The Powers of 2 Trivia Quiz

Answer: 2^10

2^10, or 1024 is the number of bytes in a kibibyte (or kilobyte). Although 2^10 is often the number of bytes thought to be in a kilobyte it is now more generally accepted to have 1000 bytes in a kilobyte (as a result of the widely used decimal number system). There are 1024 in it's binary neighbour, the kibibyte.

Answer: False

Two to the power of 0 is equal to 1 which is an odd number. Apart from this all of the remaining powers are even numbers.

Answer: 2^16

From 2^0, the powers of 2 are as follows;

1 (1 is a power of 2)
2 (2 is a power of 2)
4 (4 is a power of 2)
8 (8 is a power of 2)
16 (1 is a power of 2)
32 (2 is a power of 2)
64 (4 is a power of 2)
128 (Both 2 and 8 are powers of 2)
256 (2 is a power of 2)
512 (Both 1 and 2 are powers of 2)
1024 (1,2 and 4 are all powers of 2)
2048 (2,4 and 8 are all powers of 2)
4096 (4 is a power of 2)
8192 (1,2 and 8 are all powers of 2)
16384 (1,4 and 8 are all powers of 2)
32768 (Both 2 and 8 are powers of 2)
65536 (No digits are powers of 2)

Answer: 100....0

In binary notation, each digit (either 1 or 0) represents a power of two. When something is written in binary the rightmost digit will either be 1 or 0 in decimal form, the second will be 2 or 0, the third 4 or 0 and so on.

For a power of two to be shown in binary, only one of the binary digits can be a 1 and if it were more the number will definitley not be a power of two. Incidentally, any binary number which is a series of ones will always be one less than a power of two.

For example: 11111 = 31 = 32-1.

Answer: 31

A Mersenne Prime is a prime number which is one less than a power of two. 31 is a Mersenne prime as its only divisors are 1 and 31 and it is one less than 32. Other examples are the numbers 3 and 7. Although there appear to be an abundance of low Mersenne primes, as you go up the powers they become few and far between.

Answer: 257

A Fermat prime is a prime number which is equal to (2^(2^n))+1, where n is some natural number. Examples include 3, 5, 17 and 257. Fermat suspected that this was true for all natural n, but this was disproved by the counter example of the case when n=5 by Leonhard Euler in 1732.

Answer: 2,4,8,6

The first few powers are 2, 4, 8, 16, 32, 64, 128, 256... Clearly as you are only multiplying by two this pattern will continue. The other answers would imply that either 2*4 ended with a 6 (clearly it is 8), 2*8 ended in a 4 (again clearly 16) or 2*8 ended in a 2 (once again 16).

Answer: The number of bytes in a mebibyte

Similar to the kilobyte/kibibyte situation above, a mebibyte (denoted by MiB) is equal to 2^20 or 1,048,576 bytes (a megabyte would be a round million). The number of seconds in a year is 31,556,926 which is approximately equal to 2^25 (two to the power of 24.91 to be more precise).

The population of the UK in 2001 was around 58,800,000 which is roughly equal to 2^26 (25.81 for precision). A tebibyte is the binary equivalent of a terabyte and contains 2^40 bytes.

Answer: 64

Four can be expressed as 2^2, clearly the exponent is a power of two. Similarly sixteen is equal to 2^4, and as 4=2^2 sixteen also has a power of two as the exponent. Again 256=2^8, 8=2^3. However sixty-four is two to the power of six which is itself not a power of two. On a side note these numbers are all powers of four.

Answer: 2^13

This question is just a case of using simple indices rules. In general x^m multiplied by x^n is equal to x^(m+n). In this case x=2, m=7 and n=6 so the answer is 2^13. 2^42 is equivalent to (2^7)^6, again using indices rules. Just out of interest 117649 is equal to 7^6 :)