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Okay C t s baring math people Here we go

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Another question for you Ah pattern exists among the unit's

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digits of the powers of three is shown below right

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down there What's the unit's digit of three to the

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ninety fifth power And note the problem gives you that

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The unit's digit of nineteen thousand six hundred eighty three

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is three No one wants to figure out three to

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the ninety fifth by hand So it's calculator time Yeah

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of course This on lee seems like a good idea

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until we realized that the scientific calculator on ly presents

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answers in scientific notation which doesn't help us find ah

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units digit like this thing right here Instead we gotta

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look for a pattern in the powers of three which

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is a well with problems suggested in the first place

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The digits cycle through a one three nine and then

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seven before beginning again we see the pattern there So

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the task now is to create a formula that tells

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us which of these units digits applies to any given

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number without counting to ninety five In groups of four

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we have three to zero through the fourth and then

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through the eighth all of those have units digits of

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one notice that the exponents of these terms are all

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evenly divisible by four Meanwhile we have three of the

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first real fifth and three to the ninth and they

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all have units digits of three Well the exponents of

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thes terms aren't divisible by four but they all share

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a remainder of one when divided by four So that's

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the pattern when divided by four exponents with remainder sze

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of to have units digits of nine while exponents with

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reminders of three have units digits of seven All right

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now let's give ninety five a shot Well ninety five

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divided by four is twenty three remainder three right So

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since the remainder is three the unit's digit is seven

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So there we go The answer here is d It's

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got to be seven right See have it gets it

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gets you cleverly there in number four It's always a 00:01:58.213 --> [endTime] seven That's it That was kind of tricky

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