Determine the last digit of $5^5+6^6+7^7+8^8+9^9$.
Problem
Source:
Tags: last digits, number theory, modulo
Bowser498
19.09.2020 02:40
Taking $\pmod{10}$ for each of the terms, we get $5+6+3+6+9 \equiv \boxed{9} \pmod{10}$.
Ha_ha_ha
19.09.2020 02:41
$5^5=3125, 6^6=46656, 7^7=823543, 8^8=16777216, 9^9=387420489. \text{add them all up, we get} 405071029, \text{so our answer is}\boxed{9}$
OronSH
19.09.2020 02:43
the last digit of 5^5 is 5 since 5*5 ends with 5
the last digit of 6^6 is 6 since 6*6 ends with 6
the last digit of 7^7 is 3 since 7*7 ends with 9, 9*7 ends with 3, 3*7 ends with 1 and 1*7 ends with 7
the last digit of 8^8 is 6 since 8*8 ends with 4, 4*8 ends with 2, 2*8 ends with 6, 6*8 ends with 8
the last digit of 9^9 is 9 since 9*9 ends with 1 and 1*9 ends with 1
5+6+3+6+9=29
answer is 9
edited because im dumb
ilovepizza2020
19.09.2020 02:45
Socoobo wrote: Take everything mod 10.
We get $5+6+3+6+9=\boxed{29}$.
Please hid your sols.
littlepetwolf
19.09.2020 03:06
This is not a tricky problem but requires a little time. I think you could figure it out with some effort, but here's the solution anyway lol. XD
The last digit of $5^5$, is of course $5$
The last digit of $6^6$ requires a little more work.
$6^1$ last digit = $6$
$6^2$ last digit = $6$
$6^3$ last digit = $6$
.
.
.
.
and so on. So we know that the last digit of $6^6$ is $6$
The last digit of $7^7$ is more tricky.
$7^1$ last digit = $7$
$7^2$ last digit = $9$
$7^3$ last digit = $3$
$7^4$ last digit = $1$
$7^5$ last digit = $7$,
and thus the pattern shall continue.
Since we want to find the last digit of $7^7$, we want to find the remainder of $7/4$ because the pattern repeats in fours. the remainder of $7/4$ is 3, so we want to find the last digit of $7^3$, which is $3$. So the last digit of $7^7$ is $3.$
Same thing with $8^8$
$8^1$ last digit = $8$
$8^2$ last digit = $4$
$8^3$ last digit = $2$
$8^4$ last digit = $6$
$8^5$ last digit = $8$,
and thus the pattern shall continue.
Since we want to find the last digit of $8^8$, we want to find the remainder of $8/4$ because the pattern repeats in fours. the remainder of $8/4$ is 0, so we want to find the last digit of $8^4$, which is $6$. So the last digit of $8^8$ is $6.$
and $9^9$
$9^1$ last digit = $9$
$9^2$ last digit = $1$
$9^3$ last digit = $9$
and thus the pattern shall continue.
Since we want to find the last digit of $9^9$, we want to find the remainder of $9/2$ because the pattern repeats in twos. the remainder of $9/2$ is 1, so we want to find the last digit of $9^1$, which is $9$. So the last digit of $9^9$ is $9.$
Finaly, we do $5 + 6 + 3 + 6 + 9$ which is $29$, and the last digit of $29$ is $9,$ so our answer is $9.$
I worked hard on this so pls give me credit.
ilovepizza2020
19.09.2020 03:07
littlepetwolf wrote: This is not a tricky problem but requires a little time. I think you could figure it out with some effort, but here's the solution anyway lol. XD
The last digit of $5^5$, is of course $5$
The last digit of $6^6$ requires a little more work.
$6^1$ last digit = $6$
$6^2$ last digit = $6$
$6^3$ last digit = $6$
.
.
.
.
and so on. So we know that the last digit of $6^6$ is $6$
The last digit of $7^7$ is more tricky.
$7^1$ last digit = $7$
$7^2$ last digit = $9$
$7^3$ last digit = $3$
$7^4$ last digit = $1$
$7^5$ last digit = $7$,
and thus the pattern shall continue.
Since we want to find the last digit of $7^7$, we want to find the remainder of $7/4$ because the pattern repeats in fours. the remainder of $7/4$ is 3, so we want to find the last digit of $7^3$, which is $3$. So the last digit of $7^7$ is $3.$
Same thing with $8^8$
$8^1$ last digit = $8$
$8^2$ last digit = $4$
$8^3$ last digit = $2$
$8^4$ last digit = $6$
$8^5$ last digit = $8$,
and thus the pattern shall continue.
Since we want to find the last digit of $8^8$, we want to find the remainder of $8/4$ because the pattern repeats in fours. the remainder of $8/4$ is 0, so we want to find the last digit of $8^4$, which is $6$. So the last digit of $8^8$ is $6.$
and $9^9$
$9^1$ last digit = $9$
$9^2$ last digit = $1$
$9^3$ last digit = $9$
and thus the pattern shall continue.
