GMAT Club Olympics 2024 (Day 9): A student was given an equation : Two-part Analysis (TPA)

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18 Jul 2024, 07:46

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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.

The different possibilities are:

Let us assume x = 0, then z will always be 0 which is contradicting to stimuli, hence not possible
Let us assume x = 1, then y and z will be same, hence x = 1 not possible
Let us assume x = 2, then y = 3, 4 is possible and z = 6, 8 respectively [The reverse combo won't be possible since y > x]
Let us assume x = 3, then we don't have any value for y since from y = 4, z wll be 2 digit
And similarly from any value more than 3, we will see the same observation

So only two possible values:

x = 2, y = 3 and z = 6
x = 2, y = 4 and z = 8

Hence min = max = 2 for x

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18 Jul 2024, 07:48

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y=9 x=1 (minimum)

y=3, x=2
y= 4, x=2 (maximum)

1-1 2-2

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18 Jul 2024, 07:54

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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 z = xy; y>x ; x, y, z are different digits
If x = 0; z = 0 = x ; But they are all different digits. Therefore, x = 0 is NOT POSSIBLE
If x = 1; z = y ; But they are all different digits. Therefore, x = 1 is NOT POSSIBLE
Minimum x = 2;

If x = 3; Minimum y = 4; Minimum z = xy = 12; Since z is a digit < 10; z = 12 is NOT POSSIBLE
Therefore, Maximum x = 2

Minimum x	Maximum x
2	2

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18 Jul 2024, 08:00

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Take x=1, y=2,3,4...9, but z=x (we need all diff)
Hence X_min cannot be 1

Take x=2, y=3,4,...9
but xy=single digit
Hence only (2,3), (2,4) are possible

Take x=3, y=4,5...9
but 3x4=12 (not single digit) hence X_min=X_Max=2

Ans=Min=2, Max=2

B, B

bb good choice

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18 Jul 2024, 08:11

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o solve the problem, we need to find the minimum and maximum possible values for
x given the equation
x×y=z, where x, y, and z are different digits and y>x.
We are to select the least possible value of x for "Minimum x" and the maximum possible value of x for "Maximum x."

Finding the Minimum Value of x:
To find the minimum value of x, we need to ensure that
y>x and the product z is a single-digit number (since z must also be a digit). Let's test the smallest possible values for x:
x=1

If y>1, the smallest y can be is 2.
Testing the smallest values for y:
1×2=2 (valid, but z=2, which is not different from y)
1×3=3 (valid, but z=3, which is not different from y)
1×4=4 (valid, but z=4, which is not different from y)
1×5=5 (valid, but z=5, which is not different from y)
1×6=6 (valid, but z=6, which is not different from y)
1×7=7 (valid, but z=7, which is not different from y)
1×8=8 (valid, but z=8, which is not different from y)
1×9=9 (valid, but z=9, which is not different from y)
x=2

If y>2, the smallest y can be is 3.
Testing the smallest values for y:

2×3=6 (valid, and z=6 is different from x and y)
Since 2 is valid,
x=1 doesn't work because the product is always the same as y.
Thus, the least possible value of x that works is:

Minimum x=2

Finding the Maximum Value of x:
To find the maximum value of x, we need to keep in mind
x must be less than y, and z must be a single-digit number. Let's test the largest possible values for x:

x=5

If y>5, the smallest y can be is 6.
Testing the smallest values for y:
5×6=30 (invalid because 30 is not a single digit)
5×7=35 (invalid because 35 is not a single digit)
5×8=40 (invalid because 40 is not a single digit)
5×9=45 (invalid because 45 is not a single digit)

x=4

If y>4, the smallest y can be is 5.
Testing the smallest values for y:
4×5=20 (invalid because 20 is not a single digit)
4×6=24 (invalid because 24 is not a single digit)

For x=3
If y>3, the smallest y can be is 4.
Testing the smallest values for y:
3×4=12 (invalid because 12 is not a single digit)
3×5=15 (invalid because 15 is not a single digit)
3×6=18 (invalid because 18 is not a single digit)

For x=2
If y>2, the smallest y can be is 3.
Testing the smallest values for y:
2×3=6 (valid, and z=6 is different from x and y)
2×4=8 (valid, and z=8 is different from x and y)

Thus, the largest possible value for x that keeps z as a single digit and y>x is:

Maximum x=2

In summary:
Minimum x=2
Maximum x=2

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18 Jul 2024, 08:26

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The minimum value is x = 1 (by hit & trial), and it satisfies the condition that 1*2 = 2.

