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Intern  B
Joined: 14 Jan 2017
Posts: 8

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Bunuel wrote:
4. The functions f and g are defined for all the positive integers n by the following rule: f(n) is the number of positive perfect squares less than n and g(n) is the number of primes numbers less than n. If f(x) + g(x) = 16, then x is in the range:

A. 30 < x < 36
B. 30 < x < 37
C. 31 < x < 37
D. 31 < x < 38
E. 32 < x < 38

Perfect squares: 1, 4, 9, 16, 25, 36, ..,
Prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, ...

If x = 31, then f(31) = 5 and g(31) = 10: f(x) + g(x) = 5 + 10 = 15.
If x = 32, then f(32) = 5 and g(32) = 11: f(x) + g(x) = 5 + 11 = 16.
...
If x = 36, then f(36) = 5 and g(36) = 11: f(x) + g(x) = 5 + 11 = 16.
If x = 37, then f(37) = 6 and g(37) = 11: f(x) + g(x) = 6 + 11 = 17.

Thus x could be 32, 33, 34, 35 or 36: 31<x<37.

It is given "f(n) is the number of positive perfect squares less than n" then X should be greater than 36, with this answer comes to D. ??? please correct if I am wrong & please correct.
Math Expert V
Joined: 02 Sep 2009
Posts: 59634

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thunderbird350 wrote:
Bunuel wrote:
4. The functions f and g are defined for all the positive integers n by the following rule: f(n) is the number of positive perfect squares less than n and g(n) is the number of primes numbers less than n. If f(x) + g(x) = 16, then x is in the range:

A. 30 < x < 36
B. 30 < x < 37
C. 31 < x < 37
D. 31 < x < 38
E. 32 < x < 38

Perfect squares: 1, 4, 9, 16, 25, 36, ..,
Prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, ...

If x = 31, then f(31) = 5 and g(31) = 10: f(x) + g(x) = 5 + 10 = 15.
If x = 32, then f(32) = 5 and g(32) = 11: f(x) + g(x) = 5 + 11 = 16.
...
If x = 36, then f(36) = 5 and g(36) = 11: f(x) + g(x) = 5 + 11 = 16.
If x = 37, then f(37) = 6 and g(37) = 11: f(x) + g(x) = 6 + 11 = 17.

Thus x could be 32, 33, 34, 35 or 36: 31<x<37.

It is given "f(n) is the number of positive perfect squares less than n" then X should be greater than 36, with this answer comes to D. ??? please correct if I am wrong & please correct.

It's not clear from your post as to why should x be greater than 36. The solution you quote explains that x could be 32, 33, 34, 35 or 36: 31<x<37, which is answer C.
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Joined: 24 Mar 2018
Posts: 246

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Bunuel wrote:
10. If x is not equal to 0 and x^y=1, then which of the following must be true?

I. x=1
II. x=1 and y=0
III. x=1 or y=0

A. I only
B. II only
C. III only
D. I and III only
E. None

Notice that if x=-1 and y is any even number, then $$(-1)^{even}=1$$, thus none of the options must be true.

Bunuel
If we consider only I
i.e x=1
1 to power anything will be one so shouldn't the OA be A ?
Math Expert V
Joined: 02 Sep 2009
Posts: 59634

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teaserbae wrote:
Bunuel wrote:
10. If x is not equal to 0 and x^y=1, then which of the following must be true?

I. x=1
II. x=1 and y=0
III. x=1 or y=0

A. I only
B. II only
C. III only
D. I and III only
E. None

Notice that if x=-1 and y is any even number, then $$(-1)^{even}=1$$, thus none of the options must be true.

Bunuel
If we consider only I
i.e x=1
1 to power anything will be one so shouldn't the OA be A ?

The question asks: which of the following MUST be true, not COULD be true. Now, ask yourself is x = 1 ALWAYS true? Doesn't the solution you quote, gives an example when x = 1 is NOT true?
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Manager  S
Joined: 24 Mar 2018
Posts: 246

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Bunuel wrote:
teaserbae wrote:
Bunuel wrote:
10. If x is not equal to 0 and x^y=1, then which of the following must be true?

I. x=1
II. x=1 and y=0
III. x=1 or y=0

A. I only
B. II only
C. III only
D. I and III only
E. None

Notice that if x=-1 and y is any even number, then $$(-1)^{even}=1$$, thus none of the options must be true.

