2023 TO THE POWER OF 2023

This is a puzzle that appeared on March 27th 2023 as a post in a blog titled PUZZLE A DAY. The challenge is to find the last digit of 2023 to the power of 2023? The clue provided is that there is a pattern to be found. Work out the last digit of 2023 to the power of 1, 2, 3, 4 and 5.

I used SageMathCell to generate the numbers for 2023 raised to the powers 1 to 12. The results were (permalink):

\(2023^1 \rightarrow 2023\)
\(2023^2 \rightarrow 4092529\)
\(2023^3 \rightarrow 8279186167\)
\(2023^4 \rightarrow 16748793615841\)
\(2023^5 \rightarrow 33882809484846343\)
\(2023^6 \rightarrow 68544923587844151889\)
\(2023^7 \rightarrow 138666380418208719271447\)
\(2023^8 \rightarrow 280522087586036239086137281\)
\(2023^9 \rightarrow 567496183186551311671255719463\)
\(2023^{10} \rightarrow 1148044778586393303510950320473649\)
\(2023^{11} \rightarrow 2322494587080273653002652498318191927\)
\(2023^{12} \rightarrow 4698406549663393600024366004097702268321\)

The repeating pattern 1, 3, 9, 7 of final digits is apparent. This is not surprising when we consider that it is only the final digit that we are interested in and that 3 raised to the same powers produces the same pattern:

\(3^1 \rightarrow 3\)

\(3^2 \rightarrow 9\)

\(3^3 \rightarrow 27\)

\(3^4 \rightarrow 81\)

\(3^5 \rightarrow 243\)

\(3^6 \rightarrow 729\)

\(3^7 \rightarrow 2187\)

\(3^8 \rightarrow 6561\)

\(3^9 \rightarrow 19683\)

\(3^{10} \rightarrow 59049\)

\(3^{11} \rightarrow 177147\)

\(3^{12} \rightarrow 531441\)

Every power that is a multiple of 4 ends in a 1 and so all we need to do is to divide 2023 by 4 which leaves a remainder of 3. Thus 2023 to the power 2023 is three positions ahead of the 1 and so the final digit must be 7. In general, any number that ends in 3, when raised to consecutive powers, will follow this same 1, 3, 9, 7 pattern just as 3 and 2023 do.

Generalising, we can look at numbers ending in digits 0 to 9. Here is the pattern for integer powers greater than zero:

• 0 --> numbers ending in 0 will always end in 0
• 1 --> numbers ending in 1 will always end in 1
• 2 -->  numbers ending in 2 will follow a 2, 4, 8, 6 pattern
• 3 -->  numbers ending in 3 will follow a 1, 3, 9, 7 pattern
• 4 --> numbers ending in 4 will follow a 4, 6 pattern
• 5 --> numbers ending in 5 will always end in 5
• 6 --> numbers ending in 6 will always end in 6
• 7 --> numbers ending in 7 will follow a 1, 7, 9, 3 pattern
• 8 --> numbers ending in 8 will follow a 2, 6, 8, 4 pattern
• 9 --> number ending in 9 will follow a 1, 9 pattern

So a question like what is final digit of 2028 raised to the power 2028 is easily answered. Let's look at the powers of 2028 from 1 to 12:

\(2028^1 \rightarrow 2028\)
\(2028^2 \rightarrow 4112784\)
\(2028^3 \rightarrow 8340725952\)
\(2028^4 \rightarrow 16914992230656\)
\(2028^5 \rightarrow 34303604243770368\)
\(2028^6 \rightarrow 69567709406366306304\)
\(2028^7 \rightarrow 141083314676110869184512\)
\(2028^8 \rightarrow 286116962163152842706190336\)
\(2028^9 \rightarrow 580245199266873965008154001408\)
\(2028^{10} \rightarrow 1176737264113220401036536314855424\)
\(2028^{11} \rightarrow 2386423171621610973302095646526799872\)
\(2028^{12} \rightarrow 4839666192048627053856649971156350140416\)

All multiples of 4 end in 6 and if we divides 2028 by 6 we get 0 and so 2028 raised to the power 2028 must end in 6 as well. This is just my way of looking at the problem and there are surely other approaches.

Overall the PUZZLE A DAY site looks interesting, providing as it does a little mathematical challenge each day.