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Math Expert V
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If x and y are both integers greater than 1, is xy>100 ?  [#permalink]

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Difficulty:   95% (hard)

Question Stats: 26% (02:20) correct 74% (01:49) wrong based on 163 sessions

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If x and y are both integers greater than 1, is xy > 100?

(1) x has exactly 7 unique factors.

(2) y has exactly 9 unique factors.

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If x and y are both integers greater than 1, is xy>100 ?  [#permalink]

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Bunuel wrote:
If x and y are both integers greater than 1, is xy > 100?

(1) x has exactly 7 unique factors.

(2) y has exactly 9 unique factors.

Given situation:- $$x>1, y>1$$ (x and y are integers)
Stem:- Is $$xy>100$$

St1:- x has exactly 7 unique factors
Or, we can write in the prime factorization form,$$x=a^p$$, where p+1=7 or, p=6 (I have taken x=a^p not x=a^p*b^q..., because we are given 7 factors ,if we consider more than one prime exponent, we can't get a multiplication result of 7 by multiplying more than one integer values(where none of the integers are 1)) or $${(p+1)(q+1)...}\neq7$$
So,$$x=a^6$$
Now $$minimum(x)=min(a^6)=2^6=64$$ (we can't consider a=1 since it yields x=1^6=1)
Therefore, $$x*y=64*2=128>100$$ (The lowest integer value of y greater than 1 is 2)
For all other positive integer values of a>2,we have x > 100. Subsequently $$xy>100$$.

Sufficient.

St2:-y has exactly 9 unique factors
With the same reasoning as stated in st1, we have $$y=a^p*b^q$$, where (p+1)(q+1)=9=3*3.
So, p+1=3 and q+1=3
Or, p=2 and q=2
So, $$min(y)=min(a^2*b^2)=2^2*3^2=36$$
Therefore, $$xy=2*36=72<100$$----------------(a) (when x=2)
and $$xy=3*36=108>100$$-----------------------(b) (when x=3)
From (a) and (b), st2 is insufficient since st2 is inconsistent with the question stem.

Ans. (A)
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##### General Discussion
MBA Section Director V
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If x and y are both integers greater than 1, is xy>100 ?  [#permalink]

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Any integer with odd number of unique factors is a perfect square

If x and y are both integers greater than 1

To determine whether xy > 100 ?

Statement 1

x has exactly 7 unique factors.

=> x is a perfect square

=> x $$\geq$$ 64 since no perfect squate less than 64 has 7 number of factors

=> Since y is an integer and y > 1 => xy $$\geq$$ 128

Statement 1 is sufficient

Statement 2

y has exactly 9 unique factors.

=> y is a perfect square

=> y $$\geq$$ 36 since no perfect square less than 36 has 9 factors

=> Since x is an integer and x > 1 => xy $$\geq$$ 72

Statement 2 is not sufficient

Hence option A
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Re: If x and y are both integers greater than 1, is xy>100 ?  [#permalink]

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Correct me if I'm wrong but doesn't 24 also have 7 factors, making statement 1 insufficient? I am a bit unsure about this. Hence I went ahead with option C.
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Re: If x and y are both integers greater than 1, is xy>100 ?  [#permalink]

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nithinjohn wrote:
Correct me if I'm wrong but doesn't 24 also have 7 factors, making statement 1 insufficient? I am a bit unsure about this. Hence I went ahead with option C.

Hi nithinjohn ,

$$24=2^3*3^1$$
So the no of factors=(3+1)*(1+1)=8

And the factors are 1,2,3,4,6,8,12, and 24.

https://gmatclub.com/forum/divisibility ... 74998.html
You may visit above link for more clarity.

FINDING THE NUMBER OF FACTORS OF AN INTEGER
First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.

The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$

Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.

This is available in the link provided.
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Re: If x and y are both integers greater than 1, is xy>100 ?  [#permalink]

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Bunuel wrote:
If x and y are both integers greater than 1, is xy > 100?

(1) x has exactly 7 unique factors.

