An integer between 1 and 300, inclusive, is chosen at random

Raised to the power 2 : 1 ; 2 ; 3 ; 4 ; 5 ; 6 ; 7 ; 8 ; 9 ; 10 ; 11 ; 12 ; 13 ; 14 ; 15 ; 16 ; 17 :: 17

Raised to the power 3 : 2 ; 3 ; 5 ; 6 :: 4

Raised to the power 5 : 2 ; 3 :: 2

Raised to the power 7 : 2 :: 1


Note : We don't consider numbers raised to even powers greater than 2 since they have already been accounted for when considering squares.


Total = 24

Probability = 24/300 = 2/25

Answer : C
_________________

Basically we need to find how many m^n (where n>1) are between 1 and 300 inclusive.

For n=2 --> m^2<300 --> m<18, so there are 17 such numbers: 1^2=1, 2^2=4, 3^2=9, 4^2=16, ..., 17^2=289;

For n=3 --> m^3<300 --> m<7, so there are 6 such numbers: 1^3=1, 2^3=8, 3^3=27, 4^3=64, 5^3=125, 6^3=216. 1^3=1=1^2 and 4^3=64=8^2 have already been counted so, that leaves only 4 numbers;

Skip n=4, since all perfect fourth power numbers are also perfect squares;

For n=5 --> m^5<300 --> m<4, so there are 3 such numbers: 1^5=1, 2^5=32, 3^5=243. 1^5=1=1^2 has already been counted so, that leaves only 2 numbers;

Skip n=6 for the same reason as n=3;

For n=7 --> m^7<300 --> m<3, so there are 3 such numbers: 1^7=1, 2^7=128. 1^7=1=1^2 has already been counted so, that leaves only 1 numbers.

Total: 17+4+2+1=24.

The probably thus equals to 24/300=2/25.

Answer: C.
_________________

correct answer 25/300 or 1/12. Did you copy the answers correctly

basically it boils down to counting all the numbers that have a square less than 300
all the numbers that have a cube less than three hundred
all the numbers raised to a 4 less than three hundred
so on and so forth

I double checked the answers, have copied it correctly. your explanation is right.But OA is different,

hint: when counting the squares, 2^6 = 8^2, 2^8 = 16^2, these shd be counted as one.

..........
Have to count fast.
have to..
this diagram can help to learn the essentials:

i just hate this type of questions. Understanding the question alone takes time, and after getting what the prompt says, I have to squeeze my brain to find a clever way to count without missing any possibility. It's just time consuming and really really tiring

Bunuel,
I tried counting it differently.
Instead of chopping it up according to degree of the number, I went by the numbers themselves:
2 : n can be - 1,2,3,5,6,7 . 3 : n can be - 1,2,3,5 . 5 : n can be - 1,2,3 . for the numbers 6,7,10,11,12,13,14,15,16,17 n can be 2.
so we have 5+4+3+10*2 = 31 and not 24 like you got.
Why am I getting a different result?
Can you see what I'm counting twice?

Dear ronr34, it's hard for me to understand what have you written there. Below are 24 numbers which satisfy the requirement:

1 = 1^(n > 1)
4 = 2^2
8 = 2^3
9 = 3^2
16 = 2^4 = 4^2
25 = 5^2
27 = 3^3
32 = 2^5
36 = 6^2
49 = 7^2
64 = 2^6 = 4^3 = 8^2
81 = 3^4 = 9^2
100 = 10^2
121 = 11^2
125 = 5^3
128 = 2^7
144 = 12^2
169 = 13^2
196 = 14^2
216 = 6^3
225 = 15^2
243 = 3^5
256 = 2^8 = 4^4 = 16^2
289 = 17^2
_________________

Hi Bunuel,
What I meant is that instead of counting the degrees and listing out the numbers that are the options, I listing each number and looked to what degree we can multiply it.
Looking at 2 - the degrees that are possible are : 1,2,3,5,6,7
Looking at 3 - the degrees that are possible are: 1,2,3,5
Looking at 5 - the degrees that are possible are: 1,2,3
Looking at 6 - the degrees that are possible are: 1,2,3
Looking at 7 - the degrees that are possible are: 1,2,3
Looking at 10 - the degrees that are possible are: 1,2
Looking at 11 - the degrees that are possible are: 1,2
Looking at 12 - the degrees that are possible are: 1,2
Looking at 13 - the degrees that are possible are: 1,2
Looking at 14 - the degrees that are possible are: 1,2
Looking at 15 - the degrees that are possible are: 1,2
Looking at 17 - the degrees that are possible are: 1,2
Summing up all the degrees: 6+4+3+3+3+2+2+2+2+2+2+2 = 33 -> 33 number.
What's wrong with that?

