Factors- I am losing it

[amit803]

x is a positive number. If 9

x + 9x+1 + 9x+2 + 9x+3 + 9x+4 + 9x+5 = y, is y divisible by 5?
1) 5 is a factor of x.

2) x is an integer.

Official Answer -

SPOILER: D

[Itsmyturn]

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

[amit803]

Anyone ?

[Fiver]

x is a positive number. If 9

x + 9x+1 + 9x+2 + 9x+3 + 9x+4 + 9x+5 = y, is y divisible by 5?

1) 5 is a factor of x.

2) x is an integer.

Official Answer -

SPOILER: D

5 is a factor of x means x=5*a, where a is some integer. If so does it not imply that x has to be an integer?
My pick is A.

[Makumajon]

I guess the question is

9^x+9^(x+1)+9^(x+2)+9^(x+3)+9^(x+4)+9^(x+5)
=9^x(1+9+9^2+9^3+9^4+9^5)
=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.
=9^x(*0).

*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.

[adt29]

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?

[sandeep_chads]

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?

I didn;t get it. Why would you refute B.

You were right all through out.

(9^x) X (10y) = multiple of 10 only if
we can prove that 9^x will not result in a fraction or an irrational number.

Both A and B prove the statement therefore D is the right answer

[mxplusb]

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?

Sure, if x=1/2 then y is divisible by 5. But try x=1/4. So, for a non-integer x, y might be divisible by 5, or it might not be. You can't tell, so you need condition #2.