 Problem of the Week

Problem D and Solution

More Power, Mr. Scott!

Problem

Mr. Scott likes to pose interesting problems to his Mathematics classes. Today, he started with the expression $$6^{2020}+7^{2020}$$. He stated that the expression was not equivalent to $$13^{2020}$$ and that he was not interested in the actual sum. His question to his class and to you is, “What are the final two digits of the sum?” Solution

Solution 1
Let’s start by examining the last two digits of various powers of 7.

$$7^1=\phantom{16\, 8}{\mbox{\bf 07}}$$ $$7^2=\phantom{117\, 6}{\mbox{\bf 49}}$$ $$7^3=\phantom{823\, }3{\mbox{\bf 43}}$$ $$7^4=\phantom{5\ 76}24{\mbox{\bf 01}}$$
$$7^5=16\,8{\mbox{\bf 07}}$$ $$7^6=117\,6{\mbox{\bf 49}}$$ $$7^7=823\,5{\mbox{\bf 43}}$$ $$7^8=5\,764\,8{\mbox{\bf 01}}$$

Notice that the last two digits repeat every four powers of 7. If the pattern continues, then $$7^9$$ ends with 07, $$7^{10}$$ ends with 49, $$7^{11}$$ ends with 43, $$7^{12}$$ ends with 01, and so on. We can simply compute these powers of 7 to verify this for these examples, but let’s justify why this pattern continues in general. If a power ends in “07”, then the last 2 digits of the next power are the same as the last 2 digits of the product $$07 \times 7 = 49$$. That is, the last 2 digits of the next power are “49”. If a power ends in “49”, then the last 2 digits of the next power are the same as the last two digits of the product $$49\times 7 = 343$$. That is, the last two digits of the next power are “43”. If a power ends in “43”, then the last 2 digits of the next power are the same as the last two digits of the product $$43\times 7 = 301$$. That is, the last two digits of the next power are “01”. Finally, if a power ends in “01”, then the last 2 digits of the next power are the same as the last two digits of the product $$01\times 7 = 07$$. That is, the last two digits of the next power are “07”. Therefore, starting with the first power of 7, every four consecutive powers of 7 will have the last two digits 07, 49, 43, and 01.
We need to determine the number of complete cycles by dividing 2020 by 4. Since $$2020\div 4=505$$, there are 505 complete cycles. This means that $$7^{2020}$$ is the last power of 7 in the $$505^{\textrm{th}}$$ cycle and therefore ends with 01.
Next we will examine the last two digits of various powers of 6.

$$6^1={\mbox{\bf 06}}$$ $$6^2=\phantom{117\,6}{\mbox{\bf 36}}$$ $$6^3=\phantom{1823\ }2{\mbox{\bf 16}}$$ $$6^4=\phantom{00\ 77}12{\mbox{\bf 96}}$$ $$6^5=\phantom{60\ \ 46}77{\mbox{\bf 76}}$$ $$6^6=\phantom{362\ \ 4}46\,6{\mbox{\bf 56}}$$
$$6^7=279\,9\mbox{\bf 36}$$ $$6^8=1\,679\,6\mbox{\bf 16}$$ $$6^9=10\,077\,6\mbox{\bf 96}$$ $$6^{10}=60\,466\,1\mbox{\bf 76}$$ $$6^{11}=362\,797\,0\mbox{\bf 56}$$

