x=10^100 --Eqn1

x^x=10^k --Eqn 2

We can solve it in 2 ways.

Substitute x value in eqn2

(10^100)^(10^100)=10^k

10^(100*10^100)=10^k[as (a^b)^c=a^bc]

10^(10^2*10^100)=10^k

10^(10^102)=10^k

k=10^102

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Another approach

x=10^100 --Eqn1

x^x=10^k --Eqn 2

Raise to power x on both sides on eqn 1

x^x=10^100x

-->10^100x=10^k

k=100x=100*10^100=10^2*10^100=10^102

Hope this clarify

Great explanation. Thank you ! I have one more...

If x and j are integers and (12^x)*(4^2x+1)=(2^j)*(3^2), what is the value of j?

A) 5
B) 7
C) 10
D) 12
E) 14

So 4^2x+1 = (2^2)^2x+1 = 2^4x+2. Can set 4x+2 = j. I'm not sure what do do with 12^x. Any suggestions?

(12^x)*(4^2x+1)=(2^j)*(3^2)

12^x can be written as 3^x*4^x

3^x*4^x*(4^2x+1)=(2^j)*(3^2)

3^x*4^(3x+1)=(2^j)*(3^2)

3^x*2^(6x+2)=(3^2)*(2^j)

comparing both sides we get,

x=2 and 6x+2=j

12+2=j; j=14

Option E

If the square root of the product of three distinct positive integers is equal to the largest of the three numbers, what is the product of the two smaller numbers?

(1) The largest number of the three distinct numbers is 12.
(2) The average of the three numbers is 20/3


(1) is sufficient. If the largest number is 12, then the product of the smaller two numbers must also be 12.

(2) I tried picking some numbers for this but I couldn't find a more robust approach.

Any suggestions?

mattocks wrote:If the square root of the product of three distinct positive integers is equal to the largest of the three numbers, what is the product of the two smaller numbers?

(1) The largest number of the three distinct numbers is 12.
(2) The average of the three numbers is 20/3


(1) is sufficient. If the largest number is 12, then the product of the smaller two numbers must also be 12.

(2) I tried picking some numbers for this but I couldn't find a more robust approach.

Any suggestions?

st 1 is sufficient ..Agreed!

coming to st 2 :

a+b+c = 20

(2,6,12) & 3,4,12) both these pairs will be suffice. But for the sum of a,b,c = 20, only (2,6,12) fits in.

So st 2 is also sufficient.

since the stem asks only find the product of other 2 smaller numbers, pick D.

If they would have asked what are the other 2 smaller numbers I would have picked C.

Mattocks,

To help you come up with a way to choose a, b, and c values for statement two consider this

if sqrt(abc)=c, A*B=C

Now instead of picking the A and B values, pick the C value. I'll choose 10.

A*B=10

My only options for A and B are 1, 10 and 2, 5. Can't be 1, 10 because they have to be distinct. I go with 2,5, but then A+B+C=17 so 10 doesn't work. They are less than 20 so we need C to be greater (if you are unsure whether C MUST be greater, I recommend that you start even lower, like at c=6 so that you can feel the patterns. I started at C=3 just to get a sense of what the pattern was and quickly moved up to 8.)

Try C=11
A*B=11 A and B can only be 1,11 which doesn't work

C=12
A*B=12 A and be can be 1,12; 2,6; 3;4. First set doesn't sum to, second set does, third set doesn't.

We have an answer.

I tried c=13 and c=14 to be safe. C=13 can't work because a,b,c must be unique and with c=14 a,b must be 7 and 2 which is now GREATER than 20 so C>14 is not going to work.

A*

mattocks wrote:If x=10^100 and x^x=10^k, k=?.

A) 10^5
B) 10^102
C) 10^99
D) 10^200
E) 10^98

I'm really not sure how to attack this problem. Any suggestions would be greatly appreciated.

Another way to tackle this is using logarithm (even though it is out of scope for the GMAT, it does make life a lot easier when exponents are used). For instance

x = 10^100
log x = 100log10 = 100 (log10=1)

x^x = 10^k
xlogx = klog10 = k
=> 100x = k

Now, x = 10^100
=> 100 x (10^100) = k
=> 10^2 x 10^100 = k
=> 10^102 = k