GMAT Club World Cup 2022 (DAY 9): If x and y are integer

Asked: If \(x\) and \(y\) are integers, \(m = 3^{x-y}*5^{2y-1}*7^{4-x}\) and \(n = 105^y\), what is the value of \(y\)?

\(m = 3^{x-y}*5^{2y-1}*7^{4-x}\)
\(n = 105^y = 3^y5^y7^y\)

(1) \(n\) is NOT a multiple of \(m\).
y < x- y: 2y <x ; y < x/2
or
y <2y-1; y > 1
or
y > 4-x; x + y > 4
Case 1: y=2; x=5
Case 2: y=3: x=7
NOT SUFFICIENT

(2) \(m\) is a multiple of \(n\).
x-y>=y; x>=2y
2y-1>=y; y>=1
4-x>=y; x+y<=4
Since y is an integer y = 1 is the only solution since if y=2; x=4; x+y>4
x = 2 or 3
SUFFICIENT

IMO B

Answer B for me

Explanation in the picture
+ some further explanations For (2) (lack of paper 😅)

If x= 0, then y <=0 and y>=1 and y<=4 => no solution for that

If x=1, then y <=1/2 and y>=1 and y<=3 => no solution for that too

If x=2,then y <=1 and y>=1 and y<=2 => one solution y=1

If x=3 then y <=3/2 and y>=1 and y<=1 => one solution y=1

If x=4 then y<=2 and y>=1 and y<=0 => no solution for that...

So (2) is sufficient to conclude.

Then answer B.



Posted from my mobile device

m =\(3^{x-y} * 5^{2y-1}*7^{4-x}\)

n = \(3^y * 5^y * 7^y\)

Statement 1

n is NOT a multiple of m.

Inference : n / m is not an integer

\(\frac{n }{ m}\) = \(\frac{3^y * 5^y * 7^y }{ 3^{x-y} * 5^{2y-1}*7^{4-x}} \)

\(\frac{n }{ m}\) = \(3^{2y-x} * 5^{1-y} * 7^{y+x-4}\)

We see that that value of y is tied up to the value of x. All we know that any of the value of y can suffice as long as the resultant is a fraction.

Hence this option is not sufficient.

Statement 2

m is a multiple of n

Inference : m / n is an integer

\(\frac{m }{ n}\) = \(\frac{ 3^{x-y} * 5^{2y-1}*7^{4-x} }{ 3^y * 5^y * 7^y} \)

\(\frac{m }{ n}\) = \(3^{x-2y} * 5^{y-1} * 7^{4-x-y}\)

Now for this to be an integer

\(x-2y\geq{0}\) ---------1

\(y-1\geq{0}\) ----------- 2

\(4-x-y\geq{0}\) ----------3

Adding 1 and 3

\(4-3y\geq{0}\)
\(y\leq{\frac{4}{3}}\)

From 2
\(y\geq{1}\)

Therefore only possible value of y = 1

Sufficient

IMO B

\(m=3^{x−y}∗5^{2y−1}∗7^{4−x}\)
\(n=105^y\)
So, \(n=3^y*5^y*7^y\)

Statement 1: n is NOT a multiple of m.
So, either 1. y < x-y; or 2. y < 2y-1; or 3. y < 4-x; or 4. any combination of these 3 points
y can take multiple values in this case. Eg. 2 or 3 or 4 etc.
Hence, statement 1 is not sufficient.

Statement 2: m is a multiple of n.
So, \({x-y}\geq{y}\) and \({2y-1}\geq{y}\) and \({4-x}\geq{y}\)
In this case, x can be 2 or 3 but y will always be 1, because both x and y are integers.
Hence, this statement alone is sufficient.

Answer is B.

[quote="Bunuel"]If \(x\) and \(y\) are integers, \(m = 3^{x-y}*5^{2y-1}*7^{4-x}\) and \(n = 105^y\), what is the value of \(y\)?

