What is the rightmost non zero digit of the number (30)^2720 and how?Please explain.
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30^2720 = 3^2720 * 10^2720
As 10^2720 produces all zeroes from right side, the last digit (unit digit) of 3^2720
is the right most non zero digit of 30^2720.
To find the unit digit of 3^2720:
( refer to “Magic of Numbers )
As 2720 is multiple of 4, we consider 4th digit of the unit digit pattern of 3n.
The unit digit pattern of 3n is 3,9,7,1
Answer is (1)
≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡
Another way to solve this is =
30^1=3
30^2=9
30^3=7
30^4=1
The pattern repeats from here.
2720/4 gives the remainder 0
Right most digit is 1.
30^2720 = 3^2720 * 10^2720
As 10^2720 produces all zeroes from right side, the last digit (unit digit) of 3^2720
is the right most non zero digit of 30^2720.
To find the unit digit of 3^2720:
( refer to “Magic of Numbers )
As 2720 is multiple of 4, we consider 4th digit of the unit digit pattern of 3n.
The unit digit pattern of 3n is 3,9,7,1
Answer is (1)
≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡
Another way to solve this is =
30^1=3
30^2=9
30^3=7
30^4=1
The pattern repeats from here.
2720/4 gives the remainder 0
Right most digit is 1.
Answered by
6
Answer:
30^2720= (3)^2720 ×( 10)^2720
As 10^2720 have always 0 as it's unit digit in every condition
Now, 3^2720
3^1= 3
3^2=9
3^3=27
3^4=81
3^5=243 ( It is repeating the unit digit again after 4 consecutive powers)
So 3 has Cyclicity of 4
2720/4= 0
3^0= 1
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