Exponents..

[Guest]

Hi! This problem took me a while to work through and I know there must be a quicker way to solve it. Any suggestions on would be much appreciated!

[(8^2)(3^3)(2^4)]/(96^2) =

A. 3
B. 6
C. 9
D. 12
E. 18

the answer is A

[Guest]

Hi! This problem took me a while to work through and I know there must be a quicker way to solve it. Any suggestions on would be much appreciated!

[(8^2)(3^3)(2^4)]/(96^2) =

A. 3
B. 6
C. 9
D. 12
E. 18

the answer is A

(64)(27)(16)/(96)(96)=3. that's the way how I did the problem and I was be able to figure out the answer under l min.

[Guest]

Hi! This problem took me a while to work through and I know there must be a quicker way to solve it. Any suggestions on would be much appreciated!

[(8^2)(3^3)(2^4)]/(96^2) =

A. 3
B. 6
C. 9
D. 12
E. 18

the answer is A

We can think this as (2^6 * 3^3* 2^4)/(2^10*3^2)....... hence on deduction we have 3 left after cancelling the common term in numerator & denominator

[jwinawer]

Hi! This problem took me a while to work through and I know there must be a quicker way to solve it. Any suggestions on would be much appreciated!

[(8^2)(3^3)(2^4)]/(96^2) =

A. 3
B. 6
C. 9
D. 12
E. 18

the answer is A

We can think this as (2^6 * 3^3* 2^4)/(2^10*3^2)....... hence on deduction we have 3 left after cancelling the common term in numerator & denominator

This is a great solution. The key is finding ways to quickly cancel while making minimal calculations. A great way to do this, as demonstrated in the comment above, is to break down each exponent into prime factors. Since 8 is 2^3, 8^2 is (2^3)^2, or 2^6. Doing this for each term works out such that everything cancels, leaving just 3. Nice solution!

[RonPurewal]

yeah.

as an add-on to jon's solution, let me give you the following general advice:
on pretty much ANY problem involving powers of specific integers, IMMEDIATELY reducing those integers to their prime factorizations is a solid opener.
although individual problems may have cute little exponent tricks that help you get through them faster, you are GUARANTEED to see the connections between the different numbers in the problem if you break them all down into primes at the outset.