What is the value of integer z?

I choose option C:

Statement 1:
Since it is given that (Y -1) is positive integer, Y cannot be 1 or lesser than 1. Y >= 2

for Y = 2, Z = 0 as 2/1 => quotient 2 remainder 0
for Y >= 2 z = 1

Z=0, Z=1 insufficient.

Statement 2:
Statement does not say anything about Z.

insufficient.

Statement 1 & 2:
from 1
we have two scenario: if y = 2 than z = 0; if y >=2 z = 1
from 2
y not equal to 2

so y >2 and in this scenario z = 1.

Answer choice C.

OA please?

Ans C.
statmnt 1: insufficient when y=2; however y cant be equal to 1 as y-1= 0 NOT a positive integer.
statmnt 2: Insuff.
combining together 1 & 2 we get z=1

Answer is E

1)z is the remainder when positive integer y is divided by positive integer y-1
let y=2
so y = 1(2) + z or reminder 0
let y = 1
however if y = 1 then the denominator becomes 0 and thus Undefined or z need not be equal to 0
so INSUFFICIENT eliminate A,D

2)y is not a prime number

Says nothing about C


Insufficent eliminate B

Now both together again will give you the possiblity of Y having 1 or 4 or a grater non prime number
however any number greater than 1 gives reminder as 1 and the number 1 alone gives a undefined value
so both together also fails
eliminate C

so the Ans is E

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The answer is A as y and z are two consec numbers and has to be 2 and 1. There are no two number which will give an integer as reminder when divided.

statement 1 is sufficient - Y is a positive number so 1, 2,3,......... and Y-1 is also a Positive number, therefore minimum value of Y can be either 2, 3, 4, etc.
... and remainder always be 1, So value of Z is 1
statement 2- is not sufficient because there is no relation between Y an Z here.

Hence Answer will be A, Sorry for the previous answer....I just missed the the 2nd point.

A is the answer as the statement 1 says. y and y-1 both are positive, if y is divided by y-1 then remainder is always 1. so z=1.
statement 2 is insufficient

Ans - A

Remainder theorem-

when y /y-1
remainder = f(1)

f(y) = y
substituting the value of y =1

Therefore answer will always be 1.

Answer: C

According to statement 1, y could be 2 or 4, which gives us different values of z; not sufficient
(2) No relation between z and y

1+2:

y should be greater than 3; therefore, y and y-1 will always have the remainder of 1.

Answer C

What is the value of integer \(z\)?

(1) \(z\) is the remainder when positive integer y is divided by positive integer \((y - 1)\)

(2) \(y\) is not a prime number[/quote]

Lets take easier statement first.

Statement 2 just tells us that Y is not 2, 3, 5, or any other odd prime but it doesn't tell us anything about Z.

Stmt 1: Lets take Y to be 1, 2 or 3, Y-1 will be 0, 1 or 2. Y could be 0 also however since statement explicitly mentions that Y is positive integer and 0 is non negative integer, Y can not be 0.

1/0- Division by 0 is undefined in GMAT
2/1= 0 remainder
3/2= 1 remainder

Thus we have multiple answers, hence not sufficient.

Together: We know that Y is not prime, hence Y = 2 is ruled out and any other integer more then 2 divided by Int-1 will give a remainder of 1. Hence together sufficient. Answer is C.

Answer A:

Since z is the remainder when positive integer y is divided by positive integer (y - 1), which is 1 short of y, therefore, z =1

Lets take easier statement first.

Statement 2 just tells us that Y is not 2, 3, 5, or any other odd prime but it doesn't tell us anything about Z.

Stmt 1: Lets take Y to be 1, 2 or 3, Y-1 will be 0, 1 or 2. Y could be 0 also however since statement explicitly mentions that Y is positive integer and 0 is non negative integer, Y can not be 0.

1/0- Division by 0 is undefined in GMAT
2/1= 0 remainder
3/2= 1 remainder

Thus we have multiple answers, hence not sufficient.

Together: We know that Y is not prime, hence Y = 2 is ruled out and any other integer more then 2 divided by Int-1 will give a remainder of 1. Hence together sufficient. Answer is C.[/quote]

Looking at statement 1 -

any number can be represented in the form of divisor and remainder as below:

A = K(y-1) +z .............for all the values of K
at k=1
A=y-1 +Z....................A=y hence z=1

As A=y K can never be more than 1
at y=1
1 =1-1+z...............z=1
2=2-1+z...............z=1
3=3-1+z...............z=1

Hence statement 1 is sufficient.

We have a winner!

Winner:

AbhiJ

Official Explanation:

Answer is C

This question is a classic “Why Are You Here?” question. Statement 1 may look sufficient if you pick numbers (8 and 7; 15 and 14; 5 and 4 → the remainder is always 1). But what about 2 and 1? There’s no remainder there, so z would equal 0. This is why statement 2 is so important. Alone, it’s useless...so why is it there? Statement 2 tells us that (y - 1) cannot be 1, because y cannot be 2. This takes away that one flaw with statement 1, and means that both statements together are sufficient. But of more strategic importance, statement 2 should alert you to that problem with statement 1. Most test-takers miss that flaw when looking at statement 1 alone, but astute test-takers will reconsider once statement 2 is presented.

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The most important thing in this problem is avoid to overlook that Y and Y-1 are consecutive non negative integer.
The division of 2 consecutive number is always 1 except in 1 case (2,1) - (1,0) non considered - therefore:
1 - Not suff. solution are 0 and 1
2 - Clearly Not suff.

1&2 Suff and Z=1

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