number is smallest three digit palindrome whose digits are all non-zero, and that is not divisible by any of its digits.

#### Show answer

#### Hide answer

The number can contain no 1 (as all numbers are multiples of 1). It's first and last digits cannot be 2 (as it would then be even and so a multiple of 2).

The lowest three-digit palindrome containing no 0 or 1 and not ending in 2 is **323**. This is not a multiple of 2 or 3 and so is the solution.

## 9 December

Today's number is the number of numbers between 10 and 1,000 that contain no 0, 1, 2 or 3.

#### Show answer

#### Hide answer

There are 6 available choices for each digit, so there are 6^{2} and 6^{3} such two- and three-digit numbers. 6^{2}+6^{3}=**252**.