What is the value of integer z?

Question

(1) z is the remainder when positive integer y is divided by positive integer (y - 1)
(2) y is not a prime number

gmat6nplus1 · Answer

statement 1: y is at least one. if y=2 then (y-1) is a factor of y and reminder z is zero. If y=13 then (y-1) is not a factor of y and reminder z is one.

statement 2: clearly not sufficient

1+2) we can eliminate the case in which y=2. z will always be one.

Answer C

peachfuzz · Answer

(1) z is the remainder when positive integer y is divided by positive integer (y - 1)
y could be 8. In this case the remainder would be 1
y could also be 2. Here the remainder is 0.
Insufficient.

(2) y is not a prime number
Nothing is indicated about integer z.
Insufficient.

Combined, the remainder will always be 1 as the denominator will always 1 less than the numerator.
Sufficient.

Answer: C

prasun9 · Answer

z = ?

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

Consecutive positive integers y and (y-1) are always co-primes and if y is divided by (y-1) then the remainder will always be 1
y = (y - 1) * 1 + 1

we can try out a few numbers
y = 2 , (y-1) = 1, remainder = 1
y = 5 , (y-1) = 4, remainder = 1
y = 10, (y-1) = 9, remainder = 1
y = 2561, (y-1) = 2560, remainder = 1

Hence z = 1
So Stmt (1) is sufficient.

(2) y is not a prime number
This gives no information about z
Hence Stmt (2) is insufficient

So the answer is  A

EMPOWERgmatRichC · Answer

Hi prasun9,

You should double check your work. You were using a math theory that actually does NOT fit every example that you listed.

Here's the error:

What's the remainder when 2 is divided by 1? (hint: it's NOT 1). Knowing this, how would your answer change?

GMAT assassins aren't born, they're made,
Rich

CrackverbalGMAT · Answer

This is an value DS question, where you got an unique value for z.
1). Statement 1 is insufficient.
 Suppose say y=40 then y-1 is 39. And remainder is 1.
If y=16 then y-1 is 15 and again remainder is 1.
But if we take y=2 then y-1 is 1. And the remainder is zero because all the numbers are divisible by 1.
2). Statement 2 is insufficient.
This doesn’t anything about z.
Together sufficient, 
Statement 2 tells that 2 is not possible. Because its a prime number.
So obviously when you divide any number by its predecessor . the quotient is 1 and remainder 1.
So z is 1.
Together sufficient.
So the answer is C.

Bunuel · Answer

VERITAS PREP OFFICIAL SOLUTION

Solution: 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. 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.

hak15 · Answer

why can't y be 1, as its not a prime???Thus 1/0---inconclusive. or 9/8,10/9--remainder 1..Thus E
@Bunuel@VeritasPrepKarishma Please clarify

Bunuel · Answer

y cannot be 1 because in this case y-1=0 and division be 0 is not allowed.

hak15 · Answer

Despite that , Isn't there possibility of distinct answer, leading us to E

Bunuel · Answer

No.

(1) says that z is the remainder when positive integer y is divided by positive integer (y - 1). If y=1, no integer can be the remainder when y=1 is divided by y-1=0 because division by 0 is NOT allowed.

nitesh50 · Answer

Hi

Can anyone clarify why are we not considering negative remainders?
For example:
3=2*1+1
or
3=2*2-1
Hence even if we combine the two statements we get 2 remainders: +1/-1.

IMO ans should then be E.

VeritasPrepBrian 
 VeritasKarishma 
 gmatbusters 
 Gladiator59 
 Bunuel 
 chetan2u

Bunuel · Answer

Here is how Official Guide defines remainder:

If  x  and  y  are  positive integers , there exist unique integers  q  and  r , called the  quotient  and  remainder , respectively, such that  y =divisor*quotient+remainder= xq + r  and   0\leq{r}<x .

As you can see, a remainder according to the GMAT is always positive or 0.

KarishmaB · Answer

"Remainder" by definition is what "remains". So after you evenly divide out the groups, what remains is the remainder. The concept of "negative remainders" is theoretical. It helps us solve some kind of questions easily as discussed here: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/0 ... -the-gmat/ When GMAT talks about remainders, it is just talking about 0 or positive remainders.

Tamalmallick13 · Answer

It is again a flawed question. If something is divided by 0 then it becomes undefined not that we are not allowed. This leads us to choice E. So this question creates ambiguity and GMAT does not like ambiguity.

Bunuel · Answer

There is nothing wrong with the question. Also, I think your doubt is already answered on the previous page:
(1) says that z is the remainder when positive integer y is divided by  positive integer  (y - 1). If y=1, no integer can be the remainder when y=1 is divided by y-1=0 because division by 0 is NOT allowed.

To elaborate more, if y = 1, then y-1=0 and 0 is not a positive integer as (2) says. Also, 1 divided by 0 cannot give the remainder of z, no matter what z is, because you cannot divide by 0. So, if y = 1, then (2) is not true.

BeerZ · Answer

Hey Guys,

i have a question regarding the remainder z. 
I understand that there is no remainder if y=2.

But I have another issue in understanding that the remainder is =1 on all other occasions. 
The remainder gets smaller the higher number we pick, right?
So it would be 3/2 = 1+1/2; 5/4 = 1+1/5
Sorry if this appears simple. Just starting out with the gmat prep and i struggle a bit with the language as well

EMPOWERgmatRichC · Answer

Hi BeerZ,

When dealing with a 'remainder' question, the end result will NOT be a fraction. A remainder is really about 'leftover pieces.'

For example, with 7/4..... how many times does 4 divide evenly into 7? One time... with 3 leftover pieces (re. 7/4 = 1 r 3)

The information in Fact 1 tells us to divide a positive number ("Y") by another positive number that is one less ("Y-1"), so obviously the difference between those two numbers is 1.

For example:
5 and 4
7 and 6
100 and 99

In each of these examples, how many times does the smaller number divide evenly into the larger number? And how many leftover pieces are there?

5/4 = 1 r 1
7/6 = 1 r 1
100/99 = 1 r 1

That's why all of those results have the SAME remainder. The only exception under these conditions is 2 and 1... since 2/1 = 2 r 0.
 
GMAT assassins aren't born, they're made,
Rich

AVMachine · Answer

what would be the answer when y is 1;

when y is 1, y-1 = 0; then solution does not exist.

That shouldn't be the answer E.

Bunuel · Answer

Your doubt has already been addressed here: https://gmatclub.com/forum/what-is-the- ... l#p3520447

What is the value of integer z?

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