Unit Digit

For us to be able to identify the units digit of a product, we only have to look at the units digit of the product of the units digit of the factors. (To demonstrate this, if I have $25 \times 25$ , I will only have to look at the units digit of the factors, $5 \times 5$ , which will give me $25$ with a last digit of $5$ . If we check on the sample problem, $25 \times 25$ , we would arrive at $625$ , which confirms our finding that $5$ is the last digit.) Let us group our set of odd numbers from $1$ to $79$ into five sets according to their units digit. Each of these five subsets have eight elements each:

$S_1: \{1, 11, 21, 31, 41, 51, 61, 71\} \\ S_3: \{3, 13, 23, 33, 43, 53, 63, 73\} \\ S_5: \{5, 15, 25, 35, 45, 55, 65, 75\} \\ S_7: \{7, 17, 27, 37, 47, 57, 67, 77\} \\ S_9: \{9, 19, 29, 39, 49, 59, 69, 79\} \\$

Let us observe the patterns done by the units digit of all these numbers as the exponent grows to $79$ .

For $S_1$ :

If we multiply any number with a last digit of $1$ by itself, no matter how many times, we will have $1$ in the units digit. This means that the $79^{th}$ power of any number that belongs to $S_1$ is $1$ . The units digit of the sum of all elements in $S_1$ is $1 \times 8 = \textbf{8}$ .

For $S_3$ :

We will only use $3$ as our calculations with this number can represent the whole set. The units digit of product of the sole factor $3$ is $3$ . Multiplying it with $3$ (for which we will have $3^2$ ) will give us a units digit of $9$ . Multiplying it once more with $3$ will give us $2\textbf{7}$ . If we continue this until the fifth iteration...

power of 3 || units digit
         3 ||           3
       3x3 ||           9
     3x3x3 ||           7  | 9 x 3 is 27
   3x3x3x3 ||           1  | 7 x 3 is 21
 3x3x3x3x3 ||           3  | Start of loop

...we would end up back to a unit digit of $3$ . We have a loop in the series of last digits; if we continue to multiply with $3$ , we will just go back and forth with the numbers $\{3, 9, 7, 1\}$ . The loop repeats every four iterations. If we continue multiplying until the $79^{th}$ term, we will have a $7$ as our last digit and this is true for any number in $S_3$ . The units digit of the sum of all elements in $S_3$ is $7 \times 8 = 5\textbf{6}$ .

For $S_5$ :

Just like the elements of $s_1$ , if we multiply any number with a last digit of $5$ by itself, no matter how many times, we will (always) have $5$ in the units digit, even after seventy-nine iterations. The units digit of the sum of all elements in $S_5$ is $5 \times 8 = 4\textbf{0}$ .

For $S_7$ :

The elements in $S_7$ show a pretty similar behavior with the elements on $S_3$ :

power of 7 || units digit
         7 ||           7
       7x7 ||           9  | 7 x 7 is 49
     7x7x7 ||           3  | 9 x 7 is 63
   7x7x7x7 ||           1  | 3 x 7 is 21
 7x7x7x7x7 ||           7  | Start of loop

The loop also repeats every four iterations. If we continue multiplying until the $79^{th}$ term, we will have a $3$ as our last digit and this is true for any number in $S_7$ . The units digit of the sum of all elements in $S_7$ is $3 \times 8 = 2\textbf{4}$ .

Lastly, $S_9$ :

Unlike $S_3$ and $S_7$ , the loop starts early with $S_9$ :

power of 9 || units digit
         9 ||           9
       9x9 ||           1  | 9 x 9 is 81
     9x9x9 ||           9  | Start of loop

The loop repeats only at every two iterations with odd-numbered iterations, including the $79^{th}$ iteration, having a last digit of $9$ . The units digit of the sum of all elements in $S_9$ is $9 \times 8 = 7\textbf{2}$ .

Thus, the units digit of the sum of all of the units digits of all subsets is $8 + 6 + 0 + 4 + 2 \equiv \boxed{0} \pmod {10}$ .

(1+7+5+7)*8=0

1^79=1 for 2- 2^1=2 2^2=4 2^3=8 2^4=16 2^5=32 unit digit is following a loop-: 2,4,8,6,then again 2,4,8,6 again and again so 2^79 unit digit ====== 79%4=3 unit digit of 2^79=8 we have to find 3^79================ 3^1==3 3^2==9 3^3==27 3^4==81 we saw that repeating of this pattern for unit digit 3,9,7,1 again 3,9,7,1 unit digit of 3^79===== 79%4=3 unit digit =7 for 5^79=== same process== unit digit =5 for 7^79---------- unit digit==3 for 9^79--------- unit digit=9 for 11^79 unit digit=1 for 13^79======== unit digit=7 .. . . . . . . . . . .. we found that

1to 80----------------------unit digit 1^79+3^79+7^79+9^79=1+7+5+3+9=25
11^79+13^79+15^79+17^79+19^79=1+7+5+3+9=25

last digit of problem==25*8=200 ans=0

1^79 + 3^79 + ... + 79^79 3^79 unit digit is same as 3^3 (as 79/4 remainder is 3)

it is same as 1^3 + 3^3 + ... + 79^3

sum of cubes of odd numbers = n² (2n² – 1) = 40²(2 × 40² – 1) = 1600(3199) gives 0 in unit place