1
$\begingroup$

Define the sequences $a_1, a_2,...$ and $*b_1, b_2,...*$ by $a_1 = b_1 = 7$ and $$a_{n+1} = {a_n}^7, \\ b_{n+1} = 7^{b_n}$$ for $n\ge 1$.

Find the last digits of $a_{2009}$, and of $b_{2009}$. What about the last two digits or more?

I know the order of 7 mod 100 is 4. Im not sure if that helps but how will we work this out?

$\endgroup$
  • $\begingroup$ For the $\{a_n\}$, say...have you written out the first few terms $\pmod {100}$? $\endgroup$ – lulu Apr 2 at 17:54
  • 1
    $\begingroup$ In both cases $a_n,b_n$ are of the form $$7^{4m+3}$$ $\endgroup$ – lab bhattacharjee Apr 2 at 17:55
2
$\begingroup$

Part 1:$$a_1\equiv(mod10)7 $$$$a_2=7^7\equiv(mod10)(-3)^7\equiv3$$$$a_3=a_2^7\equiv(mod10)3^7\equiv7$$$$.$$$$.$$$$.$$$$a_{2009}\equiv(mod10)7$$'''''''''''''''''''''$$a_1\equiv(mod100)7$$ by a simple calculation we have $7^4\equiv(mod100)1$ $$a_2\equiv(mod100)a_1^7\equiv7^7\equiv7^3$$ $$a_3\equiv(mod100)a_2^7\equiv(7^3)^7\equiv7^{21}\equiv7(7^4)^5\equiv7$$ $$a_4\equiv(mod100)7^3$$ $$$$ and then we have: $$a_{2009}\equiv(mod100)7 $$''''''''''''''''''''''$$$$ Part 2:$$$$ We need to show $b_n\equiv(mod4)3$$$$$ Proof by induction:$$b_1\equiv(mod4)3$$ $$b_k\equiv(mod4)3 \Longrightarrow b_{k+1}\equiv(mod4)3$$$$b_{k+1}=7^{b_k}\equiv(mod4)(-1)^{4k+3}\equiv-1\equiv3$$$$$$now we have:$$b_n\equiv(mod100)7^{b_{n-1}}\equiv7^{4k+3}\equiv7^3\equiv43$$$$$$$$b_n\equiv(mod10)7^{4k+3}\equiv7^3\equiv3$$

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Hint: We have

$$a_n=7~\widehat~~7~\widehat~~(n-1)$$

$$b_n=7~\widehat~~ b_{n-1}=7~\widehat~~7~\widehat~~b_{n-2}=\dots$$

and in the case of finding the last two digits, the order of $7$ mod $100$ is $4$ and the order of $7$ mod $4$ is $2$, so

$$a_n\equiv7~\widehat~~(7~\widehat~~[(n-1)\bmod2]\bmod4)\pmod{100}$$

$$b_n\equiv7~\widehat~~(7~\widehat~~[b_{n-2}\bmod2]\bmod4)\pmod{100}$$

and the same strategy can be applied to find more digits.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.