Puzzles
8 December
Angel wrote out a muliplication square for the numbers from 1 to 3 (the table has the numbers 1 to 3 in the top row and left column, then every other
entry is equal to the number at the top of its column multiplied by the number at the left of its row):
The sum of the numbers in the bottom row is 18. The sum of all the numbers in the table is 36.
Angel then wrote out another multiplication square with the numbers from 1 to \(n\). The sum of all the numbers in the new table is 2025. What is the
sum of the numbers in the bottom row of the new table?
Show answer
Hide answer
For a multiplication table from 1 to \(n\),
the sum of all the numbers in the top row is \(1+...+n\),
the sum of all the numbers in the second row is \(2(1+...+n)\),
the sum of all the numbers in the third row is \(3(1+...+n)\),
..., and the sum of all the numbers in the \(n\)th row is \(n(1+...+n)\).
The sum of all the numbers is therefore \((1+...+n)(1+...+n)=n^2(n+1)^2/4\) and the sum of the numbers in the bottom row is \(n^2(n+1)/2\).
If the sum of all the numbers is 2025, then \(n=9\) and so the sum of the numbers in the bottom row if 405.
7 December
Carol organised a knockout competition in December 2024, which 6 people entered. There were 2 matches in the first round with the remaining two players given byes
(so they went into the next round without playing a match). The second round was made up of two semi-finals, then one final match was played to decide the winner.
In total 5 matches were played.
This year, Carol is organising the competition again, but it has become a lot more popular: 355 people have entered.
While planning the tournament, she can decide which rounds to give people byes in.
What is the smallest number of matches that could be included in the tournament?
Show answer
Hide answer
In each match, one person is knocked out, so for a tournament with \(n\) players, you'll always need \(n-1\) matches (as everyone except the winner must be knocked out).
Therefore there will be 354 matches however Carol arranges the byes.
6 December
Put the digits 1 to 9 (using each digit exactly once) in the boxes so that the sums are correct. The sums should be read left to right and top to bottom ignoring the usual order of operations. For example, 4+3×2 is 14, not 10.
Today's number is the product of the numbers in the red boxes.
| + | | × | | = 20 |
| + | | × | | – | |
| + | | × | | = 26 |
| × | | ÷ | | + | |
| × | | – | | = 28 |
=
32 | | =
2 | | =
11 | |
5 December
The number 36 is equal to two times the product of its digits.
What is the only (strictly positive) number that is equal to four times the product of its digits?
Show answer
Hide answer
We know the answer has three digits: we are looking for a number ABC that is equal to 4×A×B×C. Dividing by A, we see that 4×B×C > 100 and so B×C > 25. This
means that the only possible options for BC are 65, 66, 74, 75, 76, 77, 84, 85, 86, 87, 88, 93, 94, 95, 96, 97, 98 and 99.
Trying these options with different values for A, we find that 384 works.
4 December
Some numbers can be written as the sum of four consecutive numbers, for example: 142 = 34 + 35 + 36 + 37.
What is the mean of all the three-digit numbers that can be written as the sum of four consecutive numbers?
Show answer & extension
Hide answer & extension
The sum of four consecutive numbers starting at \(n\) is \(n + (n + 1) + (n + 2) + (n + 3) = 4n+6=4(n+1)+2\). This tells us that the numbers
that are the sum of four consecutive numbers are the numbers that are two more than multiples of 4. The smallest three digit number of this form is 102 and the largest is 998: the mean of these
two is 550. The second smallest and second largest are 106 and 994: the mean of these is also 550. All the numbers can be paired up in this way except for 550 itself, and so the mean of all the
numbers is 550.
Extension
What is the mean of all 3-digit numbers that can be written as the sum of \(k\) consective numbers, for different values of \(k\)? When is the answer an integer and when is it not?
3 December
Holly picks the number 513, reverses it to get 315, then adds the two together to make 828.
Ivy picks a three-digit number, reverses it, then adds the two together to make 968. What is the smallest number that Ivy could have started with?
Show answer
Hide answer
If Ivy's number is ABC, then A+C must be 8 or 18 (as this makes this final digit of the sum correct). A+C cannot be 18 of the sum would be bigger than 1000, so A+C must be 8.
The middle digit tells us that 2×B must be 6 or 16. As the first digit of the sum is 9 and not 8, there must be a 1 carried over into the hundreds column, so 2×B = 16 and B = 8.
Overall, we know that Ivy's number is A8C, with A+C = 8. The smallest possible number Ivy could have started with is therefore 187.
2 December
Eve writes down the numbers from 1 to 10 (inclusive). In total she write down 11 digits.
Noel writes down the number from 1 to 100 (inclusive). How many digits does he write down?
Show answer
Hide answer
Noel writes down 9 one-digit numbers, 90 two-digit numbers and 1 three-digit number. In total, these include 1×9 + 2×90 + 3×1 = 192 digits.
1 December
Some numbers contain a digit more than once (eg 313, 111, and 144). Other numbers have digits that are all different (eg 123, 307, and 149).
How many three-digit numbers are there whose digits are all different?
Show answer
Hide answer
There are 9 options for the first digit of the number (1 to 9). There are then 9 options for the second digit (0 to 9 excluding the digit already used),
and 8 options for the third digit (0 to 9 excluding the two digits already used). In total, this makes 9×9×8 = 648 numbers whose digits are all different.
