As per my understanding, the question stem translates into
y = 6(9x) + (1+2+3+4+5)
y = 54x +15
Based on this only A can be the answer
not sure how you got D
I guess the question is
=9^x(1+9+*1+*9+*1+*9), where * represents one or more digit. I have not shown other digits because the Q asks about divisibility by 5. Therefore, it is enough to see whether units’ digit is 0 or 5.
*0 is divisible by 5. But for the whole 9^x(*0) to be divisible by 5, we must show that x is a positive integer. The Q-stem tells that x is positive. So, we only need to show whether x is an integer.
(1) 5 is a factor of x. So, x is an integer. Sufficient.
(2) x is an integer. Sufficient.
I did the problem this way.
Factored out 9^x. So we have the equation simplified to:
9^x (1+ 9^1 + 9^2 + 9^3 + 9^4 + 9^5) = y.
Since odd powers of 9 end in 9, the sum of ( 9^1 + 9^2 + 9^3 + 9^4 + 9^5 ) will end in 9. If you add a '1' to that, you get some number that ends in '0'.
Now if x=1/2 which is not an integer, 9^(1/2) = 3. And 3 times a number that ends in 0, will give you a number that is divisible by 5.
So doesn't statemtnet 2 get refuted? What am I missing here?
Hey Harvard, I am right here!!
rep me if I made some sense
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