Since we want to find the last digit of $9^9$, we want to find the remainder of $9/2$ because the pattern repeats in twos. the remainder of $9/2$ is 1, so we want to find the last digit of $9^1$, which is $9$. So the last digit of $9^9$ is $9.$
Finaly, we do $5 + 6 + 3 + 6 + 1$ which is $21$, and the last digit of $21$ is $1,$ so our answer is $1.$
I worked hard on this so pls give me credit. But it was wrong
littlepetwolf
19.09.2020 03:08
oh. I'm so sorry. I tried my best.
ilovepizza2020
19.09.2020 03:09
littlepetwolf wrote: oh. I'm so sorry. I tried my best. I would review the 9 part. But good job!
littlepetwolf
19.09.2020 03:10
yeah I saw, that was silly of me. sorry! It's fixed!
ilovepizza2020
19.09.2020 03:12
littlepetwolf wrote: yeah I saw, that was silly of me. sorry! It's fixed! Good job! Keep doing this and you will have a bright future!
riemann-1
19.09.2020 04:20
Keith50 wrote: Determine the last digit of $5^5+6^6+7^7+8^8+9^9$. It only involve the basic concept of cyclcity of of the unit digit
CoolJupiter
19.09.2020 04:32
def cooljupiter(n): return n ** n result = 0for k in range(5, 10): result += cooljupiter(k)print(result % 10)def cooljupiter(n): return n ** n result = 0 for k in range(5, 10): result += cooljupiter(k) print(result % 10)RunResetPop Out /
jasperE3
19.09.2020 05:15
This should be moved to MSM.
$5^{5}+6^{6}+7^{7}+8^{8}+9^{9}=405071029\rightarrow\boxed9$
ilovepizza2020
19.09.2020 16:33
jasperE3 wrote: This should be moved to MSM.
$5^{5}+6^{6}+7^{7}+8^{8}+9^{9}=405071029\rightarrow\boxed9$
Nice Bash, but there are better ways...
juliankuang
19.09.2020 16:40
CoolJupiter wrote: def cooljupiter(n): return n ** n result = 0for k in range(5, 10): result += cooljupiter(k)print(result % 10)def cooljupiter(n): return n ** n result = 0 for k in range(5, 10): result += cooljupiter(k) print(result % 10)RunResetPop Out / This method is definitely applicable to similar problems in math contests.
CoolJupiter
19.09.2020 17:45
juliankuang wrote: CoolJupiter wrote: def cooljupiter(n): return n ** n result = 0for k in range(5, 10): result += cooljupiter(k)print(result % 10)def cooljupiter(n): return n ** n result = 0 for k in range(5, 10): result += cooljupiter(k) print(result % 10)RunResetPop Out / This method is definitely applicable to similar problems in math contests. Yeah, like Purple Comet
Zanderman
19.09.2020 20:48
Powers of $5$
We find the pattern that any power of $5$ other than $5^0$ will end in a $5$, so we know that $5^5 \equiv 5 \pmod{10}$.
Powers of $6$
We find that the pattern in powers of $6$ also always ends in a $6$ except for when it's $6^0$, so we know that $6^6 \equiv 6 \pmod{10}$.
Powers of $7$
We find the pattern in powers of $7$ goes: $1, 7, 9, 3$, so we find that $7^7 \equiv 7^3 \equiv 3 \pmod{10}$.
Powers of $8$
The pattern in powers of $8$ goes: $8, 4, 2, 6$, so $8^8 \equiv 8^4 \equiv 6 \pmod{10}$.
Powers of $9$
The pattern is powers of $9$ goes $1, 9$, so we know that $9^9 \equiv 9^1 \equiv 9 \pmod{10}$
And summing: $5+6+3+6+9 \equiv \boxed{9} \pmod{10}$
fungarwai
20.09.2020 13:12
$5^5+6^6+7^7+8^8+9^9\equiv 1+0+1+0+1\equiv 1\pmod{2}$ By Fermat's little, $a^4\equiv 1\pmod{5}$ when $(a,5)=1$ $5^5+6^6+7^7+8^8+9^9\equiv 0+1^2+2^3+1+(-1)^1\equiv 4\pmod{5}$ By CRT, $5^5+6^6+7^7+8^8+9^9\equiv 5\times 1 -4\times 4\equiv 9\pmod{10}$
littlepetwolf
21.09.2020 01:04
littlepetwolf wrote: This is not a tricky problem but requires a little time. I think you could figure it out with some effort, but here's the solution anyway lol. XD
The last digit of $5^5$, is of course $5$
The last digit of $6^6$ requires a little more work.
$6^1$ last digit = $6$
$6^2$ last digit = $6$
$6^3$ last digit = $6$
.
.
.
.
and so on. So we know that the last digit of $6^6$ is $6$
The last digit of $7^7$ is more tricky.
$7^1$ last digit = $7$
$7^2$ last digit = $9$
$7^3$ last digit = $3$
$7^4$ last digit = $1$
$7^5$ last digit = $7$,
and thus the pattern shall continue.
Since we want to find the last digit of $7^7$, we want to find the remainder of $7/4$ because the pattern repeats in fours. the remainder of $7/4$ is 3, so we want to find the last digit of $7^3$, which is $3$. So the last digit of $7^7$ is $3.$
Same thing with $8^8$
$8^1$ last digit = $8$
$8^2$ last digit = $4$
$8^3$ last digit = $2$
$8^4$ last digit = $6$
$8^5$ last digit = $8$,
and thus the pattern shall continue.
Since we want to find the last digit of $8^8$, we want to find the remainder of $8/4$ because the pattern repeats in fours. the remainder of $8/4$ is 0, so we want to find the last digit of $8^4$, which is $6$. So the last digit of $8^8$ is $6.$
and $9^9$
$9^1$ last digit = $9$
$9^2$ last digit = $1$
$9^3$ last digit = $9$
and thus the pattern shall continue.
Since we want to find the last digit of $9^9$, we want to find the remainder of $9/2$ because the pattern repeats in twos. the remainder of $9/2$ is 1, so we want to find the last digit of $9^1$, which is $9$. So the last digit of $9^9$ is $9.$
Finaly, we do $5 + 6 + 3 + 6 + 9$ which is $29$, and the last digit of $29$ is $9,$ so our answer is $9.$
I worked hard on this so pls give me credit. This i how I solved it.
OlympusHero
21.09.2020 02:37
You could use modular arithmetic to determine the value of each power $\pmod {10}$