Puzzles
8 December
The sum of three integers is 51. The product of the same three integers is 836. What is the product of largest integer and the second-largest integer?
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836=2×2×11×19, and so for the product to be correct the integers could be:
- 1, 1, and 836 (=2×2×11×19)
- 1, 2, and 418 (=2×11×19)
- 1, 11, and 76 (=2×2×19)
- 1, 19, and 44 (=2×2×11)
- 1, 4 (=2×2), and 209 (=11×19)
- 1, 22 (=2×11), and 38 (=2×19)
- 2, 2, and 209 (=11×19)
- 2, 11, and 38 (=2×19)
- 2, 19, and 22 (=2×11)
- 4 (=2×2), 11, and 19
The only one of these whose sum is 51 are 2, 11, and 38. Therefore the product of the largest and second-largest integers is 418
7 December
The picture below shows eight regular decagons. In each decagon, a red triangle has been drawn with vertices at three of the vertices of the decagon.
The area of each decagon is 240. What is the total area of all the red triangles?
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By rotating the triangles, they can be made to exactly cover the decagon.
The total area of all the triangles is therefore 240.
6 December
When 12345 is divided by today's number, the remainder is 205. When 6789 is divided by today's number, the remainder is 112.
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The first division tells that 12345 is 205 more than a mutliple of today's number, and so today's numwber is a factor of 12140.
The second division tells that 6789 is 112 more than a mutliple of today's number, and so today's number is a factor of 6677.
The only numbers that are factors of both 12240 and 6677 are 1 and 607.
5 December
How many different isosceles triangles are there whose perimeter is 50 units, and whose area is an integer number of square-units?
(Two triangles that are rotations, reflections and translations of each other are counted as the same triangle. Triangles with an area of 0 should not be counted.)
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The largest possible isosceles triangles with a perimeter of 50 units is the equilateral triangle with sides of length 50/3. The area of this triangle is 120.28 square-units.
By continuously adjusting the width of the base of the isosceles triangle, triangles with every area between 0 and 120.28 can be created. The animation below shows this.
Triangles with areas of each integer from 1 to 120 can therefore be created. Each area can be made twice: once with a tall and thin triangle, and once with a short and wide triangle.
This means that there are 240 different triangles with a perimeter of 50 units and an integer area.
4 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.
| + | | - | | = 5 |
| ÷ | | × | | × | |
| + | | - | | = 5 |
| - | | ÷ | | ÷ | |
| + | | × | | = 10 |
=
-6 | | =
18 | | =
35 | |
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| 6 | + | 4 | - | 5 | = 5 |
| ÷ | | × | | × | |
| 3 | + | 9 | - | 7 | = 5 |
| - | | ÷ | | ÷ | |
| 8 | + | 2 | × | 1 | = 10 |
=
-6 | | =
18 | | =
35 | |
The product of the numbers in the red boxes is 160.
3 December
If you write out the numbers from 1 to 1000 (inclusive), how many times will you write the digit 0?
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The numbers 10, 20, ..., 1000 end with a 0. There are 100 of these numbers.
The numbers 100, 101, ..., 109, 200, ..., 209, 300, ..., 309, ..., 908, 909, 1000 contain a 0 in their tens column. There are \(10\times9+1=91\) of these numbers.
The number 1000 has a 0 in its hundreds column.
In total, there are 100+91+1=192 zeros.
2 December
The number \(7n\) has 37 factors (including 1 and the number itself). How many factors does \(8n\) have?
There was a typo in this puzzle. It originally read "38 factors" when it was meant to say "37 factors".
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If \(p_1\), \(p_2\), etc are distinct prime numbers, then the factors of the number \(N=p_1^{a_1}p_2^{a_2}...\) all have the form \(p_1^{b_1}p_2^{b_2}...\),
where \(b_1\) is 0 or 1 or ... or \(a_1\) (and similar conditions for \(a_2\), \(a_3\), etc). There are therefore \(a_1+1\) possible values of \(b_1\), \(a_2+1\) possible values
of \(b_2), and so on. Therefore, \(N\) has \((a_1+1)(a_2+1)(...)\) factors.
From this we see that the only way to have a number with 37 factors is if it is a prime number to the power of 36, and so \(n=7^{35}\).
\(8n\) is therefore equal to \(2^3\times7^{35}\). This has 4×36=144 factors.
Extension
Which other numbers can 37 be replaced with leaving a puzzle that still has a unique solution?
1 December
The geometric mean of a set of \(n\) numbers can be computed by multiplying together all the numbers then computing the \(n\)th root of the result.
The factors of 4 are 1, 2 and 4. The geometric mean of these is 2.
The factors of 6 are 1, 2, 3, and 6. The geometric mean of these is \(\sqrt{6}\).
The geometric mean of all the factors of today's number is 22.
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The geometric mean of a number \(N\) is always \(\sqrt{N}\). You can see why this is true by considering the pairs of factors that multiply to make the number:
if \(N\) if not square and has \(k\) pairs of factors, then the product of these factors is \(N^k\), so the geometric mean is \((N^k)^{1/(2k)}=N^{1/2}\);
if \(N\) is square and has \(k\) pairs of factors plus the square root, then the product of its factors is \(N^k\sqrt{N}=N^{k+1/2}\), so the geometric mean is \((N^{k+1/2})^{1/(2k+1)}=N^{1/2}\).
Interestingly, this means that the geometric mean of the factors of a number is an integer only when the number is square.
Therefore the only number whose factors have a geometric mean of 22 is \(22^2\), or 484.
![](javascript:showlimage("advent2021-decagons.png"))
![](javascript:showlimage("advent2021-dodec-ans.png"))
![](javascript:showlimage("advent2021-isosceles.gif"))