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01 Sep 2019, 16:55
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I don't know if it's a very "GMAT-like" question ... maybe at 800 level? It's not immediately obvious how best to manipulate and brute force is too time consuming as well, I started by the same reasoning method EMPOWERgmatRichC showed (you can quickly plug in x=1,y=1 to see that x>y) but abandoned it because it's way too much to keep straight and honestly I doubt many people besides Q51 geniuses could reason through that in any sensible amount of time under test conditions... If I saw a question like this on the test I would just guess and move on.
Anyways, a bit of a different solution to those above:
\((5^x)-(5^y)=(2^{y-1})*(5^{x-1})\)
\((5^x)-(2^{y-1})*(5^{x-1})=(5^y)\)
\((5^{x-1})[(5^1)-(2^{y-1})]=(5^y)\)
Since we have powers of 5 on both sides the the only way for LHS to be a power of 5 is if the expression in brackets is = 1
Therefore, \([(5)-(2^{y-1})] = [5-2^{3-1}] = [5-4] = 1\)
We can see that y=3, and now we have \((5^{x-1})[1]=(5^3)\) so \(x-1 = 3\)
Now we can see that x = 4 ... 4*3 = 12, E)
17 May 2026, 11:17
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(5^x - 5^y = 2^(y-1) × 5^(x-1))
We know the RHS is positive.
So:
(5^x - 5^y > 0)
Since powers of 5 increase with exponent,
(x > y)
Now comes the key simplification step.
Notice the equation contains three powers of 5:
(5^x , 5^y , 5^(x-1))
The RHS already has (5^(x-1)), so divide the entire equation by (5^(x-1)).
The goal is to “normalize” the powers and make the structure visible.
((5^x / 5^(x-1)) - (5^y / 5^(x-1)) = (2^(y-1) × 5^(x-1)) / 5^(x-1))
Now simplify each term:
(5^x / 5^(x-1) = 5)
and
(5^y / 5^(x-1) = 5^(y-(x-1)) = 5^(y-x+1))
So the equation becomes:
(5 - 5^(y-x+1) = 2^(y-1))
Now the structure becomes much cleaner.
Right side is a pure power of 2.
Therefore the left side must also be a positive integer power of 2.
Since (x > y),
(y - x + 1 ≤ 0)
Now test possibilities.
If:
(y - x + 1 = 0)
then:
(5 - 5^0 = 5 - 1 = 4)
which is:
(2^2)
Perfect — a power of 2.
Now suppose exponent were negative.
Then we'd get things like:
(5 - 1/5 , 5 - 1/25)
These are not integers.
But RHS:
(2^(y-1))
is always an integer.
Impossible.
Positive exponent is impossible because:
(y - x + 1 > 0 ⇒ y ≥ x)
which contradicts:
(x > y)
So the only possibility is:
(y - x + 1 = 0)
Thus:
(x = y + 1)
And since:
(5 - 1 = 4)
we get:
(2^(y-1) = 4 = 2^2)
So:
(y - 1 = 2)
(y = 3)
Then:
(x = 4)
Therefore:
(xy = 4 × 3 = 12)
(12)
We know the RHS is positive.
So:
(5^x - 5^y > 0)
Since powers of 5 increase with exponent,
(x > y)
Now comes the key simplification step.
Notice the equation contains three powers of 5:
(5^x , 5^y , 5^(x-1))
The RHS already has (5^(x-1)), so divide the entire equation by (5^(x-1)).
The goal is to “normalize” the powers and make the structure visible.
((5^x / 5^(x-1)) - (5^y / 5^(x-1)) = (2^(y-1) × 5^(x-1)) / 5^(x-1))
Now simplify each term:
(5^x / 5^(x-1) = 5)
and
(5^y / 5^(x-1) = 5^(y-(x-1)) = 5^(y-x+1))
So the equation becomes:
(5 - 5^(y-x+1) = 2^(y-1))
Now the structure becomes much cleaner.
Right side is a pure power of 2.
Therefore the left side must also be a positive integer power of 2.
Since (x > y),
(y - x + 1 ≤ 0)
Now test possibilities.
If:
(y - x + 1 = 0)
then:
(5 - 5^0 = 5 - 1 = 4)
which is:
(2^2)
Perfect — a power of 2.
Now suppose exponent were negative.
Then we'd get things like:
(5 - 1/5 , 5 - 1/25)
These are not integers.
But RHS:
(2^(y-1))
is always an integer.
Impossible.
Positive exponent is impossible because:
(y - x + 1 > 0 ⇒ y ≥ x)
which contradicts:
(x > y)
So the only possibility is:
(y - x + 1 = 0)
Thus:
(x = y + 1)
And since:
(5 - 1 = 4)
we get:
(2^(y-1) = 4 = 2^2)
So:
(y - 1 = 2)
(y = 3)
Then:
(x = 4)
Therefore:
(xy = 4 × 3 = 12)
(12)
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