Ending Digits of Numbers Raised to Various Powers

 

A table of the numbers 1 through 10 each raised to the powers of 1 through 5. This table is shown below.

 

Table of numbers 1 to 10 raised to different powers

For this post, I would like to focus on just the ending digit of each result of the exponentiations. The following table lists the ending digit of each result and now we can explore some of the patterns that arise for the different powers. When we get the results for Power 5, we see why we only have to consider this explorations for just 5 powers as the pattern will repeat for powers higher than 5.

One additional note is that each row of this table applies not only to the numbers 1 through 10 but also to any number, higher than 10 that has the same last digit. For example, the results for raising the number 12 to various powers will have the same ending digit as 2 raised to the same powers. For this reason, the first column of the second table is named “Starting Number’s Last Digit” and the last row’s name has been changed from “10” to “0.”

Table of last digits resulting from exponentiation

Discussion of results: 

Power 0

The table doesn't list the results of raising numbers to the power of 0. All numbers raised to the power of 0 equal 1. One result that will likely become another post on its own is zero raised to the power of zero. There is disagreement on the result (see link). Some think this should equal 1; some believe it should undefined as with division by zero.

Power 1

These results are a bit trivia but are included for completeness of the exercise. Any number raised to the power of 1 is itself and therefore the last digit of the result will match the last digit of the starting number. A shorthand way of expressing this result is 1à1, 2à2, etc. For Power 1, this is the identity function, for all i, i à i.

Power 2

Squaring numbers ending in 1, 5, 6 and 0 give results that will end with 1, 5, 6, and 0, respectively. This pattern continues for all powers. Results of squaring numbers 2 and 8 will end in 4. For the numbers ending in 3 and 7 yield 9. Squaring numbers that end in 9 yields 1. Another observation is there are no numbers that when squared which will have an ending digit of 2, 3, 7 or 8. Summarized in shorthand expressions:

Identity function for 1, 5, 6, 0
2à4, 8à4, 3à9, 7à9
Result exclusions: 2, 3, 7, 8

Power 3

Unlike Power 2 in which it was impossible to obtain results with certain end digits (2, 3, 7 or 8), we observe cubes that end in all digits 0 through 9. Cubing numbers ending in 1, 4, 5, 6, 9 and 0 give results that will end with 1, 4, 5, 6, 9 and 0, respectively. For the other four digits: numbers ending in 2 yield 8 and those ending in 8 yield 2; numbers ending in 3 yield 7 and those ending in 7 yield 3.

Summary
Identity function for 1, 4, 5, 6, 9 and 0
Inverse function for 2, 3, 7, 8: 2à8, 8à2, 3à7, 7à3
No exclusions

Power 4

Reviewing Power 2 we observed that resultant ending digits excluded 2, 3, 7 and 8 and with Power 3 all digits were possible. With Power 4, there are more exclusions. All numbers raised to the 4^th power will only end in 0, 1, 5 or 6. There are no numbers when raised to the 4^th power that end in 2, 3, 4, 7, 8 or 9. As with the other powers, raising numbers ending in 1, 5, 6 and 0 will yield the same end digit as the starting number. Along with 6, numbers ending in 2, 4 or 8 will also yield a result having an end digit of 6. Similarly, numbers ending in 1, 3, 7, 9 yield 1.

Summary
Identity function for 1, 5, 6 and 0
(1, 3, 7, 9) à 1
(2, 4, 6, 8) à 6
Result Exclusions: 2, 3, 4, 7, 8 and 9

Power 5

Power 5 completes the full cycle of the power pattern. All numbers when raised to the 5^th power will have an ending digit that is the same as the starting number’s last digit. This is the same as Power 1 which resulted in the identity function for all ending digits. Another observation is that we could continue this exercise for powers higher than 5. The patterns and results we observed for Powers 2 to 5 will repeat for Powers 6 to 9, Powers 10 to 13 and higher.

The Ramanujan Machine

In an earlier post, we were introduced to University of Illinois math professors Wolfgang Haken and Kenneth Appel, who used a computer to prove the four-color map theorem. They were the first people to use a computer to prove a long-establish conjecture.

Recently, a team from Technion-Israeli Institute of Technology, used artificial intelligence and computer technology to create a "conjecture generator." Conjectures are mathematical statements that are generally considered to be true but have not been proven. Well known conjectures included in this blog are the Collatz Conjecture, Goldbach Conjecture and the Twin Prime Conjecture. The research team named the generator, the Ramanujan Machine after the great self-taught Indian mathematician, Srinivasa Ramanujan.

