Puzzles

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 $-a^2/4=-160000$, 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 $(x-\alpha )(x-\beta )(x-\gamma )(x-\delta )(x-ϵ)=0$ gives $x^5+\cdots =\alpha \beta \gamma \delta ϵ$. 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?

The vertices of this triangle will be every other vertex of the hexagon: and other triangle can be made larger by moving one of its vertices closer to a vertex of the hexagon. The area of this triangle is 476.

There is 1 number whose digits are all non-zero and whose digits add up to 1 (1).

There are 2 numbers whose digits are all non-zero and whose digits add up to 2 (2, and 11).

There are 4 numbers whose digits are all non-zero and whose digits add up to 3 (3, 12, 21, and 111).

There are 8 numbers whose digits are all non-zero and whose digits add up to 4 (4, 13, 31, 112, 121, 211, and 1111).

The amount of numbers is doubling each time. You can justify this by noticing that every number whose digits are all non-zero and whose digits add to $n+1$ can be made from a number adding to $n$ by either adding 1 to the final digit or appending a 1 onto the end of the number.

Therefore, there are 128 numbers whose digits are all non-zero and whose digits add up to 8.

There is 1 number whose digits are all non-zero and whose digits add up to 1. There are 2 numbers whose digits are all non-zero and whose digits add up to 2. There are 4 numbers whose digits are all non-zero and whose digits add up to 3. There are 8 numbers whose digits are all non-zero and whose digits add up to 4. There are 16 numbers whose digits are all non-zero and whose digits add up to 5. There are 32 numbers whose digits are all non-zero and whose digits add up to 6. There are 64 numbers whose digits are all non-zero and whose digits add up to 7. There are 128 numbers whose digits are all non-zero and whose digits add up to 8. There are 256 numbers whose digits are all non-zero and whose digits add up to 9.

Are there 512 numbers whose digits are all non-zero and whose digits add up to 10?

The product of the numbers in the red boxes is 315.

The last three digits of $5^1$ to $5^7$ are:

This pattern continues: the last three digits of 5 to the power of any odd number (greater than 2) is 125, and the last three digits of 5 to the power of any even number (greater than 3) is 625. The last three digits of $5^{2,022,000,000}$ are therefore 625.