Given that N = a^3b^4c^5 where a, b and c are distinct prime numbers

oh wow...I spent a good amount of time on this one yesterday before finally getting a headache and hoping Bunuel or Karishma would chime in...I now realize I was attempting to solve a completely different problem

As it is formatted, it looks like the question (and relative format of the answer choices) is:
N = a^(3b)^(4c)^5

and not:
N = (a^3)(b^4)(c^5)



many thanks Bunuel!

I was intimidated by the exponential when I started reading this question. As it turns out, it is an LCM question.

Even if you don not recognize that it is an LCM question, you can solve it in 30-45 seconds as shown below:

\(N = a^3b^4c^5\)

Now you need to find more powers of a,b,c such that the 5th root, 3rd root and the square root are all integers (this is what the question means by "such that it becomes a perfect square, a perfect cube as well as a perfect fifth power?" )

For this to happen, N needs to be multiplied by another number (= the correct option to the question asked).

Given that \(N = a^3b^4c^5\) ---> Start with the 5th power or root (as you will see, this is the most restrictive of the 3 required roots).

For a number to have an integer value of a "n"th root, it needs to be of the form : \(N = a^n\) ---> \(N^{1/n} = a\) (if you have any other power then you may not get an integer value when you take the 'n' th root.)

Applying the above logic to the question at hand,

\(N = a^3b^4c^5\) , we need to make powers of a,b,c multiple of 5 when these options are multiplied by N. Also \(x^a * x^b = x^{a+b}\)

Thus, Options A and B are out as they will not provided power of 'a' as a multiple of 5 when these options are multiplied by N. No need to check the other powers for these 2 options.

We have options C-E remaining. check now for 3rd root or cube power!! For this we need to make sure that the powers of a,b,c are multiples of 3 when these options are multiplied by N.

Options C and D are thus eliminated as they do not provide power of 'a' as a multiple of 3 when these options are multiplied by N.

E is thus the final answer.

Although, this text looks intimidating, it took me not more than 30 seconds to figure out the correct answer.

The powers of a, b and c have to be divisible by 2, 3 and 5 for N to be a perfect square, perfect cube and perfect 5th power.
LCM of 2, 3 and 5 is 30
The smallest integer N = (a^30)(b^30)(c^30)

Answer: E

Given: N=a^3*b^4*c^5
We need to make this a perfect square, a perfect cube as well as a perfect fifth power.

LCM of 2, 3 and 5 = 30
Therefore we need to have the powers of a, b and c as 30
Option E gives us the term that on multiplication will make N as (a^30)(b^30)(c^30)
Correct Option : E

We need to turn N = a³b⁴c⁵ into a perfect square, cube and a fifth power.
Hence whatever are the powers of a, b and c, they should all the divisible by 2, 3, and 5
Or in other words, the powers should be the LCM of 2, 3 and 5 = 30 (Since we need the smallest number)

Hence the multiplication factor should be: a²⁷b²⁶c²⁵

Correct Option: E

Answer:E
N1=a^3*b^4*c^5 should be perfect square, a perfect cube as well as a perfect fifth power
Perfect square: ^2
Perfect cube: ^3
Perfect fifth: ^5
to combine all, we need a common factor 2x3x5 = 30
then N3=a^30*b^30*c^30 is the perfect of the three numbers.
N1*N2=N3
N2 will be a^27*b^26*c^25

Great Question this one.
Perfect square => Powers of all primes=> Multiple of 2.
Perfect cube=>Powers of all the primes => Multiple of 3.
Perfect fifth power => Powers of all the primes => Multiple of 5

Hence powers of all the primes must be a multiple of all->2,3,5 => LCM=30
Smallest multiplication => \(a^{27}*b^{26}*c^{25}\)
Hence E.

Lets understand some fundamentals:

1. In order to make an integer a perfect square, we would need the power to be - Multiples of Two

2. In order to make an integer a perfect cube, we would need the power to be - Multiple of Three

5. In order to make an integer a perfect fifth power, we would need the power to be - Multiple of Five

Considering all of the above, the power should have the LCM of \(2 * 3 * 5 = 30\)

So we are looking for - \(N = a^{30}b^{30}c^{30}\)

We already have - \(N = a^3b^4c^5\)

We would N to be multiplied by - \(N = a^{30-3}b^{30-4}c^{30-5}\)

= \(a^{27}b^{26}c^{25}\)

Hence, Answer is E

Hi All,

This question is built around a few Number Properties involving exponents.

For a number to be a perfect square, all of the "exponent terms" must be EVEN.

for example….
25 is a perfect square because 25 = 5^2
16 is a perfect square because 16 = 4^2 = 2^4

For a number to be a perfect cube, all of the "exponent terms" must be A MULTIPLE OF 3.

8 is a perfect cube because 8 = 2^3
64 is a perfect cube because 64 = 4^3 = 2^6

For a number to be a perfect fifth power, all of the "exponent terms" must be A MULTIPLE OF 5.

32 is a perfect fifth power because 32 = 2^5
1024 is a perfect fifth power because 1024 = 4^5 = 2^10

Here, we need each exponent to become a multiple of 2, 3 and 5. The prompt refers to the SMALLEST number, so we need the Least Common Multiple of 2, 3 and 5 ……which is 30. The correct answer will multiply by the given prompt to equal 30.

Since we're starting with (A^3)(B^4)(C^5), we'll need to multiply with something that will end with (A^30)(B^30)(C^30).

Final Answer:

GMAT assassins aren't born, they're made,
Rich

The exponents of a perfect square are all multiples of 2. The exponents of a perfect cube must be all multiples of 3. And the exponents of a perfect fifth power must be all multiples of 5. Since we need N to be a perfect square, perfect cube, and perfect fifth power, we need each exponent to be a multiple of the LCM of 2, 3, and 5, which is 30. Hence, we know that the smallest number of each exponent must be 30.

The original exponent of a is 3; so we need 27 more a’s to get to 30: a^3 x a^27 = a^30

The original exponent of b is 4; so we need 26 more b’s to get to 30: b^4 x b^26 = b^30

The original exponent of c is 5; so we need 25 more c’s to get to 30: c^5 x c^25 = c^30

Therefore, we need to multiply N = a^3b^4c^5 by a^27 x b^26 x c^25 to make it become a perfect square, a perfect cube and a perfect fifth power.

Answer: E

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