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

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Given that N = a^3b^4c^5 where a, b and c are distinct prime  [#permalink]

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Updated on: 28 Feb 2014, 20:43
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34
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35% (medium)

Question Stats:

69% (01:28) correct 31% (01:40) wrong based on 483 sessions

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Given that $$N = a^3b^4c^5$$ where a, b and c are distinct prime numbers, what is the smallest number with which N should be multiplied such that it becomes a perfect square, a perfect cube as well as a perfect fifth power?

A. $$a^3b^4c^5$$
B. $$a^5b^4c^3$$
C. $$a^2b^3c^5$$
D. $$a^7b^6c^5$$
E. $$a^{27}b^{26}c^{25}$$

While this problem on the surface is an exponent problem, it turns out to be even more of a divisibility, least-common-multiple problem. To make N a perfect square, powers of a, b and c must be even; to make N a perfect cube, the powers must be multiples of 3; to make N a perfect fifth power, they must be multiples of 5. This implies that each of the powers of a, b and c must be a multiple of 30 (because 30 is the LCM of 2, 3 and 5) and since we are looking for the smallest number, the powers should each be 30. Hence we need to multiply N by a^27b^26c^25 to make the powers of each of a, b and c equal to 30.

Hi, I want to know how I can recognise that question is asking LCM please.

Originally posted by goodyear2013 on 27 Feb 2014, 11:09.
Last edited by MacFauz on 28 Feb 2014, 20:43, edited 2 times in total.
Renamed the topic and edited the question.
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Given that N = a^3b^4c^5 where a, b and c are distinct prime  [#permalink]

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28 Feb 2014, 00:20
8
6
goodyear2013 wrote:
Given that N = a^3b^4c^5 where a, b and c are distinct prime numbers, what is the smallest number with which N should be multiplied such that it becomes a perfect square, a perfect cube as well as a perfect fifth power?

A. a^3b^4c^5
B. a^5b^4c^3
C. a^2b^3c^5
D. a^7b^6c^5
E. a^27b^26c^25

While this problem on the surface is an exponent problem, it turns out to be even more of a divisibility, least-common-multiple problem. To make N a perfect square, powers of a, b and c must be even; to make N a perfect cube, the powers must be multiples of 3; to make N a perfect fifth power, they must be multiples of 5. This implies that each of the powers of a, b and c must be a multiple of 30 (because 30 is the LCM of 2, 3 and 5) and since we are looking for the smallest number, the powers should each be 30. Hence we need to multiply N by a^27b^26c^25 to make the powers of each of a, b and c equal to 30.

Hi, I want to know how I can recognise that question is asking LCM please.

A perfect square, is an integer that can be written as the square of some integer. For example 16=4^2, is a perfect square. Notice that the powers of the primes of a perfect square are even.
A perfect cube, is an integer that can be written as the cube of some integer. For example 27=3^3, is a perfect cube. Notice that the powers of the primes of a perfect cube are multiples of 3.
A perfect fifth power, is an integer that can be written as the fifth power of some integer. For example 32=2^5, is a perfect fifth. Notice that the powers of the primes of a perfect fifth power are multiples of 5.

According to the above, a number to be a perfect square, a perfect cube as well as a perfect fifth power, must have its primes in power of 30 (the least common multiple of 2, 3, and 5), so it must be a perfect 30th power.

The least positive integer $$a^3b^4c^5$$ should be multiplied by, in order the powers to be multiples of 30 is $$a^{27}b^{26}c^{25}$$: $$(a^3b^4c^5)(a^{27}b^{26}c^{25})=a^{30}b^{30}c^{30}=(abc)^{30}$$.

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Re: Given that N = a^3b^4c^5 where a, b and c are distinct prime  [#permalink]

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28 Feb 2014, 20:31
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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!
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Re: Given that N = a^3b^4c^5 where a, b and c are distinct prime  [#permalink]

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03 Mar 2014, 03:12
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We require to find LCM of 2, 3 & 5 which is 30... as it should be perfect square AND cube AND Fifth

So, 30 should be the power of each expression a, b, & c

Answer = E (30-3) (30-4 (30-5)
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Re: Given that N = a^3b^4c^5 where a, b and c are distinct prime  [#permalink]

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02 Aug 2015, 06:03
I was intimidated by the exponential when I started reading this question. As it turns out, it is an LCM question.
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Given that N = a^3b^4c^5 where a, b and c are distinct prime  [#permalink]

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02 Aug 2015, 06:18
evdo wrote:
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.
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Re: Given that N = a^3b^4c^5 where a, b and c are distinct prime  [#permalink]

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15 Jun 2016, 20:27
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
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Re: Given that N = a^3b^4c^5 where a, b and c are distinct prime  [#permalink]

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09 Jul 2016, 10:22
i dont get this question at all:((
what like which concept is being used here? the question is confusing like what am i supposed to find and how?
how is this LCM? i dont get what this is

here is my understanding of the question:

N= a raise to the power 3, b raise to the power 4 and c raise to the power 5

A, B and C are all PRIME numbers. Then out of nowhere, whats the SMALLEST NUMBER with which N should be multiplied? I just dont get it.

so im doing something like N multiply by x (smallest number ) = a number that is perfect sq, perfect cube, perfect fifth power? like where the hell do i go from here?
and none of the options give that NUMBER that we need to find. How do i understand this its soo confusing?!!
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Re: Given that N = a^3b^4c^5 where a, b and c are distinct prime  [#permalink]

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08 Jul 2017, 13:15
1
goodyear2013 wrote:
Given that $$N = a^3b^4c^5$$ where a, b and c are distinct prime numbers, what is the smallest number with which N should be multiplied such that it becomes a perfect square, a perfect cube as well as a perfect fifth power?

A. $$a^3b^4c^5$$
B. $$a^5b^4c^3$$
C. $$a^2b^3c^5$$
D. $$a^7b^6c^5$$
E. $$a^{27}b^{26}c^{25}$$

While this problem on the surface is an exponent problem, it turns out to be even more of a divisibility, least-common-multiple problem. To make N a perfect square, powers of a, b and c must be even; to make N a perfect cube, the powers must be multiples of 3; to make N a perfect fifth power, they must be multiples of 5. This implies that each of the powers of a, b and c must be a multiple of 30 (because 30 is the LCM of 2, 3 and 5) and since we are looking for the smallest number, the powers should each be 30. Hence we need to multiply N by a^27b^26c^25 to make the powers of each of a, b and c equal to 30.

Hi, I want to know how I can recognise that question is asking LCM please.

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 - $$N = a^{30-3}b^{30-4}c^{30-5}$$

= $$a^{27}b^{26}c^{25}$$

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Re: Given that N = a^3b^4c^5 where a, b and c are distinct prime  [#permalink]

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23 Jan 2018, 12:39
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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).

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Re: Given that N = a^3b^4c^5 where a, b and c are distinct prime  [#permalink]

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03 May 2018, 15:25
goodyear2013 wrote:
Given that $$N = a^3b^4c^5$$ where a, b and c are distinct prime numbers, what is the smallest number with which N should be multiplied such that it becomes a perfect square, a perfect cube as well as a perfect fifth power?

A. $$a^3b^4c^5$$
B. $$a^5b^4c^3$$
C. $$a^2b^3c^5$$
D. $$a^7b^6c^5$$
E. $$a^{27}b^{26}c^{25}$$

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.

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Re: Given that N = a^3b^4c^5 where a, b and c are distinct prime &nbs [#permalink] 03 May 2018, 15:25
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