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

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Re: If x and y are both integers greater than 1, is xy>100 ? [#permalink]
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]
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]
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]
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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Re: If x and y are both integers greater than 1, is xy>100 ? [#permalink]
VeritasKarishma - is there a blog post discussing this concept which is needed to solve this question ? (on the blog post : quarter wit blog post)
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If x and y are both integers greater than 1, is xy>100 ? [#permalink]
Bunuel VeritasKarishma - i was wondering as I did this problem - if i could have this take-away

if the total number of factors is odd = perfect square
-- if the total number of factors is odd and number of factors is a prime number of factors (ex : 7 factors) = this perfect square has only one type of prime number in its prime box
-- if the total number of factors is odd and number of factors happens to be non-prime number of factors (Ex :9 factors) = this perfect square has $$\geq{1}$$ type of prime number in its prime box

if the total number of factors is even = non perfect square
-- If the total number of factors is 2 = number has to be Prime. Number of types of prime numbers in the prime box = 1 type only ($$2^{1}$$ or $$3^{1}$$ or $$7^{1}$$ or $$29^{1}$$- all have 2 factors only )
-- If the total number of factors is 4 = Number of types of prime numbers in the prime box $$\geq{1}$$ type ($$2^{3}$$ or $$2^{1}$$ * $$3^{1}$$, both have 4 factors each )
-- If the total number of factors is 8 = Number of types of prime numbers in the prime box $$\geq{1}$$ type ($$2^{7}$$ or $$2^{1}$$ * $$3^{3}$$ or $$2^{1}$$ * $$3^{1}$$ * $$5^{1}$$, all three have 8 factors each )

Other take-aways

** If the total number of factors is 1 = number has to be 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.

Check this first:

(1) x has exactly 7 unique factors.

Total number of factors is given by (p+1)(q+1)...
If (p+1)(q+1)... = 7 = 1 * 7, this can only be obtained when p = 6.
The smallest value of x can be 2^6 = 64.
Now, since y is an integer greater than 1, it will be 2 or more. No matter what y is then, xy will certainly be more than 100.
Sufficient.

(2) y has exactly 9 unique factors.
If (p+1)(q+1)... = 9 = 1*8 or 3*3
The smallest value y can taken is when p and q are 2 each such that y = 2^2 * 3^2 = 36.
Then if x is 2, xy < 100.
If x = 3, xy > 100.
Not sufficient.

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