The Third Smallest Trivia Quiz

Answer: 5

A prime number has exactly 2 factors. Historically mathematicians have gone back and forth on whether 1 should be prime, but notice that it only has 1 factor so our current definition excludes it. (See if you can spot the later statements that would not be true if 1 was prime!) The smallest primes are 2, 3, 5 and 7.

There are infinitely many primes: to show this, imagine somebody presents you with a complete list of finitely many primes. Multiply all their numbers together and add 1 to make a new number, N. Then, N can't be a multiple of anything on their list. For instance, if 2 is on their list then N is odd, so not a multiple of 2. If 3 is on their list then N is 1 more than a multiple of 3. And so forth. Therefore, the list you were given (and any finite list) is incomplete.

Answer: 6

Draw a row with 1 dot, then a row below with 2 dots, then a row below with 3 dots. You have a triangle made up of 6 dots. By adding more rows, or just by calculating 1+2+3+...+n, you can construct all the triangular numbers: 1, 3, 6, 10, 15, 21 and so on.

There are infinitely many triangular numbers, of the form n(n+1)/2 where n is a positive integer.

Answer: 7

If you divide a number by 6, you can only get remainders of 0, 1, 2, 3, 4 or 5. For instance, dividing 27 by 6 gives "4 remainder 3" under this analysis. If the remainder is 0, 2 or 4 then the number must be even. If it's 0 or 3 then it must be a multiple of 3. That leaves us with the numbers that have remainder 1 or 5.

With remainder 1, we have 1, 7, 13, 19, ...
With remainder 5, we have 5, 11, 17, 23, ...
Notice that we can substitute any positive integer n into either 6n+1 or 6n+5 to get a number of these forms, so there are infinitely many of them.

This reasoning is the basis of modular arithmetic.

Answer: 10

The smallest such number, 6, has the factors 1, 2, 3 and 6. For 10, the factors are 1, 2, 5 and 10.

There are two ways to generate a number with exactly 4 factors. Firstly, you can multiply 2 different prime numbers together (e.g. 5*7 = 35). This number is a "semiprime" with 4 factors: 1, itself, and the 2 prime numbers you used to make it.

Secondly, you can take a prime number and cube it (e.g. 4^3 = 64). If you call your prime number p then the factors are 1, p, p^2 and p^3.

Because there are infinitely many prime numbers, either method can generate infinitely many numbers with exactly 4 factors. These numbers begin: 6, 8, 10, 14, 15, 21, 22, 26, 27.

Answer: 11

The sequence goes 1, 10, 11, 21, 31, ..., 91, 100, 101, 102, 103, ..., 199, 201, 210, ...

To show there are infinitely many such numbers, think of some pattern that will extend forever: for instance, 1, 10, 100, 1000, ... (powers of 10).

As your upper limit gets larger, the percentage of numbers that contain a digit 1 increases: for instance, 19% of numbers from 1 to 100 contain a 1. Up to 1000, it is 27.1%. Up to a million, it is around 47%. In fact, as the upper bound increases it approaches 100%. However, there are still infinitely many numbers that don't contain a 1 e.g. 2, 22, 222, 2222, ... (Infinity is complicated!)

Answer: 17

17 is 2 more than 15, which is 3*5. The sequence is linear and infinite: 7, 12, 17, 22, 27, ... In general, these numbers have the form 5n+2.

0 is a multiple of 5, but not a positive number (nor is it negative).

Answer: 20

'Two', 'twelve' and 'twenty' all contain a 'w'. This sequence is infinite: for instance, any number with a 2 in the ones place value ends in 'two' ('thirty-two', 'one hundred and two' etc.).

Answer: 25

Take a prime number and square it: that number has exactly 3 factors, 1, itself and the prime number. For instance, 5^2 = 25 and 25 has the factors 1, 25 and 5.

In fact, there are no other numbers that have exactly 3 factors. In general, though, the numbers with an odd number of factors are the square numbers. As there are infinitely many prime numbers, we can square them to get infinitely many numbers with 3 factors.

Answer: 34

The Fibonacci sequence starts 1, 1, 2, 3, 5, 8, 13, 21, 34, ... To get the next number in the sequence, take the most recent 2 numbers and add them together (e.g. 13 + 21 = 34). The sequence sometimes starts with 0 and 1 (and can even be extended backwards into negative numbers), but this doesn't affect our answer.

The Fibonacci numbers containing a 3 begin 3, 13, 34. It's a bit tricky to see that there's infinitely many of them, but the last digits of the Fibonacci sequence form a new sequence that repeats every 60 numbers, so in fact there are infinitely many Fibonacci numbers with 3 as the last digit.

Answer: 39

The sequence is 13, 26, 39, 52, ... It should be obvious that this sequence never ends (you can keep adding 13), but a mathematician might prefer the explanation that these are the numbers of the form 13n, and you can substitute n for any positive integer to get a multiple of 13.

You might consider 0 to be a multiple of 13, but it is not a positive number (or negative!).

Answer: 81

1^2 = 1 and squaring again gives 1 again.
2^2 = 4 and 4^2 = 16.
3^2 = 9 and 9^2 = 81.

These numbers are precisely the fourth powers i.e. n^4 for some positive integer n, which shows that the sequence is infinite.

Answer: 100

These numbers are the powers of 10 i.e. 1, 10, 100, 1000, ..., and so there are infinitely many of them.

"Digit sum" is the name for the result of adding all the digits of a number and it is studied in number theory. A number is a multiple of 3 if and only if its digit sum is a multiple of 3. This rule works with 9, too, but not for other numbers. One funny consequence of this is that you can take any multiple of 3 and shuffle its digits around, and the number remains a multiple of 3.

Answer: 288

This sequence starts 88, 188, 288, 388, ..., 788, 808, 818, ..., 878, 881, 882, 883, ..., 887, 889, 898, 988, ...

To see that it's infinite, notice that it contains 88, 880, 8800, 88000, 880000, ...

For a similar underlying reason as our previous question about numbers that contain the digit 1 (which approach 100% of positive integers), the proportion of numbers that contain exactly two 8s approaches 0% - because most numbers have many more 8s in them! At least, this is the idea informally. Mathematicians use lots of strange phrases like "almost all" when speaking of infinities that need formal definitions to back them up.

Answer: 729

These numbers are those of the form n^6, or "sixth powers", beginning:

1^6 = 1 (1^2 and 1^3)
2^6 = 64 (8^2 and 4^3)
3^6 = 729 (27^2 and 9^3)
4^6 = 4096 (64^2 and 16^3)

There are infinitely many of them.

Answer: 2048

A "power of 2" is a number of the form 2^n, for any non-negative integer n. The sleight of hand "non-negative" (rather than "positive") includes n = 0, where 2^0 = 1.

The powers of 2 begin: 1, 2, 4, 8, 16, 32, 64, 128, 256, ... Notice that the last digits (excluding 1!) repeat in the pattern 2, 4, 8, 6.

So, every fourth number in the sequence starting with 8 will end in an 8: 8, 128, 2048, 32768, ... To get the next number, you can multiply the previous one by 16 (as 2^4 = 16). These are the numbers of the form 2^(4n-1), of which there are infinitely many.