(1) \(n\) is NOT a multiple of \(m\).
(2) \(m\) is a multiple of \(n\).

n=105^y=(3*5*7)^y=3^y*5^y*7^y
m=3^{x-y}*5^{2y-1}*7^{4-x}
1)As per statement 1 n is not a multiple of m
and as x and y both are integers, lets consider x=4 and y=4
Then m=5^7
and n=3^4*5^4*7^4
Thus n is not multiple of m
Again if we consider x=2 and y=2
Then m=5^3*7^2
And n=3^2*5^2*7^2
In this case also n is not a multiple of m. Thus we are not able to find the exact value of y
Thus 1 is insufficient
2)For the second case m is a multiple of n
Thus for m=3^{x-y}*5^{2y-1}*7^{4-x}
x-y>0 ,2y-1>0 and 4-x>0 - All of these have y
Now as 4>x and x>y Thus 2 values of x are possible {2,3}
Again 2y>1 Thus the only possible value of y=1
Thus 2 is sufficient.
hence B is the answer.

\(m = 3^{x-y}*5^{2y-1}*7^{4-x}\) and \(n = 105^y\)

Statement 1
"\(n\) is NOT a multiple of \(m\)."
Case-1
Let x=2 and y=2
Then \(m=3^{(2-2)}*5^{(4-1)}*7^{(4-2)}\)=\(5^3*7^2\)
and \(n=105^2=3^2*5^2*7^2\)
Clearly n is NOT a multiple of m

Case-2
Let x=4 and y=4
Then \(m=3^{(4-4)}*5^{(8-1)}*7^{(4-4)}\)=\(5^7\)
and \(n=105^4=3^4*5^4*7^4\)
Clearly n is NOT a multiple of m

So for two cases when n is not a multiple of m we are getting two different values of y.
Hence INSUFFICIENT


Statement 2
"\(m\) is a multiple of \(n\). "
We can put in possible values of integers.
Since \(m = 3^{x-y}*5^{2y-1}*7^{4-x}\)
We can have x={2,3} and y={1,2) [none of the powers of 3, 5 and 7 can be 0 or negative for m to be a multiple of n]

Case-1
x=2, y=1
\(m=3^{(2-1)}*5^{(2-1)}*7^{(4-2)}\)=3*5*7^2 -> is a multiple of (105)^1

Case-2
x=3, y=1
\(m=3^{(3-1)}*5^{(2-1)}*7^{(4-3)}\)=3^2*5*7 -> is a multiple of (105)^1

Case-3
x=3, y=2
\(m=3^{(3-2)}*5^{(4-1)}*7^{(4-3)}\)=3*5^3*7 -> is NOT a multiple of (105)^2

Thus when m is a multiple of n, the only possible value of y=1
Hence SUFFICIENT

ANS B

Answer choice (B) IMO

Answer: B

If x and y are integers, m=3^(x−y)∗5^(2y−1)∗7^(4−x) and n=105^y, what is the value of yy?

n = 105^y
= 3^y * 5^y * 7^y

m = 3^(x-y) * 5^(2y-1) * 7^(4-x)
=> m = n * 3^(x-2y) * 5^(y-1) * 7^(4-x-y) ....Eq 1.

(1) n is NOT a multiple of m.
For this to be true, from Eq. 1
3^(x-2y) * 5^(y-1) * 7^(4-x-y) > 1
For, y = 1 and x = 1,
3^(-1) * 5^0 * 7^2 = 49/3 > 1
i.e. m = n*(49/3)...true to the statement. y = 1 is a possibility.
For, y = 2 and x = 1
3^(-3) * 5^(1) * 7^(1) = 35/27 > 1
i.e. m = n*(35/27) ...true to the statement, y = 2 is also a possibility.
Since we can have multiple values of y,
Not sufficient.


(2) m is a multiple of n.
For this to be true, from Eq. 1
3^(x-2y) * 5^(y-1) * 7^(4-x-y) >= 1 and integer.
=> y-1 >= 0, x-2y >=0 and 4-x-y >=0
=> y >= 1
When y>=1, x >=2y => x >=2
When y >=1 and x >=2,
then 4 >= x+y
If y =1, then for x = 2, 3, above expression will be true.
If y = 2, then x >= 2y i.e. x >= 4, x+y > 4 and above expression will not be true.
Hence, y = 1 is the only possible value.
Sufficient.

Given information: x & y are integers. \(m = 3^{x-y}*5^{2y-1}*7^{4-x}\) and \(n = 105^y\).
n = 105^y = 3^y * 5^y * 7^y
To determine: y=?

(1) \(n\) is NOT a multiple of \(m\).

