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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
4 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 · 4 years, 4 months ago

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Take a look at this. \(4^1=4\) , \(4^2=16\), \(4^3=64\), \(4^4=256\). In other words, \(4^{odd}= ...4\) and \(4^{even}= ...6\). So, \(4^{odd}+4^{even}=...10\) Now we are going to take the terms in pairs. \(1+(4^1+4^2)+...+(4^{2011}+4^{2012})\) We know that terms inside the brackets are going to have \(0\) as the rightmost digit. And adding all of them up together will have \(0\) as the rightmost digit. So if we add the \(1\) that is outside the bracket (the leftmost \(1\)), the last digit's going to be \(1\). Hope this helps! Mursalin Habib · 4 years, 4 months ago

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

This avoids the slight issue of \( 4^0 =1 \), while claiming that \( 4^{even} = \ldots 6 \). Calvin Lin Staff · 4 years, 4 months ago

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@Calvin Lin Thank you for pointing it out. Mursalin Habib · 4 years, 3 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 · 4 years, 3 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 · 4 years, 3 months ago

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Consider the following: \(4^{2k}\equiv 4{\pmod{10}}\) and \(\\4^{2k+1}\equiv 6{\pmod{10}}\) \[\] Therefore \( 1+4^1+4^2+4^3...4^2012 \equiv \pmod{10} \) Observe all the \(4's\) and \(6's\) pair up and we get the remainder \(1\) mod 10. So the rightmost digit is \(1\) Vikram Waradpande · 4 years, 4 months ago

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@Vikram Waradpande Latex error: It should be \(1+4^1+4^2+4^3...4^{2012}≡ 1+4+6+4+6.....\pmod{10}\) Vikram Waradpande · 4 years, 4 months ago

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@Vikram Waradpande right Superman Son · 4 years, 3 months ago

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