The maximum value is x = 2 as for x = 3, the minimum condition, i.e. y = x+1 exceeds, as 3*4=12 which is not less than 10.

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18 Jul 2024, 08:30

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x, y, z all are different DIGITS & NOT Numbers.

So 0 to 9 is our domain.

0 & 1 cant be any of X, Y, Z as it would make X=Z or Y = Z

So next lowest value is 2.

So X min = 2.

Y min also is 2. So Xmax can be 4 . ( Xmax 5 or 6 will take Z to 10/12 which makes it out of digits. )

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18 Jul 2024, 08:39

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IMO - x=1 (min) and x=2 (max)

Minimum X = 1. This has to be 1 to satisfy 2 conditions y>x and x, y, and z are different digits

Maximum X =2. Y can take 3,4. This is the maximum possible as z has to be a digit

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18 Jul 2024, 08:47

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If y=1then z and x will be less than 1,also if y is more than 1 then z will be more than 1 but x will be less than 1.Implies maximum of x is 1.If y is less than 1 then x will be more than 1 , which does not satisfy the equation. Therefore minimum of x is 1 also.

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18 Jul 2024, 08:51

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Bunuel

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We must keep in mind that x, y, and z are different digits. Since they must be different, x or y cannot be 1 or we'll get a duplicate. Furthermore, x must be the lower value.

Then, we can check if x can be 2:

$2*3=6$ -> possible as this fits within all our criteria.

Let's see what would happen if x would be 3:

$3*4=12$ -> not possible as z is not a digit here.

As we continue to increase x, we will continue to increase z and no value will fit.

That means the only possible value for x is 2.

Minimum x: 2
Maximum x: 2

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18 Jul 2024, 08:51

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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.

by digits value of x,y,z has to be 0 to 9
y>x
x cannot be 1
minimum is 2 value of x
and maximum is also 2 as with x =3 z possible will be 2 digits ..
IMO both options 2,2 is correct

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18 Jul 2024, 08:54

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Minimum x=2 Since 2*3=6
x =1 is not possible because y=z but they are supposed to be different

Maximum x=2 Since 2*3=6
3*4=12 which is not possible since they all are digits

Ans B, B

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18 Jul 2024, 08:58

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Given x, y and z are different digits and y>x

For smallest value of x
0 and 1 not possible since x and z will be the same.
when x =2
y can be 3 or 4 so that z will be either 6 or 8 which are all different digits.

Hence smallest possible x = 2

For smallest value of x
Lets try x = 3
When x = 3, smallest possible y and z will be when y = 4, here z becomes 12 which is more than a digit.
When x is more than 2 then z will be multiple digitis, hence not possible.

Hence, maximum value of x is also 2

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18 Jul 2024, 09:00

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Bunuel

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Student was given equation x*y = z
x, y and z are differnt digits
x< y

min. value of x and max. possible value of x
2(x)*3(y)= 6

so maximum is 3

if x = 3
3*4= 12 but z cannot be more than 10

so maximum value of x = 2

if x =1 , y could be anything from 2 to 9
so minimum value of 1

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18 Jul 2024, 09:00

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Answer: Min: x=2, Max: x=2;

I don't know if my reasoning is correct but digit means 0-9. x,y and z are different digits and y>x.

Min: x can't be 0 or 1 because 0*y=0 and thus x=z or 1*y=y and thus y=z. Both cases are not possible.
Thus x = 2. (2*3=6 or 2*4=8)

Max: x can't be 3 because otherwise, y would be a minimum of 4 and 3*4=12 and z cannot be 12 because 12 is not a digit.
Therefore x can only be 2.

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18 Jul 2024, 09:07

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Bunuel

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Minimum value of x will be definitely 1.
Maximum value will be 2 as 2 into 3=6 and y is greater than x.

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18 Jul 2024, 09:39

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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.

x * y = z
x, y, z are diff
min value of x
1*n= n
so min value can't be 1
since y>x so it max value of x can't be 6
we are left with 2,3,4,5
x=2; y=3; z=6
x=2, y=4 ; z=8
x=2, y=5 ; z=10
x=2, y=6 ; z=12
min value =2

Max value :
5
Ans : BE