Notice that the last two digits repeat every five powers of 6 starting with the $$2^{\textrm{nd}}$$ power of 6. This pattern can be justified using an argument similar to the one above for powers of 7. So $$6^{12}$$ ends with 36, $$6^{13}$$ ends with 16, $$6^{14}$$ ends with 96, $$6^{15}$$ ends with 76, $$6^{16}$$ ends with 56, and so on. Starting with the second power of 6, every five consecutive powers of 6 will have the last two digits 36, 16, 96, 76, and 56.
We need to determine the number of complete cycles in 2020 by first subtracting 1 to allow for 06 at the beginning of the list and then dividing $$2020-1$$ or 2019 by 5. Since $$2019\div 5=403$$ remainder 4, there are 403 complete cycles and $$\tfrac{4}{5}$$ of another cycle. Since $$403 \times 5 = 2015$$, $$6^{2015+1}=6^{2016}$$ is the last power of 6 in the $$403^{\textrm{rd}}$$ cycle and therefore ends with 56.
To go $$\frac{4}{5}$$ of the way into the next cycle tells us that the number $$6^{2020}$$ ends with the fourth number in the pattern, namely 76. In fact, we know that $$6^{2017}$$ ends with 36, $$6^{2018}$$ ends with 16, $$6^{2019}$$ ends with 96, $$6^{2020}$$ ends with 76, and $$6^{2021}$$ ends with 56 because they would be the numbers in the $$404^{\textrm{th}}$$ complete cycle.
Therefore, $$6^{2020}$$ ends with the digits 76.
The final two digits of the sum $$6^{2020}+7^{2020}$$ are found by adding the final two digits of $$6^{2020}$$ and $$7^{2020}$$. Therefore, the final two digits of the sum are $$01+76=77$$.
Solution 2
From the first solution, we saw that the last two digits of powers of 7 repeat every 4 consecutive powers. We also saw that the last two digits of powers of 6 repeat every 5 consecutive powers after the first power of 6.
Let’s start at the second powers of both 7 and 6. We know that the last two digits of $$7^2$$ are $$49$$ and the last two digits of $$6^2$$ are $$36$$. When will this combination of last two digits occur again? The cycle length for powers of 7 is 4 and the cycle length for powers of 6 is 5.
The least common multiple of 4 and 5 is 20. It follows that 20 powers after the second power, the last two digits of the powers of 7 and 6 will end with the same two digits as the second powers of each. That is, the last two digits of $$7^{22}$$ and $$7^2$$ are the same, namely 49. And, the last two digits of $$6^{22}$$ and $$6^2$$ are the same, namely 36. The following table illustrates this repetition.

Powers $$7^2$$ $$7^3$$ $$7^4$$ $$7^5$$ $$7^6$$ $$7^7$$ $$7^8$$ $$7^9$$ $$7^{10}$$ $$7^{11}$$ $$7^{12}$$ $$7^{13}$$ $$7^{14}$$ $$7^{15}$$ $$7^{16}$$ $$7^{17}$$ $$7^{18}$$ $$7^{19}$$ $$7^{20}$$ $$7^{21}$$ $$7^{22}$$
Last 2 digits 49 43 01 07 49 43 01 07 49 43 01 07 49 43 01 07 49 43 01 07 49
Powers $$6^2$$ $$6^3$$ $$6^4$$ $$6^5$$ $$6^6$$ $$6^7$$ $$6^8$$ $$6^9$$ $$6^{10}$$ $$6^{11}$$ $$6^{12}$$ $$6^{13}$$ $$6^{14}$$ $$6^{15}$$ $$6^{16}$$ $$6^{17}$$ $$6^{18}$$ $$6^{19}$$ $$6^{20}$$ $$6^{21}$$ $$6^{22}$$
Last 2 digits 36 16 96 76 56 36 16 96 76 56 36 16 96 76 56 36 16 96 76 56 36

Since 2000 is a multiple of 20, we then know that the $$2022^{\textrm{nd}}$$ power of 7 will end with 49 and that the $$2022^{\textrm{nd}}$$ power of 6 will end in 36.
Working backwards through the cycle of the last two digits of powers of 7, it follows that the $$2021^{\textrm{st}}$$ power of 7 ends in 07 and that the $$2020^{\textrm{th}}$$ power of 7 ends in 01.
Working backwards through the cycle of the last two digits of powers of 6, it follows that the $$2021^{\textrm{st}}$$ power of 6 ends in 56 and that the $$2020^{\textrm{th}}$$ power of 6 ends in 76.
The final two digits of the sum $$6^{2020}+7^{2020}$$ are found by adding the final two digits of $$6^{2020}$$ and $$7^{2020}$$. Therefore, the final two digits of the sum are $$01+76=77$$.