If x and y are integers greater than 1, is x a multiple of y?

FACTS ESTABLISHED BY STEMS

• \(x\) is an integer
• \(y\) is an integer
• \(x > 1\)
• \(y > 1\)

QUESTION TRANSLATED

Since both \(x\) and \(y\) are integers the question "is \(x\) a multiple of \(y\)?" really asks if

\(x = q \cdot y + r\)

where \(q\) is an integer (the quotient) and \(r\) is the remainder when you divide \(x\) by \(y\) so that \(r = 0\) which means that

\(x = q \cdot y + 0\)

which implies that

\(x = q \cdot y\)

So the question translates to:

is \(x = q \cdot y\) when \(x\), \(q\) and \(y\) are all integers, and when \(x>1\) and \(y > 1\)?

STATEMENT 1

\(3y^2 + 7y = x\)

can be rewritten as

\(y(3y + 7) = x\)

however, from the question stems we know that \(y\) is an integer, which implies \(3y\) is an integer and that \(3y+7\) is also an integer:

\(y \cdot integer = x\)

which can be rewritten as

\(x = y \cdot integer\) or better yet, as

\(x = integer \cdot y\)

which if you notice is an equation of the form \(x = q \cdot y\) since \(q\) is an integer which implies that \(x\) is indeed a multiple of \(y\).

Therefore, Statement A is sufficient.

STATEMENT 2

the statement "\(x^2 - x\) is a multiple of \(y\)" must be rewritten as "\(x (x - 1)\) is a multiple of \(y\)" which implies that

either \(x\) is a multiple of \(y\) or \(x-1\) is a multiple of \(y\)

the sub-statement \(x\) is a multiple of \(y\) implies that \(x = q \cdot y\) which would answer the question, but we still need to figure out if the second sub-statement answers the question with the same answer as the first sub-statement.

the second sub-statement \(x-1\) is a multiple of \(y\) implies that

\(x-1 = q \cdot y\)

which can be rewritten as

\(x = q \cdot y + 1\)

which is an equation of the form \(x = q \cdot y + r\) when \(r=1\) which implies that when \(x\) is divided by \(y\) we get a remainder of 1, meaning that \(x\) is NOT a multiple of \(y\).

Since both sub-statements are contradictory, this means that when \(x^2 - x\) is a multiple of \(y\), \(x\) is not always a multiple of \(y\). meaning that we cannot determine whether \(x\) is always a multiple of \(y\) according to Statement 2 alone.

Therefore, Statement 2 is not sufficient.

ANSWER

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

for(2)

x(x-1) is multiple of y.
so either x is divisible by y or x-1 is divisible by y.
if x-1 is divisible by y, then x is not divisble by y.

plug in y=8, x=16--->16*15 is divisible by 8 --> yes x is a multiple of y.
plug in y=8,x=17----> 17*16 is still divisible by 8 --> But x is not a multiple of y.

So insufficient.

Hi Bunuel,

Two questions:
1) If x = 2 and y = 2, does that mean that x is a multiple of y? I always thought that x needed to be 1 multiple greater, meaning, x=4 and y=2 etc?
2) In situations like these, I usually stumble with either starting off with factoring(like you did above) or some other method. What do you suggest?
3) Lastly, the way I solved statement 1 was -- I saw it as (y)(multiple of y) = x, and the rule states that if i multiply two numbers that are multiples of y, then the result will be a multiple of y. Is that correct to assume?(I know that we can assume that with addition, right?)

1. The least multiple of a positive integer is that integer itself. So, the least multiple of 2 is 2.
2. I suggest the way I used.
3. 3y+7 is not necessarily a multiple of y but it does not need to be. y*integer=x implies that x is a multiple of y.

Can you please explain statement 2 again with better clarity? Bunuel

Not sure what to explain. From (2) x could be a multiple of y (for example, x = 2 and y = 2) but x could also not be a multiple of y (for example, x = 3 and y = 2).

1:
y(3y+7)=x
y*K (integer) is x, so x is a multiple of y

2:
x(x-1)=y
so x*K(integer) is y, so y is a multiple of X, but not the opposite

Attached is a visual that should help.

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