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GMAT Club Olympics 2024 (Day 9): A student was given an equation

1 2 3 4

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18 Jul 2024, 09:40

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Bunuel

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x, y and z are different digits between 0 to 9 (both inclusive).
Let us check for possible values: x * y = z
If x = 0, then 0 * y = 0, means z would also be equal to 0 but x,y, and z are different digits. So, not a possible case.
If x = 1, then 1 * y = y, means z would also be equal to y but x,y, and z are different digits. So, not a possible case.
If x = 2, and let y = 3 ( as y>x), 2 * 3 = 6........a possible case satisfying the equation.
or x = 2, and let y = 4 ( as y>x), 2 * 4 = 8........a possible case satisfying the equation.
If x=3, and y = 4, then 3 * 4 = 12...the value of z cannot be 12 as its value can only be between 0 to 9 (both inclusive).

Hence, at only x=2, the two possible scenarion exists.
Minimum x = 2
Maximum x = 2

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18 Jul 2024, 09:43

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Minimum=Maximum = 2

Minimum cannot be 1 coz --> 1*2=2, but x,y,z should be unique so
next best is 2.

Maximum
If x=3
Y= min.=4= 3*4=12 --> not a single digit

so x max=2 itself.

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18 Jul 2024, 09:52

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A student was given an equation, x * y = z, where x, y, and z are different digits and y > x. Select for Minimum x the least possible value of x, and select for Maximum x the maximum possible value of x.

We have 2 constraints at hand.
1. x, y, and z are different digits
2. y>x

If we take x as smallest possible digit i.e. x=0, z will also be same i.e. z=0. Not allowed
If we take x as next smallest possible digit i.e. x=1, z will also be same as y i.e. z=y. Not allowed
If we take x as next smallest possible digit i.e. x=2, 2*y=z.
If y=3>2, z=6
If y=4>2, z=8
Thus X minimum is 2.

Now, if we take x=3, then y has to ≥ 4.
If y=4, z=12 but that is not a digit.
Thus not allowed.

Post this analysis, we can conclude that x can not take any value other than 2.
Minimum x = Maximum x = 2
Option B is correct for both columns.

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18 Jul 2024, 10:48

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If x were to be 1
then x* y will always be y. But the digits x,y and z all are different digits.

Therefore, minimum value of x =2

For maximum value of x, assume x =3

Since y> x, y can be 4, but the product of 3 and 4 is 12 which is a 2 digit number whereas all three numbers have to be single digits.

Therefore, even the maximum value of x has to be 2 because of the above constraint.

Therefore, Minimum x = 2 and Maximum x =2

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18 Jul 2024, 11:14

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A student was given an equation, x * y = z, where x, y, and z are different digits and y > x. Select for Minimum x the least possible value of x, and select for Maximum x the maximum possible value of x.

Ans.
x, y, z are different digits and belong to set (0,1,2,3,4,5,6,7,8,9)

x * y = z (and y > x)
2 *3 = 6 , x=2
2 *4 = 8 , x=2

Hence x=2 is the only possible value of x

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18 Jul 2024, 11:22

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Solution: Given that,

x * y = z
x, y and z are different single digits
y > x

Minimum value of x
x cannot be = 0
as x and z will be 0 for any value of y and they have to be different.

if x = 1
y = z as 1 * y = z
This is not possible

if x = 2
2 * y = z
since y > x
say y = 3
z = 6

if y = 4
z = 8
Hence, the minimum value of x = 2

Maximum value of x
if x = 3
3 * y = z
As y > x
say y = 4
z = 12
which is not possible
Hence, the maximum value of x = 2

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18 Jul 2024, 11:26

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Minimum x = -2
Maximum x = -2

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18 Jul 2024, 12:33

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Given - A student was given an equation, x * y = z, where x, y, and z are different digits and y > x.
To find - Select for Minimum x the least possible value of x, and select for Maximum x the maximum possible value of x. Make only two selections, one in each column.

As given
x, y, and z are different digits and y > x.
x cant be 0 or 1 because it will make x = z or y = z that is not possible because all are different digits.
and all are digits so max value for z can be 9
Hence, x also cant be 3 because only 3*3 gives us 9.
Plus as we know y > x that will make y atleast 4,
and 3*4 = 12 which is not applicable as z has to be in single digit.
Hence values for x cant be greater than 2.
Therefore, the only value for x is possible = 2

Minimum x = Maximum x = 2
Option B for both.

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18 Jul 2024, 12:35

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We're given x * y = z, where x, y and z are distinct digits and y>x.

Let's check with digits:

1 X 2 = 2........ 1 multiplied by any digit gives the same digit in product, hence condition is not met. Thus, ruling out all cases of 1.

2 X 3 = 6........ this is giving us distinct digits and 3>2, fulfilling both the condition. Thus, minimum value of x is 2.
2 X 4 = 8........ giving us another valid option.

3 X 2 = 6....... although gives us 3 distinct digits, but y<x. thus, ruling out all cases of 4.

4 X 2 = 8........... although gives us 3 distinct digits, but y<x. thus, ruling out all cases of 4.

5 X 2 = 10........ giving us a 2-digit product, hence all cases of 5 being ruled out. Similarly, for all digits beyond 5 as well.

Therefore, the maximum value of x is also 2.

