# Question

Find the last digit for

7^401-3^444

GEP P6 question

For incremental power, the last digit will repeats in cycle of 4.

Power   > last digits

71              => 7             remainder of 1
72             => 9             remainder of 2
73                => 3             remainder of 3
74                 => 1             remainder of 0, multiple of 4.

75          => 7
76          => 9
77              => 3         last digit repeat after a cycle of 4.
78             => 1

For the below, 3 to the power, is the same, repeat in cycle of 4

31          = >3             remainder of 1

32          => 9            remainder of 2

33         => 7             remainder of 3

34          =>1           remainder of 0, multiple of 4.

So just use the power, divide by 4 and get the remainder. Use the remainder to match the above to find out the last digit.

7401 => 401/4 => Remainder is 1.  Last digit will be 7

3444=> 444/4 => Remainder is 0,  last digit will be 1.

Last digit of 7401 – 3444  is 7-1 => 6 ##

Is this really a P6 question?  It feels more like math olympiad type questions.  Are P6s taught on these concepts?

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OMG ! I think question this is rather “chim” if it is P6 level.

Well done ! Ender.

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Thanks so much for the answer.

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