Question

(437x)^939 where the last digit is 3 find x

Answer 1

(437x)*(939)
To get last digit 3
we write x=7
Multiply two unit places
Because x*9=7×9= 63 unit place is 3

Answer 2

The possible number of x are
3 & 7

1.cyclicity of 3 are --> 3,9,7,1,3,9.............

The unit digit 3 repeats after 4 th time
That means the unit digit of 3^n will be there only when n is divisible be 4

But 939 is not divisible by 4, hence the value of x cannot be 3.

2. Cyclicity of 7 are --> 7,9,3,1,7,9.............

The unit digit 3 repeats after 3 th time
That means the unit digit of 7^n will be three only when n is divisible be 3

As 939 is not divisible by 3, hence the value of x Will be 7.

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