Puzzles
15 December
The odd factors of 2025 are 1, 3, 5, 9, 15, 25, 27, 45, 75, 81, 135, 225, 405, 675 and 2025. There are 15 of these factors and 15 is itself an odd factor of 2025.
What is the smallest three-digit number whose number of odd factors is itself an odd factor of the number?
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In order for the number of odd factors of the number to be odd, it must be an odd square number or a power of two times an odd square number.
128 is 1 (an odd square) times 128 (a power of two) and no smaller three digit number is of this form.
12 December
Mary uses the digits 1, 2, 3, 4, 5, 6 and 7 to make two three-digit numbers and a one-digit number (using each digit exactly once). The sum of her three numbers is 1000.
What is the smallest that the larger of her two three-digit numbers could be?
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The sum of the units digits of the three numbers must be 10 (as the units digit of the result must be 0 and the number are not big enough to allow the sum to be 20). Similarly, the sum of the
two tens digits must be 9, and the sum of the two hundreds digits is 9 (so that 10s are made when the 1 is carried).
The> hundreds digits could be 2 and 7, 3 and 6, or 4 and 5. As we're looking to make the larger number as small as possible, we should pick 4 and 5.
The tens digits could be 2 and 7 or 3 and 6. We pick 2 and 7 and as 2 is the smallest possible number here.
Finally, the units digits are 1, 3 and 6. We use 1 for our number as it's smallest, leading to 521.
10 December
2025 is the smallest number with exactly 15 odd factors.
What is the smallest number with exactly 16 odd factors?
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If the numbers \(p_1\), \(p_2\), ..., \(p_n\) are odd prime numbers and \(i_1\), \(i_2\), ..., \(i_n\) are positive integers, then the number \(p_1^{i_1}p_2^{i_2}...p_n^{i_n}\)
has \((i_1+1)(i_2+1)...(i_n+1)\) odd factors.
If \((i_1+1)(i_2+1)...(i_n+1)=16\) the the possible values for the \(i\)s are:
- 15
- 7 and 1
- 3 and 3
- 3, 1 and 1
- 1, 1, 1 and 1
These options lead to:
- 315 = 14348907
- 37×51 = 10925
- 33×53 = 3375
- 33×51×71 = 945
- 31×51×71×111 = 1155
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?
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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?
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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?
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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?
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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.