(2) y has exactly 9 unique factors.

xy>100 and both x and y>1. MEANS both are at least 2 and therefore if we can prove at least one is >50, it would be sufficient

1) x has exactly 7 unique factors
a) 7=1*7, no other possibilities so it is a^6
b) odd number of factor means X is perfect square
c) if it were just 3 factors.. it meant a perfect square of prime number
So least value =2^6=64
So minimum value of xy is 2*64=128>50
Sufficient

2) y has 9 unique factors.
a) higher number of factors necessarily does not mean LARGER number
b) 9=1*9....so least value 2^8=256>50 ....yes
c) 9=3*3.... So type a^2*b^2 least value = 2^2*3^2=4*9=36>50....No
So xy>100 and xy<100 possible
Insufficient

A
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Re: If x and y are both integers greater than 1, is xy>100 ?  [#permalink]

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Hi guys,

im just starting out on this topic, but i am very confused.
The Tasks states "unique" factors, so finding the first 7 UNIQUE factors and multiply them results always in >100 when y>=1 ?
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Re: If x and y are both integers greater than 1, is xy>100 ?  [#permalink]

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Bunuel wrote:
If x and y are both integers greater than 1, is xy > 100?

(1) x has exactly 7 unique factors.

(2) y has exactly 9 unique factors.

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 2 variables ($$x$$ and $$y$$) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Visit https://www.mathrevolution.com/gmat/lesson for details.

Conditions 1) & 2)
Since the integer x is $$p^6$$ type of integer greater than 1 by condition 1), where $$p$$ is a prime integer, the smallest possible number of is $$2^6 = 64$$.
Since the integer y is $$p^8$$ or $$p^2*q^2$$ types of integers greater than 1 condition 2), where $$p$$ and $$q$$ are prime numbers, the smallest possible number of y is $$2^2*3^2 = 36$$.
Then, the smallest possible value of $$xy$$ is $$64*36 > 100$$ and the answer is 'yes'.
Thus, both conditions together are sufficient.

Since this question is an integer question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
Since the integer x is $$p^6$$ type of integer greater than 1 by condition 1), where $$p$$ is a prime integer, the smallest possible number of is $$2^6 = 64$$.
Since the smallest possible number of $$y$$ is 2, the smallest possible number of $$xy$$ is 128, which is greater than 100 and the answer is yes.
That's why condition 1) is sufficient.

Condition 2)
Since the integer y is $$p^8$$ or $$p^2*q^2$$ types of integers greater than 1 condition 2), where $$p$$ and $$q$$ are prime numbers, the smallest possible number of y is $$2^2*3^2 = 36$$.
If $$x = 2$$ and $$y = 36$$, then $$xy = 72$$ is less than 100 and the answer is "no".
If $$x = 3$$ and $$y = 36$$, then $$xy = 108$$ is greater than 100 and the answer is "yes".
Since condition 2) doesn't yield a unique answer, it is not sufficient.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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Re: If x and y are both integers greater than 1, is xy>100 ?  [#permalink]

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Oh now I get it.
Nice explanation.
Thanks.

chetan2u wrote:
Bunuel wrote:
If x and y are both integers greater than 1, is xy > 100?

(1) x has exactly 7 unique factors.

(2) y has exactly 9 unique factors.

xy>100 and both x and y>1. MEANS both are at least 2 and therefore if we can prove at least one is >50, it would be sufficient

1) x has exactly 7 unique factors
a) 7=1*7, no other possibilities so it is a^6
b) odd number of factor means X is perfect square
c) if it were just 3 factors.. it meant a perfect square of prime number
So least value =2^6=64
So minimum value of xy is 2*64=128>50
Sufficient

2) y has 9 unique factors.
a) higher number of factors necessarily does not mean LARGER number
b) 9=1*9....so least value 2^8=256>50 ....yes
c) 9=3*3.... So type a^2*b^2 least value = 2^2*3^2=4*9=36>50....No
So xy>100 and xy<100 possible
Insufficient

A

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GMAT:[640 Q44, V34, IR4, AWA5] Re: If x and y are both integers greater than 1, is xy>100 ?   [#permalink] 17 Aug 2019, 04:17
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