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

Question

An integer between 1 and 300, inclusive, is chosen at random. What is the probability that the integer so chosen equals an integer raised to an exponent that is an integer greater than 1?

A. 17/300
B. 1/15
C. 2/25
D. 1/10
E. 3/25

sriharimurthy · Accepted Answer

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

Bunuel · Answer

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.

Asifpirlo · Answer

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

mhngo · Answer

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

ronr34 · Answer

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?

Bunuel · Answer

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

ronr34 · Answer

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?

Bunuel · Answer

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!

mathivanan · Answer

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.

GMATinsight · Answer

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!

EMPOWERgmatRichC · Answer

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:  C

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

GMATinsight · Answer

The video solution of this question is as mentioned below

https://www.youtube.com/watch?v=xWR1VpyCKP0

Answer: Option C

Sa800 · Answer

hello why arent negative integers considered ...for example -2 raised to the 2 or 4th power

Bunuel · Answer

We need to count the integers x from 1 to 300, inclusive, such that x = m^n for some integer m and some integer n > 1.

Negative integers do not need to be considered separately, because if n is even, then (-m)^n = m^n, so they produce the same positive value of x that has already been counted. For example, (-2)^2 = 4 and 2^2 = 4, but both expressions give the same single value of x, namely 4. So allowing negative bases does not create any new values of x; it only repeats perfect powers that are already included.

Adit_ · Answer

This is basically counting exercise:
Take powers of 2:
From 2 to 17^2 thus 16 numbers.

Take powers of 3:
Here we have to be careful to not double count 4^3 = 8^2 = 64 covered in previous steps.
Apart from that we have from 2^3 to 6^3 = 5 values - 1   = 4 numbers.

Take powers of 5 now:
2^5 = 32
3^5 = 243
Total = 2

Take 2^7 now:
 = 128 
Total = 1

Adding all: 16+4+2+1 = 23.
We also have to include 1 itself as 1^anything = 1.
Thus 23+1 = 24
Prob = 24/300 = 2/25.

Answer:  Option C

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards