2. What is the units digit of (17^3)^4-1973^{3^2}?
A. 0
B. 2
C. 4
D. 6
E. 8


Units digit for 17^3 = 3
Units digit for (17^3)^4 = 3^4 = 1
Units digit for 1973^9 = 3
Units digit for xxx1 - xxxx3 = 8

Option

4. What is the value of 5+4*5+4*5^2+4*5^3+4*5^4+4*5^5?
A. 5^6
B. 5^7
C. 5^8
D. 5^9
E. 5^10


a(r^n - 1) / (r-1)
5+4*5+4*5^2+4*5^3+4*5^4+4*5^5
1+4+4*5+4*5^2+4*5^3+4*5^4+4*5^5
1 + 4(5^6 -1)/4
5^6

Option

5. If x=23^2*25^4*27^6*29^8 and is a multiple of 26^n, where n is a non-negative integer, then what is the value of n^26-26^n?
A. -26
B. -25
C. -1
D. 0
E. 1

n=0 because x is not a multiple of 26

n^26 - 26^n = 0-1 = -1

Option

6. If x=\sqrt[5]{-37} then which of the following must be true?
A. \sqrt{-x}>2
B. x>-2
C. x^2<4
D. x^3<-8
E. x^4>32


x^5 = -37

-2^5 = -32
-3<x<-2

A. Eliminated
B. Eliminated
C. x^2<4 = -2<x<2 Eliminated
D. x^3<-8 Correct


Option

8. If x is a positive number and equals to \sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}}}, where the given expression extends to an infinite number of roots, then what is the value of x?
A. \sqrt{6}
B. 3
C. 1+\sqrt{6}
D. 2\sqrt{3}
E. 6



x= √(6+√(6+√(6+...
x^2 = 6 + x
x^2-x-6=0
(x-3)(x+2) = 0
x = 3 or -2
Value of sqrt cant be -ve
Option

10. Given that 5x=125-3y+z and \sqrt{5x}-25-\sqrt{z-3y}=0, then what is the value of \sqrt{\frac{45(z-3y)}{x}}?
A. 5
B. 10
C. 15
D. 20
E. Can not be determined

5x-125=z-3y

sqrt{5x}-25=\sqrt{z-3y}
(sqrt{5x}-25)^2=z-3y

(sqrt{5x}-25)^2=5x-125
\sqrt{5x}=15
x=5*9=45

so the expression \sqrt{\frac{45(z-3y)}{x} = \sqrt{45(15-25)/45}=\sqrt{-10}

E is the answ
_________________

7. If x=\sqrt{10}+\sqrt[3]{9}+\sqrt[4]{8}+\sqrt[5]{7}+\sqrt[6]{6}+\sqrt[7]{5}+\sqrt[8]{4}+\sqrt[9]{3}+\sqrt[10]{2}, then which of the following must be true:
A. x<6
B. 6<x<8
C. 8<x<10
D. 10<x<12
E. x>12

my answ to this q is intuitive.

everyone knows that sqroot10=3.16
\sqrt[3]{9} is approx \sqrt[3]{2^3} >2 (but less than 3)
the rest of roots are more than 1 and less than 2

so we have 3.16+2,..+1+1+1+1+1+1+1 = approx 12.16

E is the answ
_________________

9. If x is a positive integer then the value of \frac{22^{22x}-22^{2x}}{11^{11x}-11^x} is closest to which of the following?
A. 2^{11x}
B. 11^{11x}
C. 22^{11x}
D. 2^{22x}*11^{11x}
E. 2^{22x}*11^{22x}


let 11=a

so we have

(2a^(2ax)-2a^2x)/(a^(ax)-a^x)=2*((a^(ax)-a^x)*(a^(ax)+a^x))/(a^(ax)-a^x)=2(a^(ax)+a^x)

2(a^(ax)+a^x)=2*(11^(11x)+11^x)
_________________

Welcome back Bunuel!

Some nice questions. A couple of comments:

Q5 has a typo at the end: \(n^26-26^n\). I think you need to include the exponent 26 in braces {26} to get \(n^{26}-26^n\)

In Q3, unless I'm making some kind of stupid mistake, the value of the expression in the question is close to 600 (so isn't among the answers). I'm wondering if you meant the question to instead read something more like:


3. If \(5^{10x}=4,900\) and \(4^{\sqrt{y}}=25\) what is the value of \(\frac{5^{5x-1}}{2^{\sqrt{y}}\)?
A. 14/5
B. 5
C. 28/5
D. 13
E. 14

In that case, I get one of the five answer choices (A).
_________________

And it looks as though every other question has been solved except for this one. I won't solve it completely, but I'll help people get started. Here, since we are asked for an approximation only, we don't need to compute an exact value. You might notice that if x is a positive integer, \(22^{22x}\) is vastly bigger than \(22^{2x}\). Since that's true, if we only need to estimate, we can ignore the small term. We can do the same thing in the denominator.
_________________

yep, if the 3d q is -If 5^{10x}=4,900 and 4^{\sqrt{y}}=25 what is the value of \frac{5^{5x-1}}{2^{\sqrt{y}}?
A. 14/5
B. 5
C. 28/5
D. 13
E. 14""

then

5^{10x}=70^2
(5^{5x})2=70^2
5^{5x}=70

4^{\sqrt{y}}=25
2^{\sqrt{y}}=5

70/5*5=14/5 ans is A
_________________

2a^(2ax)/a^(ax)=2*a^(ax)=2*11^(11x)

ans is ?
_________________

Your substitution might be making things more difficult. If we want to estimate the value of

\(\frac{22^{22x}-22^{2x}}{11^{11x}-11^x}\)

then we can ignore the comparatively small values in the numerator and denominator; this is roughly equal to


\(\frac{22^{22x}}{11^{11x}}\)

which is equal to

\(\frac{22^{22x}}{11^{11x}} = \frac{2^{22x} 11^{22x}}{11^{11x}} = 2^{22x} 11^{11x}\)
_________________

Hello Ian. Glad to be back!