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18 Jul 2024, 12:46

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Since they are digits z must be 0-9
therefore,
x *y = z
x min = 1 * 9 = 9 => x min is 1
x max = 2*4= 9 => x max is 2
because z must be an single digit

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18 Jul 2024, 13:46

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3*3=9
2*4=8
1*7=7
2*3=6
1*5=5
2*2=4 or 1*4=4
1*3=3
1*2=2
1*1=1

Only this two combination is possible. Hence both max & min is 2

Answer 2,2

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18 Jul 2024, 14:03

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And.. here we are.
The day before the last day.
It's time to bring our highest performance.
"Money Time"

OK. Let's get to our analysis of our question:

Glance - the Question:
Question: We are dealing here with a Algebra/Logic/Min&Max problem with some constraints.
Answer Choices: We can see they areclose to each other. thus, we need to be carefull here with estimation.

Rephrase - Reading and Understanding the question:
Given: Okay, we have an equation: x * y = z
x < y
x ≠ y ≠ z
[?]: What is the Minimum value of "x" ? What is the Maximum value of "x"?

Solve:
Okay, lets think about the possibilites here:
z = 60 so x = 1 y = 60
but y ≠ z so it doesn't comply the rules.
if x = 2 y = 30 z= 60
alright we got it! x=2 is the minimum value, Our first answer.

Now with the second part:
z = 60 so x = 6 and y = 10
Thus our second answer is x=6.

THE END
I hope you liked the explanation, I have tried my best here.
Let me know if you have any questions about this question or my explanation.

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18 Jul 2024, 14:04

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This was another tricky worded question. At first it seemed easy; too easy that it made me suspicious...and for good reason I might add. Then I read the question more carefully and picked up on the details.

The key pieces of information are that x,y and z are different digits and that y>x. Really the key word here is digit. A digit is any of the numerals from 0 to 9. This changes the structure of the question significantly to what I had originally thought where I defined digit and number to be the same.

The values for x range from 1 to 6.

If x is 1, then by definition, y and z must be the same. Therefore 1 does not work.

If x is 2, then y can be 3 and z will be 6. This works. So we have established that x = 2 is the smallest value for x

If we take x is 3, then y, at the lowest, will be 4 whicih would lead to z = 12. This will not work as 12 is not a digit. This will also be the same for x is 4, 5 or 6.

Therefore, there is only one possible value for x and that is 2.

Minimum x = 2
Maximum x = 2

Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

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18 Jul 2024, 14:09

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We have x * y = z
Minimum value of x
Since all the digits x, y and are different then, x must not be 1,
Since if it is 1, whatever y will be the product z will be equal to y and that is contradictory to the fact that all digits are different, So 1 is out
there is no restriction on 2, so
Minimum value of x is 2

Maximum value of x
since x, y and z are digits and not numbers of 2 or 3 digits, then the maximum value of x should allow us to be sure that the product stay in a digit,
6 is out since , since y>x then y must be at least 7 then z=42 at least, OUT
5 is out since , since y>x then y must be at least 6 then z=30 at least, OUT
4 is out since , since y>x then y must be at least 5 then z=20 at least, OUT
3 is out since , since y>x then y must be at least 4 then z=12 at least, OUT
we're left with 2

Maximum value of x=2

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18 Jul 2024, 14:14

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A student was given an equation, x * y = z, where x, y, and z are different digits and y > x. Select for Minimum x the least possible value of x, and select for Maximum x the maximum possible value of x. Make only two selections, one in each column.

Since any number multiplied by one remain the same, X can only start from 2= Hence the MINIMUM
The maximum is constrained by the given choises so the maximum is 6

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18 Jul 2024, 16:42

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x*y = z
Since y != z then x != 1. Now let's test different values of z:
9 = 3*3 => z can't be 9 because x != y
8 = 4*2 => y=4 and x=2
7 = 7*1 => z can't be 7 because x!=1
6 = 3*2 => y=3 and x=2
4 = 2*2 => z can't be 4 because x != y
z can't be equal to 5,3,2,1 because if so x =1.
Therefore the minimum and maximum value of x is 2

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18 Jul 2024, 16:48

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x*y=z
Constrains
In order to x, y and z to be different none of them can be 1 or zero. Because a number multiplying by 1 is the number itself and a number multiplied by zero is zero
so x is at least 2 because y>x
x, y and z are digits numbers which means it can be -9 to 9 inclusive but as x is the smallest number and looking at the option they are only positive the range of value for y and z must be between 3 to 9 inclusive because the least value of x is 2( x=2)

so the possible combination for multiplying 2 different digits where y>x and getting a a digit is
2*3=6
2*4=8

so also the largest value x can take is 2 as y>x
Minimum x
x=2
Maximum x
x=2

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18 Jul 2024, 17:41

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X can't be 1 because otherwise y would equal z and that contradicts the stem.
X also can't be 3 because the only possible value for Y would also be 3 since Z is a digit.
The only possible value for X is 2 and that's it's minimum and maximum.

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18 Jul 2024, 18:02

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Given,
x*y = z
x,y,z are different
y>x

x can not be 1, as x must be <y and y cant be equal to z

so as per the given options
min x can be 2

Also x, y, z are digits, and also y>x
max x can be 2
as if x = 3, min y = 4
z = 12, Not possible
for x = 2, y can still be 3, 4

min: 2
max: 2