What is the unit digit in the product $3547^{153}\times251^{72}$?
I know that the required digit is equal to the unit digit in $7^{153}\times1^{72}$. And I also know that $1^{72}=1$, as $1$ is the multiplicative identity. But I am having trouble evaluating $7^{153}$.
Similarity,
What is the unit digit in $264^{102}+264^{103}$?
$$264^{102}+264^{103}=264^{102}×(1+264)=264^{102}\times265=\cdots$$
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$\begingroup$$251^{72} \equiv 1 \pmod{10}$
$3547^{153} \equiv 7^{153} \pmod{10}$
Cyclic sequence of $7^{n}$ is as follows.
$7^{1} = 7$
$7^{2} = 49$
$7^{3} = 343$
$7^{4} = 2401$
Lets divide $153$ by $4$ and the remainder is $1$.
Thus, the unit digit of $7^{153}$ is equal to the unit digit of $7^{1} = 7.$
Hence the unit digit of $3547^{153}\times251^{72}$ is equal to $7.$
$264^{102} \equiv 4^{102} \pmod{10}$
Cyclic sequence of $4^{n}$ is as follows.
$4^{1} = 4$
$4^{2} = 16$
Lets divide $102$ by $2$ and the remainder is $0$.
Thus, the unit digit of $4^{102}$ is equal to the unit digit of $4^{2} = 6.$
Hence the unit digit of $264^{102}+264^{103}$ is equal to $0.$
Hint:
The last digit of the increasing powers of a number must follow a periodical pattern. This is because the last digit of the power $n+1$ only depends on the last digit of the power $n$ ($k^{n+1}\bmod10=k\cdot k^n\bmod10= k(k^n\bmod10)\bmod10)$.
Once you know the period, you can find the last digit of any power easily.
E.g. $\color{green}7, 4\color{green}9, 34\color{green}3, 240\color{green}1, 1680\color{green}7, 11764\color{green}9,\cdots$
$\endgroup$ $\begingroup$Units digit of a product of numbers only depends on product of units digits
Let's say we have number 343. We can break it down into 340+3 and if we multiply it by a different number, for example $(340+3) (340+3)$ We'll have $340^2 + 2*340*3 + 3*3$ We know that $340^2$ and $2*340*3$ will not be units digits because $2*340*3 = 2*34*3*10$ and $(34*10)^2 = 34^2*10^2$ so they are not in the units digits. So we only need to look for units digit, if units digit repeat like in our case of 7, 49, 343,2401, 16807. We didn't even need to compute powers of 7, we could just multiply last digits by 7 giving: 7, 9, 3, 1, 7. So there's 4 numbers in a sequence 7, 9, 3, 1 and then they repeat. Now we need to figure out how many sequences can we fit in 157 numbers, since we alternate a units digit 157 times. We can fit 39 sequences with remainder of 1. Meaning we repeat 7, 9, 3, 1 thirty-nine times after that the last digit will be 1 because 1 is the last number in a sequence and since we place numbers back-to-back 7,9,3,1,7,9,3,1 the last digit in a series will be last digit but we have one more power of 7, since we're left with remainder of 1 after fitting all them neatly together, so we choose the next digit after 1 which is 7. If for example remainder was 2 we would choose second number in a sequence 9.
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