The key is to read the question carefully: an integer raised to an exponent that is an integer greater than 1:

Exclude all integers raised to the power of 1;
Exclude 7 to the power of 3: 7^3 = 343 > 300.
Add 1^(n > 1).
Add 2^4 = 16.
Add 2^8 = 256.
Add 3^4 = 81.

You get 24!
_________________

answer is (D)
I just counted all the powers of integers which equal less than or equal to 300.
1^2=1
2^8=256 (also 7,6,....2)
3^5=243(also 4,3,2) and so on
we get 30 such numbers.

Hi matvan,

You seem to have counted several Numbers twice

Please note that for exponent 2, The Numbers that we have are

\(1^2\), \(2^2\), \(3^2\), \(4^2\), \(5^2\), \(6^2\), \(7^2\), \(8^2\), \(9^2\), \(10^2\), \(11^2\), \(12^2\), \(13^2\), \(14^2\), \(15^2\), \(16^2\), \(17^2\)

Please note that for exponent 3, The Numbers that we have are

\(2^3\), \(3^3\), \(5^3\), \(6^3\) but Now you can't reconsider \(1^3\) and \(4^3\) as they have been counted already among the above 17 numbers

Similar Duplication must have inflated your count of such numbers from 24 to 30.

I hope it clears your mistake!

For me at least, the key to understanding this question was to carefully read what they were asking for.

So, what are all of the unique integers found among perfect squares, cubes, etc. between 1 and 300?

Perfect squares can be counted manually... 20^2 is 400 so it must be less than 20. 15^2 is 225 so plug and chug. You'll get 17^2, so that's 17 numbers for perfect squares.

Perfect cubes (just look for the highest cube first and eliminate any repeated integers). 7^3 is 7*49= 343, so it's all cubed integers below 7. 2 is 8, 3 is 9 but repeats, 4 is 64 but repeats, 5 is 125, 6 is 216

Repeat for the other powers and count the unique integers.

Hi All,

This question does require a bit of work/knowledge, but there is a Number Property that can save you some time and the prompt has a subtle hint in the answer choices that you could use to avoid some of the work:

The prompt is written as a "probability" question, but since the answers are fractions, you can work backwards and "translate" them into actual values. We're dealing with the first 300 positive integers and asked for the probability of randomly selecting a number that equals an integer raised to a power greater than 1. Here's how the answers can be rewritten:

17/300 = 17 numbers that fit the description
1/15 = 20 numbers
2/25 = 24 numbers
1/10 = 30 numbers
3/25 = 36 numbers

Having your "perfect squares" memorized will make the work go a bit faster; as you may have seen in this thread, there are 17 perfect SQUARES (and you SHOULD write them on the pad for easy reference). Finding the perfect cubes won't take too long (but there ARE some values that criss-cross with the perfect squares, so you CAN'T count them twice).

At this point, we have 17 + 4 = 21 values. Working higher (4th power, 5th power, etc.), there cannot be that many additional values that "fit", since we're dealing with a smaller and smaller sub-group each time and we've already seen that there ARE duplicates.

The Number Property that I mentioned earlier is that the "even powers" greater than 2 have all already appeared in your list (as perfect squares)

For example 2⁴= (2²)(2²) = 4²

So there's no reason to check the 4th, 6th, 8th, etc. powers since there won't be anything new.

Answers A and B are now too small and answers D and E seem way too big. Logically, the answer would have to be 24.

Final Answer:

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

The video solution of this question is as mentioned below





Answer: Option C

to me it felt more natural the cycle through the 2-9 as base and increase the exponent
2^2,3,4,5,6,7,8<300-->8
3^2,3,4,5<300-->4
and so on... this yields 24 options out of 300 numbers -->23/300 gives the right answer

All the powers less than 300 are:

1^Any
2^8
3^5
4^4
5^3
6^3
7^2
...
17^2

Sum the exponents minus 1 (except for 1^any which is counted once): 1+7+4+3+2+2+1*11 = 30


Find duplicates:

2^4 = 4^2
2^6 = 4^3 = 8^2
2^8 = 4^4 = 16^2
3^4 = 9^2

Subtract the duplicates: 30 - 6 = 24

24/300 = 2/25

By the language of the question, I can't determine what exactly it is asking for. How can you say that "Basically we need to find how many m^n (where n>1) are between 1 and 300 inclusive."


the integer so chosen = Integer which is between 1 and 300
say our exponent expression is \(m^n\)

we only know that n is in between 1 and 300, inclusive.
How can I say that \(m^n\) in in between 1 and 300.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________