Thanks Bunuel for these interesting problems.

With respect to problem number 2

"So, we have that the units digit of is 1 and the units digit of is 3. Also notice that the second number is much larger then the first one, thus their difference will be negative, something like 11-13=-2, which gives the final answer that the units digit of is 2."

I don't understand the above statement. If the last digit is 1 for the smaller number and the last digit is 3 for the larger number.
And we are trying to find different between smaller number - larger number

eg 757571 - 58299374483. We borrow one from the previous digit and make 1 as 11 and then subtract 3 from 11. So shouldn't the last digit be 8.

I am unable to think any other way. Please let me know where I am going wrong.

Thanks
gmatrant

First of all 757,571 - 58,299,374,483 = -58,298,616,912

You could consider easier cases: 11-13=-2, 11-23=-12, 21-83=-62, ...
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You talk about even and odd in this problem. Wouldn't this not be enough? Since 26 has factors of 2 and 13, even if you had added some even numbers in the product it would still not have sufficed? You would need a 13, 39 etc ? Is my reasoning correct?

The fact that the product is odd, is already enough to say that it cannot be a multiple of 26^n unless n=0.
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Hi Bunuel,

I am well versed with these exponent rules, however I am becoming confused with the above two rules stated-

If \(\sqrt{x^2}\) =\(|x|\) then -> \(\sqrt{5^2}\) = \(|5|\), which means that value can be either 5 or -5. However, as per your example, value should be 5 only. How come this is true? Please clarify.

Thanks
H

When you square \(5x= (5+\sqrt{z-3y})^2\) you should get 5x=25+10(z-3y)^1/2+(z-3y) (5^2=25 not 125).
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Hi Bunuel
Can u pls explain the step.....(5^(5x))^2 = 70^2 implies 5^5x = 70
does that mean if a^n = b^n, a=b...

Archit

It's very simple, take the square root from \((5^{5x})^2=70^2\) to get \(5^{5x}=70\).
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Bunuel, could you please kindly explain how you come up with the 70 here? Many thanks

Can you please tell me which step is unclear:

\(5^{10x}=4,900\) --> \((5^{5x})^2=70^2\) --> \(5^{5x}=70\).
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The first step with the 4,900 is clear. I do not understand the following: 5^{(5x-5)}*4^{\sqrt{y}}=5^{5x}*5^{-5}*(2^{\sqrt{y}})^2=70*5^{-5}*25^2=70*5^{-5}*5^4=70*5^{-1}=\frac{70}{5}=14[/m]

and how you get to the 70 there.

\(5^{5x}=70\). So, this step is clear.

\(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}}=5^{(5x-5)}*4^{\sqrt{y}}=5^{5x}*5^{-5}*(2^{\sqrt{y}})^2\). Is this step clear?

Next, replace \(5^{5x}\) with 70 and \(2^{\sqrt{y}}\) with 25: \(70*5^{-5}*25^2=70*5^{-5}*5^4=70*5^{-1}=\frac{70}{5}=14\).

Does this make sense?
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[/edit]


Thanks for the explanation. The part where you take 5+4 out of the full equation is still unclear to me - 5+4(5+5^2+5^3+5^4+5^5). Could you please break it down into multiple steps?

4 is factored out from 4*5+4*5^2+4*5^3+4*5^4+4*5^5, not 5+4 from the whole expression: \(4*5+4*5^2+4*5^3+4*5^4+4*5^5=4(5+5^2+5^3+5^4+5^5)\).

So, \(5+(4*5+4*5^2+4*5^3+4*5^4+4*5^5)=5+4(5+5^2+5^3+5^4+5^5)\).

Hope it's clear.
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This is how I went about solving this problem:

Given \(\sqrt{5x} - 5 - \sqrt{z-3y}= 0\):

Rearrange and square
\((\sqrt{5x})^2 = (\sqrt{z-3y}+5)^2\)
\(|5x| = z-3y + 25 + 10\sqrt{z-3y}\)

Assuming that x is positive and given \(5x=125-3y+z\), set the two equations equal to each other
\(125 - 3y + z = z-3y + 25 + 10\sqrt{z-3y}\)
\(100 = 10\sqrt{z-3y}\)
\(100 = z-3y\)

\(5x = 125-3y+z\)
\(5x = 125 + 100\)
\(x = 45\)

Thus by substitution:
\(= \sqrt{\frac{45(z-3y)}{x}}\)

\(= \sqrt{\frac{45(100)}{45}}\)

\(= \sqrt{100}\)

\(= 10\)

However, notice that I had to assume that |5x| = 5x
Why is this permissible in this problem?

Even roots from negative numbers are undefined for the GMAT. We have \(\sqrt{5x}\) in one of the expressions, which means that x is non-negative, thus |5x|=5x.

Hope it's clear.
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5^{5x}*5^{-5}*(2^{\sqrt{y}})^2=70*5^{-5}*25^2

How come you arrive to 25 from y?

\(4^{\sqrt{y}}=(2^2)^{\sqrt{y}}=2^{2*\sqrt{y}}=(2^{\sqrt{y}})^2=25^2\) --> .

P.S. Please go this post: rules-for-posting-please-read-this-before-posting-133935.html#p1096628 (Writing Mathematical Formulas on the Forum). Thank you.
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\((2^{\sqrt{y}})^2=25^2\)

Bunuel explain this part please

We are given in the stem that \(2^{\sqrt{y}}=25\), thus \((2^{\sqrt{y}})^2=25^2\).
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