Thanks for your comments. Yes, there were 2 typos:
For Q5: included the exponent 26 in braces {26} to avoid confusion;
For Q3: added the brackets which were lost in copy/past, it should read \(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}\).
_________________

Hi,
My answers are:
1 - C
2 - E
3 - E
4 - A
5 -
6 - B
7 - E
8 - B
9 -
10 - E
11 - E

Unable to solve for 5th & 9th

i believe the answer is : 7^12-3^6 ---> 11-9= 2 ---> B

note: I believe the part I wrote (which i colored red) is wrong. Can someone explain why? when an exponent is raised to another exponent then what operation should we do first? - Okey! Just read the rules about routs that Bunuel posted and found my mistake! (exponentiation by stacked symbols)

We know that:

\(5x=125-3y+z\) ---> [highlight]\(z-3y=5x-125\) (1)[/highlight]

and that:

\(\sqrt{5x}-25-\sqrt{z-3y}=0\) ---> \(\sqrt{5x}-25=\sqrt{z-3y}\) ----> \([\sqrt{5x}-25]^2=[\sqrt{z-3y}]^2\) ----> using (1): \([\sqrt{5x}-25]^2=[\sqrt{5x-125}]^2\) --->

\(5x-50\sqrt{5x}+625=5x-125\) ---> \(50\sqrt{5x}=750\) ---> \(\sqrt{5x}=15\) ---> \(5x=15*15\)--->[[highlight]\(x=45\) (2)[/highlight]

(1), (2) ---> \(z-3y=5*45-125\)--->[highlight]\(z-3y=100\) (3)[/highlight]

therefore: \(\sqrt{\frac{45(z-3y)}{x}}\) ---->using (2),(3) :\(\sqrt{\frac{45*100}{45}}\) ---> 10 ---->B

The answer should be E. 14

Here is how:

\(5^{10x}=4900\)
so \({(5^{5x})}^2=70^2\)

Take square root of both sides >

so \(5^{5x}=70\)

Also from question stem:

\(2^{\sqrt{y}} = 5^2\)

So (from the question stem)

\(4^{-\sqrt{y}} = \frac{1}{4^{\sqrt{y}}}\)
\(4^{-\sqrt{y}} = (\frac{1}{2^{\sqrt{y}}})^2 = \frac{1}{5^4}\)

Now \(5^{(x-1)^5}\)

Should be \(\frac{5^{5x}}{5^5}\)

We know \(5^{10x}=4900\)

so \(5^{10x}=70^2\)

so \(5^{5x}=70\)

So \(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}\)

is basically \(\frac{5^x}{5^5}*5^4\)

We already know the value of \(5^x\) which is \(70\)

So now it becomes \(\frac{70}{5^5}*5^4\)

Which should resolve to \(14\)

Hence Answer = E = \(14\)

Q1.

\(\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}}\)

First look and we know that this is of the type:

let a = \(\sqrt{25+10\sqrt{6}}\)

Let b = \(\sqrt{25-10\sqrt{6}}\)

let \(a+b=x\)

Then \((a+b)^2=x^2\)

So \(a^2+b^2+2ab=x^2\)

So \(a^2 = {25+10\sqrt{6}}\)
So \(b^2 = {25-10\sqrt{6}}\)
So \(a^2+b^2=50\)

Now on to \(2ab\)

\(a=\sqrt{25+10\sqrt{6}}\)
\(b=\sqrt{25-10\sqrt{6}}\)

let \(ab=y\)

then \(\sqrt{25+10\sqrt{6}}*\sqrt{25-10\sqrt{6}}=y\)
So \(({25+10\sqrt{6}})*({25-10\sqrt{6}})=y^2\)
so \(625-600=y^2\)
so \(y=5\)
so \(2ab=2y=10\)
so \(x^2=a^2+b^2+2ab=50+10=60\)
so \(x^2=60\)
so \(x=\sqrt{4*15}\)
so \(x=2\sqrt{15}\)

Hence Answer is C. \(2\sqrt{15}\)

Q2. What is the units digit of \((17^3)^4-1973^{3^2}\)?[/b]

Now we know that:

\((17^3)^4 = (17^6)^2\)

So the expression now becomes

\((17^6)^2 - (1973^3)^2\)

Now we know that \(a^2-b^2\) can be written as \((a+b)*(a-b)\)

Let \(a=17^6\) and \(b=1973^3\)

Now on to \((a+b)*(a-b)\)

\((17^6+1973^3)*(17^6-1973^3)\)

Lets see what the units digit of \(17^6\) is:
We are only concerned with \(7\):
\(7*7*7*7*7*7= ..........9\) unit digit of 9

Lets see what the units digit of \(1973^3\) is:
We are only concerned with \(3\):
\(3*3*3= ..........7\) unit digit of 7

Lets translate this into our expression for \((a+b)*(a-b)\)

\((17^6+1973^3)*(17^6-1973^3)\)

Should in digit terms give:

\((9+7)*(9-7)\)
\((6)*(2)\)

Should Yield a units digit of 2.

Hence Answer = 2