Puzzles

There are \((729-n)/2\) even numbers between \(n\) and 729, and so we want to solve \((729-n)/2=n\). The solution of this is 243.

For which odd numbers \(N\) does there exist an odd number \(n\) such that there are \(n\) even numbers between \(n\) and \(N\)?

The two pentagons at the end of the pattern each contribute 4 units to the permimeter. The remaining 202 pentagons each contribute 3 units to the perimeter. This gives a total perimeter of 2×4+202×3=614.

1D is "a square number"; 2D is "two times 1A"; 3D is "today's number": today's number is 564.

The largest deteminant will be made by making \(a\) and \(d\) as large as possible (ie 12) and \(b\) and \(c) as small as possible (ie 1). This gives a determinant of 143.

The determinant of the 3 by 3 matrix \(\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}\) is \(a(ei-fh)-b(di-fg)+c(dh-eg)\).

If a 3 by 3 matrix's entries are all in the set \(\{1, 2, 3\}\), the largest possible deteminant of this matrix is 28.

What is the largest possible determinant of a 3 by 3 matrix whose entries are all in the set \(\{1, 2, 3, ..., 12\}\)?

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.

A line \(y=\text{constant}\) that is tangent to a parabola must be a tangent to the minimum (or maximum) point.

\(y=x^2-ax\) can be rewritten as \(y=(x-a/2)^2-a^2/4\). The minimum of this will be at \(y=-a^2/4\). If \(-a^2/4=-160\,000\), then \(a\) is 800.

The largest number you can make with the digits in the red boxes is 532.

The product of all the solutions is 840.

Expanding \((x-\alpha)(x-\beta)(x-\gamma)(x-\delta)(x-\epsilon)=0\) gives \(x^5 + \dots = \alpha\beta\gamma\delta\epsilon\). As long as the coefficient of \(x^5\) is 1, the constant term when written like this will be the product of all the solutions (for this to always work, you will need to include repeated solutions and complex solutions.)

The equation \(x^8 - 19x^7 + 126x^6 - 294x^5 - 231x^4 + 1869x^3 - 1576x^2 - 1556x + 1680=0\) has eight real solutions. What is the product of all these solutions?