Srinivasa Ramanujan (1887-1920), 

by unknown author - creofire.com (archived from the original), Public Domain, https://commons.wikimedia.org/w/index.php?curid=3911499

An excerpt from the Technion website further describes the generator:

The project deals with one of the most fundamental elements of mathematics – mathematical constants. A mathematical constant is a number with a fixed value that emerges naturally from different mathematical calculations and mathematical structures in different fields. Many mathematical constants are of great importance in mathematics, but also in disciplines that are external to mathematics, including biology, physics, and ecology. The golden ratio and Euler’s number are examples of such fundamental constants. Perhaps the most famous constant is pi, which was studied in ancient times in the context of the circumference of a circle.

The Technion researchers proposed and examined a new idea: The use of computer algorithms to automatically generate mathematical conjectures that appear in the form of formulas for mathematical constants.

You can use your own computer to run the Ramanujan Machine and you might be lucky enough to have a new conjecture named after yourself. Additionally, you can contribute by proving the conjectures proposed by other users.

Technion - Israeli Institute of Technology: Ramanujan Machine.

Related post: Taxicab Numbers

Divisibility

This post deals with the divisibility of numbers. Specifically, rules that determine if one integer is divisible by another by examination of the number's digits. Prime numbers have no divisors other than one and themselves. This blog has several prior posts on prime numbers: Blogger: Posts.    

Here is a list of rules to apply to determine if a number (greater than 0) is divisible by another:

Divisibility by:

1: All integers are divisible by one so we won't need any help. However, one might ask why we didn't start with zero. Divisibility by zero is undefined. If you use a calculator or a calculator app and divide by zero, you'll get an error message. Therefore, divisibility rules start with the integer 1.

2: All even numbers are divisible by 2. Numbers greater than zero ending in 0, 2, 4, 6 or 8 are even numbers. If the number written in the binary base ends in zero, then the number is divisible by two. See the prior post (Ethiopian Multiplication), for learning how to convert a number to base 2.

3: Sum the digits in the number. If this sum is divisible by 3, the original number will be divisible by 3. If the starting number is very long and you can't determine immediately that the sum of the digits is divisible by 3, apply the rule to sum and continue until you determine the number is or is not divisible by 3. This method will be used for other divisibility rules as well.

4: One hundred and all multiples of 100 are divisible by 4. Therefore, examine the last two digits of the number. If that number formed by the last two digits is divisible by 4, then the original number is divisible by 4.

5: If the last digit is 0 or 5, the number is divisible by 5.

6: Apply the rules for 2 and 3. The number must be even and the sum of its digits are a multiple of three. Another rule (for just some multiples of 6): if the number is greater than 4 is sandwiched by two primes, then the number is a multiple of six (see the prior post: Twin Prime Sandwich).

7: For smaller numbers - drop the last digit (the ones digit) from the number and subtract twice this digit from the remaining number. The original number is a multiple of 7 if the result is a multiple of 7. Example: start with 336. Drop 6, double it and subtract from 33. 33 minus 12 is 21 which is a multiple of 7 so 336 is a multiple of 7.

For larger numbers (>1000) - starting with the right-most digits, alternately sum and subtract each block of three digits. If the result is a multiple of 7, the original number will be a multiple of 7. For example: start with 1,234,567. Sum 567-234+1 = 334 (not a multiple of 7). Try 8,641,969 to get the sum 969-641+8 = 336 which is divisible by 7 (see example above for smaller numbers).

8: Similar to the method for multiples of 4, the number 1000 and all its multiples are divisible by 8. Therefore, if the last 3 digits are a multiple of 8, the original number will be a multiple of 8. Example, start with 2104. The number 104 is divisible by 8 (8x13=104); therefore 2104 is divisible by 8 (8x263=2104). If it is not clear that the number of the last three digits is divisible by 8, divide the number by 2 and now perform the check for divisible by 4 (check to see if the number formed by the last two digits is divisible by 4). Also, as a check for divisibility by 8 is to see if you can divide the number by 2 three times (since 8 = 2 x 2 x 2). For example, 104/2=52, 52/2=26, 26/2=13.

9: Sum the digits in the number. If the resulting sum is divisible by 9, the original number will be divisible by 9. This rule was the basis of the magic trick given in a prior post (Missing Digit Trick).

10: If the last digit  of the number is 0, the number is divisible by 10. 

11: Subtract the last digit (the ones digit) from the remaining digits. If the resulting answer is divisible by 11, then so is the original number. Example: start with 1001. Subtract 1 from 100 to get 99  which is divisible by 11 (11 x 91 = 1001).

12: A number divisible by 12 must meet both rules for divisibility by 3 and 4.

At this point, the reader can see that rules for divisibility of higher compound numbers can be based on combining the divisibility rules of the number's multiples. The more difficult divisibility rules are for higher prime numbers. Wikipedia gives a good summary for numbers up to 30 here.

Credit given to Numberphile for inspiring me to add this post. Numberphile has a good video about divisibility rules and includes more magic tricks based on these rules.