Bunuel
If we consider only I
i.e x=1
1 to power anything will be one so shouldn't the OA be A ?

The question asks: which of the following MUST be true, not COULD be true. Now, ask yourself is x = 1 ALWAYS true? Doesn't the solution you quote, gives an example when x = 1 is NOT true?

Bunuel
Yeah I got that but what's wrong with C ?
X=1 than x^y=1
y=0 than x^y=1
there doesn't exsist any other case for which x^y is not equal to 1

Originally posted by teaserbae on 23 Jul 2018, 02:57.
Last edited by teaserbae on 23 Jul 2018, 03:10, edited 2 times in total.
Math Expert V
Joined: 02 Sep 2009
Posts: 59634

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teaserbae wrote:
Bunuel wrote:
teaserbae wrote:
10. If x is not equal to 0 and x^y=1, then which of the following must be true?

I. x=1
II. x=1 and y=0
III. x=1 or y=0

A. I only
B. II only
C. III only
D. I and III only
E. None

Bunuel
If we consider only I
i.e x=1
1 to power anything will be one so shouldn't the OA be A ?

The question asks: which of the following MUST be true, not COULD be true. Now, ask yourself is x = 1 ALWAYS true? Doesn't the solution you quote, gives an example when x = 1 is NOT true?

Yeah I got that but what's wrong with C ?
X=1 than x^y=1
y=0 than x^y=1
there doesn't exsist any other case for which x^y is not equal to 1

I'll copy my solution here:

Notice that if x=-1 and y is any even number, then $$(-1)^{even}=1$$, thus none of the options must be true.

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Manager  S
Joined: 24 Mar 2018
Posts: 246

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Bunuel wrote:
teaserbae wrote:
Bunuel wrote:
10. If x is not equal to 0 and x^y=1, then which of the following must be true?

I. x=1
II. x=1 and y=0
III. x=1 or y=0

A. I only
B. II only
C. III only
D. I and III only
E. None

Bunuel
If we consider only I
i.e x=1
1 to power anything will be one so shouldn't the OA be A ?

The question asks: which of the following MUST be true, not COULD be true. Now, ask yourself is x = 1 ALWAYS true? Doesn't the solution you quote, gives an example when x = 1 is NOT true?

I'll copy my solution here:

Notice that if x=-1 and y is any even number, then $$(-1)^{even}=1$$, thus none of the options must be true.

Bunuel
By III x=1 or y=0
since x= -1 so y=0
Isn't (-1)^0 = 1 ?
As 0 is even integer
Math Expert V
Joined: 02 Sep 2009
Posts: 59634

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1
teaserbae wrote:
Bunuel wrote:
teaserbae wrote:
10. If x is not equal to 0 and x^y=1, then which of the following must be true?

I. x=1
II. x=1 and y=0
III. x=1 or y=0

A. I only
B. II only
C. III only
D. I and III only
E. None

The question asks: which of the following MUST be true, not COULD be true. Now, ask yourself is x = 1 ALWAYS true? Doesn't the solution you quote, gives an example when x = 1 is NOT true?

I'll copy my solution here:

Notice that if x=-1 and y is any even number, then $$(-1)^{even}=1$$, thus none of the options must be true.

Bunuel
By III x=1 or y=0
since x= -1 so y=0
Isn't (-1)^0 = 1 ?
As 0 is even integer

I'll try this last time. The question asks: which of the following MUST be true?

None of the options is necessarily true because if x = -1, and say y = 2, then x^y = 1 and not I, not II and not III is true.
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Rice (Jones) School Moderator P
Joined: 18 Jun 2018
Posts: 314
Location: United States (AZ)
Concentration: Finance, Healthcare
GMAT 1: 600 Q44 V28 GPA: 3.36

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Hello Bunuel,

You mentioned that we can factor 17 in the solution to this problem, is that a typo?

Thanks.

Bunuel wrote:
5. Which of the following is a factor of 18!+1?

A. 15
B. 17
C. 19
D. 33
E. 39

18! and 18!+1 are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1.

Now, since we can factor out each 15, 17, 33=3*11, and 39=3*13 out of 18!, then 15, 17, 33 and 39 ARE factors of 18! and are NOT factors of 18!+1. Therefore only 19 could be a factor of 18!+1.

