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1 5 + 2 5 + 3 5 + 4 5 + 5 5 + … + 1 2 2 5 + 1 2 3 5
Find the units digit of the number above.
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8 solutions
Moderator note:
As Prasun has pointed out, you need to show that n 5 ≡ n ( mod 1 0 ) is true for all integers n .
Not quite. Euler's Theorem states that only for positive integers n which are coprime to 1 0 .
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True. Though it is possible to extend Euler's Theorem to the following:
For a , n positive integers where n is squarefree, we have:
a ϕ ( n ) + 1 ≡ a ( mod n ) .
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The extension you mentioned seems really interesting. I shall look into it later. For now, here's my take on the problem.
Lemma: For all integers n , we have n 5 ≡ n ( m o d 1 0 )
Proof: First, by Fermat's Little Theorem, we have n 5 ≡ n ( m o d 5 ) ∀ n ∈ Z . Now, we have,
∀ n ∈ Z , n 5 ≡ ⎩ ⎪ ⎨ ⎪ ⎧ { 0 , when n is even 1 , when n is odd ( m o d 2 ) n ( m o d 5 )
Using Chinese Remainder Theorem, we get,
∀ n ∈ Z , n 5 ≡ { 6 n , when n is even 5 + 6 n , when n is odd
Now, use n = 2 k for even n and n = 2 k + 1 for odd n where k ∈ Z . It'll follow that n 5 ≡ n ( m o d 1 0 ) ∀ n ∈ Z .
Hence, the lemma is established and proved.
Using this Lemma, our problem becomes,
n = 1 ∑ 1 2 3 n 5 ≡ n = 1 ∑ 1 2 3 n ≡ 2 1 2 3 ⋅ 1 2 4 ≡ 1 2 3 × 6 2 ≡ 3 × 2 ≡ 6 ( m o d 1 0 )
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@Prasun Biswas – Very nice! And more complete than my solution, I did 'cheat' by omitting a real proof for the lemma.
About the Euler extension, I'm not sure if it already has a name or if it has been studied thoroughly, I at least couldn't find much on the internet. But I did manage to write down a proof; I can post it later if people are interested!
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@Tijmen Veltman – Yes, I'd like to see the proof of that extension for myself. So, you already have one person interested in it. You can either add it here or maybe post it separately as a note.
In the meantime, I shall give it a try myself 'cause it looks really interesting. :D
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@Tijmen Veltman – That's a really neat proof! Will take me some time to understand it completely. :P
Nonetheless, I already liked and reshared that note. Do you know that I had to make a separate set just now to save that note of yours?! :D :)
Lets 1st find the digit on unit place for no.s 1 to 10 using cyclicity of no.s:-
digit on unit place 1^5=1
2^5=2
3^5=3
4^5=4
5^5=5
6^5=6
7^5=7
8^5=8
9^5=9
10^5=0
sum of which=45
on continuing it upto 120
i.e. 11^5=1
12^5=2
..
..
..
..
120^5=0
sum of 10 terms mentioned above=45
if we do so till120
i.e. 45+45+45...........................................+45 (12 times)
=0---------------1
so digit at units place of
(1^5)+ (2^5)+(3^5)+........................................ (119^5)+( 120^5)=0
now, digit at units place of
=(121^5)+ (122^5)+ (123^5)
=1+2+3
=6---------------2
now from 1 & 2
answer is 0+6=6
Moderator note:
Going through the calculation works too. But it gets tedious. You should be careful with your notations, 2 5 = 2 is simply not true. Try 2 5 ≡ 2 .
I adopted the same process. Nice solution Ambuj
Easy way to do this is realizing that all units digit for exponent number repeats after 4.
Example: 2,4,8,16,32, as you see, there is a repetition after 4 different unit digits, pattern for two is 2,4,8,6, and repeat.
Knowing this, we can immediately conclude that all the units digit of the numbers are their bases.
Then we add numbers from 1 to 10; and that gives us 55, with a units digit of 5. Because there are 12 full tens in 123, we multiply this units digit by 12 giving us 60 with a unit's digit of 0. The left over terms are 121,122,123 we add the unit digits.
We get 6
can u elaborate?
Well, it is pretty obvious that something to the fifth power last digit is the last digit of itself. Therefore, 1+2+3...+0 last digit is 5 since there is 12set if it 5×12 last digit is 0. Finally, 1+2+3=6 no need to make life more complicated.
Notice this, 1 to 120 will have 0 in the units digit because it is 120 times the units digit of 1 to 10, so all you need to calculate is the units digit of 121 -> 1, 122 -> 2, and 123 ->3, since we only need the units digit. 1 5 + 2 5 + 3 5 = 1 + 3 2 + 2 4 3 Since we only require the units digit, therefore 1 + 2 + 3 = 6
Moderator note:
Your first statement is not that obvious. Can you elaborate on it?
I think what the solution writer means here is that if we look at the unit digit of all the fifth powers, we will see that 1^5+...+120^5 has in the unit's digit the same number as the units digit in 12*(1^5+..+10^5) which is easy to calculate to 0. Hence the answer is the last digit of 121^5+122^5+123^5 which is 6.
unit digit of 1^5 is 1, 2^5 is 2, and same goes with 3, 4,5,6,7,8,9, and till 120 we get 12 that no. which end with 1,2,3,4,5,6,7,8,9&0. so the summation of unit digit till 120 is 0. now see for unit digit of 121^5, 122^5& 123^5. we get answer 6.
I solved the above question by using the options given.
I divided the whole by 5.
And used the euler totient rule. So the question becomes (1+2+3+..............+123)÷5 Now if we observe it carefully we will find that 2+123=125, 3+122=125 which is divisible by 5
Therefore we will be left with only 1 which would be the remainder after division by 5. So according to the options given if the remainder comes out to be 1 then it the last digit should be 6. If 2 then the answer would be 2 If 3 then 8 and if 4 then 4. Since the remainder is 1
therefore the answer is 6
If the remainder is 1 then last digit is 6 and not 1 itself. Why is it so?
Using fermat's little theoram
Thanks to Euler's totient theorem, we know that n 5 ≡ n ( mod 1 0 ) for all integers n . This means that, modulo 10, we have:
1 5 + 2 5 + … + 1 2 3 5 ≡ 1 + 2 + … + 1 2 3 = 2 1 2 3 ⋅ 1 2 4 = 1 2 3 ⋅ 6 2 ≡ 3 ⋅ 2 = 6 .