So, n/m is NOT an integer. Let us examine some cases with possible values

If y=1 & x=3: n/m = {3*5*7}/{3^2 * 5 * 7} = 1/3 which is not an integer
If y=2 & x=5: n/m = {3^2 * 5^2 * 7^2}/{3^3 * 5^3 * 7^-1} = 7^3/15 which is not an integer
We have 2 possible values for y: NOT SUFFICIENT

(2) \(m\) is a multiple of \(n\).

So, m/n IS AN INTEGER. Let us examine cases with possible values

If y=1 & x=2: m/n = {3*5*7^2}/{3*5*7} = 7, which is an integer
If y=2 & x=4: m/n = {3^2 * 5^3 * 7^0}/{3^2 * 5^2 * 7^2} = 5/49 NOT AN INTEGER. Cannot consider this
If y=0 & x=1: m/n = {3 * 5^-1 * 7^3}/{1} = 7^3/5 NOT AN INTEGER. Cannot consider this
If y=-2 & x=1: m/n = {3^3 * 5^-5 * 7^3}/{3^-2 * 5^-2 * 7^-2} = {3^5 * 7^5}/{5^3} NOT AN INTEGER. Cannot consider this

We have consider every possible range of y value apart from 1 and they all void the statement constraint of m/n being an integer that is why only y=1 is the possible solution in this statement. SUFFICIENT

Answer - B

x and y are integers
m = 3^(x−y) ∗ 5^(2y−1) ∗ 7^(4−x)
n = 105^y = 3^y * 5^y * 7^y

(1) n is NOT a multiple of m.
n ≠ a * m
=> x - y > y AND/OR
2y - 1 > y AND/OR
4-x > y

=> x - 2y > 0 AND/OR
y > 1 AND/OR
x + y < 4

Since there are OR conditions 'y' can have multiple values here. Hence insufficient

(2) m is a multiple of n
m = b * n
x - 2y >= 0 AND
y >= 1 AND
x + y <= 4

Solving these equations we find that the equations enclose a triangle with two vertices having integer values (2,1) and (3,1). In both cases, the value of y = 1
Sufficient

IMHO Option B

Official Solution:


If \(x\) and \(y\) are integers, \(m = 3^{x-y}*5^{2y-1}*7^{4-x}\) and \(n = 105^y\), what is the value of \(y\)?

This is a highly challenging question that involves multiple mathematical concepts like number properties, exponents, divisibility, and inequalities. It's crucial to analyze all aspects thoroughly and systematically, breaking the problem into smaller parts and understanding the relationships between the variables.

(1) \(n\) is NOT a multiple of \(m\).

The above implies that \(\frac{n}{m} \neq {integer}\):

The above expression is not an integer for different values of \(y\). For example, if \(x = y = 0\), \(3^{2y-x}*5^{1-2y}*7^{x+y-4}=3^0*5^1*7^{-4}=\frac{5}{7^4}\), which is not an integer. Alternatively, if \(x = 0\) and \(y = 1\), \(3^{2y-x}*5^{1-2y}*7^{x+y-4}=3^2*5^{-1}*7^{-3}=\frac{9}{5*7^3}\), which is not an integer either. Not sufficient.

(2) \(m\) is a multiple of \(n\).

The above implies that \(\frac{m}{n} = {integer}\):

Since \(x\) and \(y\) are integers, for the above expression to be an integer, all multiples must be integers and consequently all exponents must be non-negative. Therefore, several conditions must simultaneously be true:

Adding the last two inequalities gives \(3-x \geq 0\). Adding this to the first inequality results in \(3-2y \geq 0\), which implies \(3 \geq 2y\). From the second inequality, we have \(2y-2 \geq 0\), or \(2y \geq 2\). Therefore, considering \(3 \geq 2y\) and \(2y \geq 2\), we get \(2 \leq 2y \leq 3\). Since \(y\) is an integer, \(y\) can only be 1. Sufficient.


Answer: B

There is a fallacy in assuming that m is a multiple of n means m =n. It is m =n*(p)
I : m is not equal to np
II : m is equal to nq
thus
a)x-y>=y --> x>=2y --> x>=2
b)2y-1>=y --> y>=1
c)4-x>=y --> 4>=x+y

y=1, x=2
y=2, x=4 , not possible to have 4>=x+y

Thus only y=1