Math Expert V
Joined: 02 Sep 2009
Posts: 59634

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1
funsogu wrote:
Hello Bunuel,

You mentioned that we can factor 17 in the solution to this problem, is that a typo?

Thanks.

Bunuel wrote:
5. Which of the following is a factor of 18!+1?

A. 15
B. 17
C. 19
D. 33
E. 39

18! and 18!+1 are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1.

Now, since we can factor out each 15, 17, 33=3*11, and 39=3*13 out of 18!, then 15, 17, 33 and 39 ARE factors of 18! and are NOT factors of 18!+1. Therefore only 19 could be a factor of 18!+1.

17 is a factor of 18!:

18! = 1*2*3*...*17*18.
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Manager  B
Joined: 16 Oct 2017
Posts: 52
Location: India
GPA: 4

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Bunuel wrote:
3. How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?

A. 16
B. 27
C. 31
D. 32
E. 64

Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is 2^5=32. Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0.

Hi Bunuel
Why didn't we approach this question in the same way as done in Q.2 of this thread?
Following that approach would the solution be as follows?
6C5+6C4+6C3+6C2+6C1+1
(1 being added for the null set)

Math Expert V
Joined: 02 Sep 2009
Posts: 59634

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1
applebear wrote:
Bunuel wrote:
3. How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?

A. 16
B. 27
C. 31
D. 32
E. 64

Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is 2^5=32. Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0.

Hi Bunuel
Why didn't we approach this question in the same way as done in Q.2 of this thread?
Following that approach would the solution be as follows?
6C5+6C4+6C3+6C2+6C1+1
(1 being added for the null set)

What is the logic behind this?

It should be: 5C5 + 5C4 + 5C3 + 5C2 + 5C1 + 1 = 1 + 5 + 10 + 10 + 5 + 1 = 32.
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Manager  B
Joined: 16 Oct 2017
Posts: 52
Location: India
GPA: 4

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Bunuel wrote:
applebear wrote:
Bunuel wrote:
3. How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?

A. 16
B. 27
C. 31
D. 32
E. 64

Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is 2^5=32. Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0.

Hi Bunuel
Why didn't we approach this question in the same way as done in Q.2 of this thread?
Following that approach would the solution be as follows?
6C5+6C4+6C3+6C2+6C1+1
(1 being added for the null set)

What is the logic behind this?

It should be: 5C5 + 5C4 + 5C3 + 5C2 + 5C1 + 1 = 1 + 5 + 10 + 10 + 5 + 1 = 32.

thank you for the early reply. i now realise it was a silly mistake on my part.
Intern  B
Joined: 04 Sep 2016
Posts: 16

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Bunuel wrote:
8. The product of a positive integer x and 377,910 is divisible by 3,300, then the least value of x is:

A. 10
B. 11
C. 55
D. 110
E. 330

Given: $$\frac{377,910 *x}{3,300}=integer$$.

Factorize the divisor: $$3,300=2^2*3*5^2*11$$.

Check 377,910 for divisibility by 2^2: 377,910 IS divisible by 2 and NOT divisible by 2^2=4 (since its last two digits, 10, is not divisible by 4). Thus x must have 2 as its factor (377,910 is divisible only by 2 so in order 377,910*x to be divisible by 2^2, x must have 2 as its factor);

Check 377,910 for divisibility by 3: 3+7+7+9+1+0=27, thus 377,910 IS divisible by 3.

Check 377,910 for divisibility by 5^2: 377,910 IS divisible by 5 and NOT divisible by 25 (in order a number to be divisible by 25 its last two digits must be 00, 25, 50, or 75, so 377,910 is NOT divisible by 25). Thus x must have 5 as its factor.

Check 377,910 for divisibility by 11: (7+9+0)-(3+7+1)=5, so 377,910 is NOT divisible by 11, thus x must have 11 as its factor.

Therefore the least value of x is $$2*5*11=110$$.

Please how did you arrive at "Therefore the least value of x is 2∗5∗11=1102∗5∗11=110." ?
Manager  B
Joined: 27 Oct 2019
Posts: 54

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Bunuel wrote:
pclawong wrote:
Bunuel wrote:
7. The greatest common divisor of two positive integers is 25. If the sum of the integers is 350, then how many such pairs are possible?

