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...

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
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Comments

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

Anoopam Mishra - 8 years ago

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 - 8 years ago

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 - 8 years ago

Thank you for pointing it out.

Mursalin Habib - 8 years ago

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 - 8 years ago

Latex error: It should be $1+4^1+4^2+4^3...4^{2012}≡ 1+4+6+4+6.....\pmod{10}$

Vikram Waradpande - 8 years ago

right

superman son - 8 years ago

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 - 8 years ago

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 - 8 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$