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

Question

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.

Bunuel · Accepted Answer

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} .

Answer: E.

PareshGmat · Answer

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)

m3equals333 · Answer

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!

evdo · Answer

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

ENGRTOMBA2018 · Answer

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.

Kurtosis · 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

TeamGMATIFY · Answer

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

OptimusPrepJanielle · Answer

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

MHI · Answer

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

stonecold · Answer

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.

ydmuley · Answer

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

EMPOWERgmatRichC · Answer

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:  E

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

JeffTargetTestPrep · Answer

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

FifthVole · Answer

My claude that helped me debunk the answer, if you have poor math fundamental just as I do, you might find it helpful

Let's rebuild this entire problem from the ground up, one piece at a time, and I'll flag exactly where each question you asked fits in and what it resolved.
The question
 N = a3b4c5, where a, b, c are distinct primes. What is the smallest number N should be multiplied by so it becomes a perfect square, a perfect cube, AND a perfect fifth power all at once? 
Piece 1: What makes something a perfect square/cube/fifth power?
This is the foundational fact everything else depends on:
 A number is a perfect square if and only if every prime's exponent in its factorization is a multiple of 2.  Same logic: perfect cube → every exponent is a multiple of 3. Perfect fifth power → every exponent is a multiple of 5.
Example: 24 is a perfect square (4 is a multiple of 2). 23 is not (3 isn't a multiple of 2).
Piece 2: "Distinct primes" — your first question
You asked what this meant. Here's the resolution:  distinct just means a, b, c are different prime numbers, no repeats  — e.g., a=2, b=3, c=5, but never a=b=2.
 Why this matters to the whole problem (this wasn't just vocabulary):  if a, b, c  could  repeat — say a=b=c=2 — then a3b4c5 would collapse into 23×24×25 = 212, a single prime factor, not three independent ones. The entire method we're about to use (solving for a, b, and c's needed exponents completely separately) only works  because  they're distinct. If they weren't, we'd have one combined exponent problem instead of three independent ones. So "distinct primes" isn't a side detail — it's the condition that makes our whole strategy valid.
Piece 3: Why this is an LCM problem
We need every exponent in N to simultaneously satisfy three conditions: multiple of 2 AND multiple of 3 AND multiple of 5. The smallest number satisfying all three at once is:
$$	ext{LCM}(2,3,5) = 30$$
So  every exponent's final target is 30 . This is the pattern-recognition signal for this problem type: "needs to satisfy multiple perfect-power conditions simultaneously" → take the LCM of the required powers as your target exponent.
Piece 4: The basic algebra — your second question
You asked me to break down where "30−3," "30−4," "30−5" actually come from. Here's the one rule underlying all of it:
$$a^m 	imes a^n = a^{m+n}$$
(Multiplying the same base means  adding  the exponents — this is just repeated multiplication written compactly.)
We want to go from a3 (what's currently in N) to a30 (our target). Call the missing exponent  e :
$$a^3 	imes a^e = a^{30} \quad\Rightarrow\quad 3+e = 30 \quad\Rightarrow\quad e = 30-3 = 27$$
Same process for b (start 4, target 30 → e=26) and c (start 5, target 30 → e=25). Each is its own independent little equation, precisely  because  a, b, c are distinct (tying back to Piece 2) — multiplying by a27 never touches b or c's exponents.
Piece 5: Where N actually fits in — your third question
This is the piece that ties everything together, and it's the one you flagged as unclear. Here's the key clarification:
 N is not something we're constructing — it's given to us as a fixed, specific number: a3b4c5.  The "3," "4," "5" in every equation above (3+e=30, 4+e=30, 5+e=30) are  not arbitrary or chosen by us  — they are literally N's own exponents, read directly off the number the problem handed us.
 The misconception to debunk directly:  it might feel like we're just "picking" the target exponents and building a30b30c30 from scratch. We're not. The question specifically asks what to  multiply N by  — meaning N's existing exponents (3,4,5) are locked in as the  starting point , and we're only allowed to add to them, never erase or ignore them. If N had different exponents (say a7 instead of a3), every step downstream changes — the "30" target stays fixed (since it only depends on needing squares/cubes/fifths), but the missing amount (30−7=23 instead of 30−3=27) changes entirely, because it depends on N's specific starting exponent.
Full picture, tied together
$$\underbrace{N} {a^3b^4c^5 	ext{ (given, fixed)}} 	imes \underbrace{a^{27}b^{26}c^{25}} {	ext{solved from } 3{+}e{=}30,\ 4{+}e{=}30,\ 5{+}e{=}30} = a^{30}b^{30}c^{30}$$
Check: 30 is a multiple of 2 (square ✓), 3 (cube ✓), and 5 (fifth power ✓).
Answer: E — a27b26c25
 
 The chain of dependency, if you want the one-sentence version:  N's given exponents (3,4,5) are the  inputs ; distinctness of a,b,c is what allows solving for each one  independently ; the LCM(2,3,5)=30 is the shared  target  every exponent must reach; and basic exponent addition (ag×aa−g=ag+(a−g)... more simply, m+e=target) is the  mechanism  for finding how much more of each prime is needed. Every piece you asked about slots into one of these four roles.

Pranjals16 · Answer

Lovely question to solve.

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

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