MAYBE a familiar problem

I have several problems. One of them is: Find is the rightmost digit of 1+4^1+4^2+…4^2012 As it's my 1st discussion I started by picking only 1 problem...

Note by Sheikh Asif Imran Shouborno
6 years, 4 months ago

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as 1+4^1+4^2+…4^2012 = 1+4+6+...+6 = 1+0+....0 = 1

Anoopam Mishra - 6 years, 4 months ago

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Take a look at this. 41=44^1=4 , 42=164^2=16, 43=644^3=64, 44=2564^4=256. In other words, 4odd=...44^{odd}= ...4 and 4even=...64^{even}= ...6. So, 4odd+4even=...104^{odd}+4^{even}=...10 Now we are going to take the terms in pairs. 1+(41+42)+...+(42011+42012)1+(4^1+4^2)+...+(4^{2011}+4^{2012}) We know that terms inside the brackets are going to have 00 as the rightmost digit. And adding all of them up together will have 00 as the rightmost digit. So if we add the 11 that is outside the bracket (the leftmost 11), the last digit's going to be 11. Hope this helps!

Mursalin Habib - 6 years, 4 months ago

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A slightly cleaner way (which is still saying that same thing) is to say that for n0n\geq 0, 42n+1+42n+2=42n(4+16)=42n×20 4^{2n+1} + 4^{2n+2} = 4^{2n} ( 4 + 16) = 4^{2n} \times 20 , hence has a units digit of 0.

This avoids the slight issue of 40=1 4^0 =1 , while claiming that 4even=6 4^{even} = \ldots 6 .

Calvin Lin Staff - 6 years, 4 months ago

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Thank you for pointing it out.

Mursalin Habib - 6 years, 4 months ago

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Consider the following: 42k4(mod10)4^{2k}\equiv 4{\pmod{10}} and 42k+16(mod10)\\4^{2k+1}\equiv 6{\pmod{10}} Therefore 1+41+42+43...42012(mod10) 1+4^1+4^2+4^3...4^2012 \equiv \pmod{10} Observe all the 4s4's and 6s6's pair up and we get the remainder 11 mod 10. So the rightmost digit is 11

Vikram Waradpande - 6 years, 4 months ago

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Latex error: It should be 1+41+42+43...420121+4+6+4+6.....(mod10)1+4^1+4^2+4^3...4^{2012}≡ 1+4+6+4+6.....\pmod{10}

Vikram Waradpande - 6 years, 4 months ago

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right

superman son - 6 years, 4 months ago

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Ignoring 1, if we start with 4^1, we will be able to recognise that 4^(odd number) has the last digit as 4, for eg:-4^3 = 64 and 4^5=1024. In the same way, 4^(even number) has the last digit as 6. The sum is congruent to 0(mod10). Now adding 1 in the sequence, we will get the sum congruent to 1(mod10). So 1 is the last digit.

Siddharth Kumar - 6 years, 4 months ago

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Another method to solve this problem, is to observe that the numbers are in geometric progression. So use the sum formula of geometric progression and try to find the last digit either by simple calculation, or modular arithmetic.

Siddharth Kumar - 6 years, 4 months ago

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