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Algebra Problem

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by kmittal82 » Fri Oct 14, 2011 6:38 am
5^21 + 2^22 = 2 x 5^n x 2^n

5^21 + 2^22 = 2^(n+1) x 5 ^ n

Comparing the sides, we get n=21
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by GMATGuruNY » Fri Oct 14, 2011 6:49 am
Gerhard wrote:Can anyone assist with solving:

5^21 * 4^11 = 2 x 10^n

Solve for n.
LHS = left-hand side of the equation, RHS = right-hand side of the equation.
The problem has been posted incorrectly: the operation on the LHS is not addition but multiplication.
I've amended the question to reflect its intent.

Since the LHS has 5²¹, the RHS must also have 5²¹.

Thus:
5²¹ * 4¹¹ = 2 * (10)^n.
5²¹ * 4¹¹ = 2 * (2*5)^n.
5²¹ * 4¹¹ = 2 * 2^n * 5^n.
Thus, n=21.

No need to worry about the rest of the equation: the only way the LHS can be equal to the RHS is if both sides include 5²¹.
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by nsamkari » Sun Oct 16, 2011 3:11 am
Dear gmatguruny,

how is it come that doesn't matter if the exponent of 4 is different than 11 and the most important part is 5^21!!!!!
could you clarify please?
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by Abhishek009 » Sun Oct 16, 2011 3:31 am
nsamkari wrote:Dear gmatguruny,

how is it come that doesn't matter if the exponent of 4 is different than 11 and the most important part is 5^21!!!!!
could you clarify please?
Take it like this -


5^21 * 4^11 = 2 x 10^n

or , 5^21 * 2^22 = 2 x (2*5)^n

or , 5^21 * 2^22 = 2 x 2^n * 5^n

or, 5^21 * 2^22 = 2^(n+1) * 5^n



Thus we can say n=21.
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by GMATGuruNY » Sun Oct 16, 2011 3:33 am
nsamkari wrote:Dear gmatguruny,

how is it come that doesn't matter if the exponent of 4 is different than 11 and the most important part is 5^21!!!!!
could you clarify please?
5²¹ * 4¹¹ = 2 * 2^n * 5^n.

It is given that the LHS is equal to the RHS.
Thus, if the LHS has a prime number raised to a power, then the RHS must include the SAME PRIME NUMBER raised to the SAME POWER.
Since the LHS has 5²¹, the RHS must also have 5²¹.
Thus, on the RHS, 5^n = 5²¹.
n=21.

No need to do any more work.

For the skeptical, here's the full algebraic solution:
5²¹ * 4¹¹ = 2 * (10)^n.
5²¹ * 4¹¹ = 2 * (2*5)^n.
5²¹ * 4¹¹ = 2 * 2^n * 5^n
5²¹ * (2²)¹¹ = 2^(n+1) * 5^n
5²¹ * 2²² = 2^(n+1) * 5^n.

The LHS has 2²².
Thus, on the right hand side:
2^(n+1) = 2²²
n=21.
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by GMATbrainwashesme » Wed Aug 08, 2012 6:05 pm
I got confused because in the solving proposed above people consider the equation 5^21 PLUS 4^11 as 5^21 TIMES 4^11. Please help.
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by truplayer256 » Wed Aug 08, 2012 8:44 pm
5^(21) * 4^(11) = 2 x 10^n

5^(21) * 4^(11) = 2^(n + 1) x 5^n

2^(22) * 5^(21) = 2^(n + 1) x 5^n

Note that only n = 21 works in this case.
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by GMATbrainwashesme » Thu Aug 09, 2012 2:29 am
Thank you for your reply. Although...

5^21 PLUS 2^11 = 2 TIMES 10^n
5^21 PLUS 2^11 = 2^(n+1) TIMES 5^n --->>> or 5^21 PLUS 2^11 = 5^n TIMES 2^(n+1)

My question is because the equations don't match because on LHS it is an addition and on the RHS it is a multiplication.
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