umg wrote:
What is the last non-zero digit of this expression?
\((573)^3^2^7*(274)^3^6*(860)^x*(1525)^y\)
Ki
(1) \(x = 10, y = 18\)
(2) \(y^2 - 36y + 324= 0\)
Hi..
Q should have mentioned that x and and y are positive integers
We are looking for last non zero digit...
It will depend on the last non-zero digit of each termExample...
260*3*40 would depend on 6*3*4...
260^4*27^6*800^567 would depend on 6^4*7^6*8^567
Let's see the statements..
(1) \(x = 10, y = 18\)
X and y are given, so we will be able to find exact value of equation.
Sufficient
(2) \(y^2 - 36y + 324= 0\)
This equation comes down to (y-18)^2=0 or y=18..
But what about x? This is the tricky part.
We are looking for LAST non-zero digit of each term.
Rest all terms are known except x as y is 18
Here the term is 860^x and it will be same as 6^x.
Since the last non-zero digit is 6 here and 6 to any power will have last digit as 6 only, the value of x will not affect the last non-zero digit of entire term..
6^1=6.....
6^2=36.... Last digit 6
6^3=216.... Last digit again 6
Sufficient
D
Hi..
Yes, you are right.
Pl look at the coloured portion above.
Not only fraction but also negative integer.
It has to be ONLY POSITIVE integer