If x is an odd integer and y=x(x+1)(x-1), which of the following must also be an integer? I. y^23\cdot3 II. y^22\cdot3^2 III. y2\cdot3

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PS · Problem Solving

655-705

Roots

If $x$ is an odd integer and $y=x(x+1)(x-1)$ , which of the following must also be an integer? I. $\frac{y^2}{3\cdot3}$ II. $\frac{y^2}{2\cdot3^2}$ III. $\frac{y}{2\cdot3}$

Answer Choices

Correct answer marked below

I only

III only

I and III

I, II, and III

Correct

none of the above

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Comments

AtharvaJun 29

Took me a while but the key is that x is odd, so among x-1, x, x+1 you have two evens and one multiple of 3. That kills the exponents fast.

GauravJul 6

Wait, why two evens? If x is odd then x-1 and x+1 are consecutive even numbers, right?

MartinaJul 5

Yes exactly, and one of any three consecutive integers is divisible by 3.

SehejAug 29

Is it just me or is III the trap? I initially picked I and III because y/6 felt obviously divisible, but then I realized y has plenty of factors of 3 but only... hmm. Need to redo this.

DhanushAug 30

Same, I rushed III and then went back. The 2*3 in the denominator is sneakier than it looks.

AdyantApr 16

Pretty tough for a roots question honestly. The wording 'must also be an integer' made me second-guess whether I needed the exact number of 3s in y or just at least one.

Gracious20h ago

Quick strategy note: plug in x=3 and x=5 to sanity check each option. Saved me from overthinking the algebra.

Richardjust now

x=1 is also a nice edge case to check quickly.

What this question tests

What it tests

Your handling of square and higher roots, including simplifying radicals.

Common trap

Forgetting the ± when taking a square root to solve an equation, or assuming √(a²) = a.

If xxx is an odd integer and y=x(x+1)(x−1)y=x(x+1)(x-1)y=x(x+1)(x−1), which of the following must also be an integer? I. y23⋅3\frac{y^2}{3\cdot3}3⋅3y2​ II. y22⋅32\frac{y^2}{2\cdot3^2}2⋅32y2​ III. y2⋅3\frac{y}{2\cdot3}2⋅3y​