A. 1
B. 2
C. 3
D. 4
E. 5

We are told that the greatest common factor of two integers is 25. So, these integers are $$25x$$ and $$25y$$, for some positive integers $$x$$ and $$y$$. Notice that $$x$$ and $$y$$ must not share any common factor but 1, because if they do, then GCF of $$25x$$ and $$25y$$ will be more that 25.

Next, we know that $$25x+25y=350$$ --> $$x+y=14$$ --> since $$x$$ and $$y$$ don't share any common factor but 1 then (x, y) can be only (1, 13), (3, 11) or (5, 9) (all other pairs (2, 12), (4, 10), (6, 8) and (7, 7) do share common factor greater than 1).

So, there are only three pairs of such numbers possible:
25*1=25 and 25*13=325;
25*3=75 and 25*11=275;
25*5=125 and 25*9=225.

Bunuel,
How do we know if x and y do not share any common factor?
what is the hint?
Thank you

If x and y are NOT co-prime, the GCD of $$25x$$ and $$25y$$ will be more than 25, not 25 as given.

Why aren't we including (9,5), (11,3) and (13,1). Moreover how would this question be solved if we were told that the least common multiple of two numbers is 350??

Posted from my mobile device
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Joined: 02 Sep 2009
Posts: 59634

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ssshyam1995 wrote:
Bunuel wrote:
7. The greatest common divisor of two positive integers is 25. If the sum of the integers is 350, then how many such pairs are possible?

A. 1
B. 2
C. 3
D. 4
E. 5

We are told that the greatest common factor of two integers is 25. So, these integers are $$25x$$ and $$25y$$, for some positive integers $$x$$ and $$y$$. Notice that $$x$$ and $$y$$ must not share any common factor but 1, because if they do, then GCF of $$25x$$ and $$25y$$ will be more that 25.

Next, we know that $$25x+25y=350$$ --> $$x+y=14$$ --> since $$x$$ and $$y$$ don't share any common factor but 1 then (x, y) can be only (1, 13), (3, 11) or (5, 9) (all other pairs (2, 12), (4, 10), (6, 8) and (7, 7) do share common factor greater than 1).

So, there are only three pairs of such numbers possible:
25*1=25 and 25*13=325;
25*3=75 and 25*11=275;
25*5=125 and 25*9=225.

Why aren't we including (9,5), (11,3) and (13,1). Moreover how would this question be solved if we were told that the least common multiple of two numbers is 350??

Posted from my mobile device

1. Don't we have these pairs? If we include say (9,5) along with (5,9) do we get different two numbers?
2. If it were not given that the sum is 350, then the answer will be - infinitely many.
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Manager  B
Joined: 27 Oct 2019
Posts: 54

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Bunuel wrote:
ssshyam1995 wrote:
Bunuel wrote:
7. The greatest common divisor of two positive integers is 25. If the sum of the integers is 350, then how many such pairs are possible?

A. 1
B. 2
C. 3
D. 4
E. 5

We are told that the greatest common factor of two integers is 25. So, these integers are $$25x$$ and $$25y$$, for some positive integers $$x$$ and $$y$$. Notice that $$x$$ and $$y$$ must not share any common factor but 1, because if they do, then GCF of $$25x$$ and $$25y$$ will be more that 25.

Next, we know that $$25x+25y=350$$ --> $$x+y=14$$ --> since $$x$$ and $$y$$ don't share any common factor but 1 then (x, y) can be only (1, 13), (3, 11) or (5, 9) (all other pairs (2, 12), (4, 10), (6, 8) and (7, 7) do share common factor greater than 1).

So, there are only three pairs of such numbers possible:
25*1=25 and 25*13=325;
25*3=75 and 25*11=275;
25*5=125 and 25*9=225.

Why aren't we including (9,5), (11,3) and (13,1). Moreover how would this question be solved if we were told that the least common multiple of two numbers is 350??

Posted from my mobile device

1. Don't we have these pairs? If we include say (9,5) along with (5,9) do we get different two numbers?
2. If it were not given that the sum is 350, then the answer will be - infinitely many.

Regarding the LCM part I meant we just changes HCF from the question and make it LCM. Then how will we proceed with the question?? Re: Fresh Meat!!!   [#permalink] 22 Nov 2019, 04:44

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