chetan2u wrote:
What is the units digit of \((2x)^8\)?
(1) The unit's digit of \(x^2\) is 9.
(2) The unit's digit of \(x^3\) is 7.
When we talkalk of units digit, we should know the following rules of cyclicity..
every digit has cyclicity when it comes to last/unit's digit...each digit surely repeats the units digit after every 4th power..
i)
1, 5 and 6 and 0 repeat with each power..1^1 or 1^2 will always give 1 and similarly for 5,6,0
ii)
4,9 repeat after increase of two powers..4^1 or 4^3 or 4^5 will all leave 4 as last digit and 4^2 or 4^4 will leave 6 as last digits
so cyclicity is 4,6,4,6..
for 9 it is 9,1,9,1,9...
iii)
remaining 2,3,7,8 repeat after every 4th powerso 2^1 , 2^(1+4), 2^(1+4+4) will leave 2 in each case ..cyclicity is 2,4,8,6,2,4,8,6..
for 3 it is 3,9,7,1,3,9,..
for 7 it is 7,9,3,1,7,9,3,1...
for 8 it is 8,4,2,6,8,4,2,6..
So now we can look at the question..
Units digit of (2x)^8...
our answer will depend on xx^8 will have same units digit as x^4...
(1) The unit's digit of \(x^2\) is 9.
Now x has to be odd and we can see from above that 7^2 and 3^2 leave 9 as the units digit.
Easy to think that different possibilities of x and take it aa insufficient,but we are looking for x^4..
So here is another point to remember..
All odd digits have same units digit, that is 1, when raised to power 4Thus both 3 and 7 raised to 4 will have 1 units digit
Sufficient
(2) The unit's digit of \(x^3\) is 7
Again we can see that x can only be 3 and x^4 =7^4 that is the units digit is 1.
Sufficient.
D
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