If a, b, and c are integers and abc ≠ 0, is a^2 – b^2 a multiple of 4?

(1) a = (c – 1)^2 --> a=c^2-2c+1. Not sufficient, since no info about b.

(2) b = c^2 – 1. Not sufficient, since no info about a.

(1)+(2) \(a^2-b^2 = (a-b)(a+b)=(c^2-2c+1-c^2+1)(c^2-2c+1+c^2-1)=(2-2c)(2c^2-2c)=4(1-c)(c^2-c)\) --> a^2-b^2 is a multiple of 4. Sufficient.

Answer: C.

Hope it's clear.
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Hi mun23

(1) a = (c – 1)^2 ALONE

(1) is INSUFFICIENT because this statement do not give any information about b

(2) b = c^2 – 1 ALONE

(2) is INSUFFICIENT because this statement do not give any information about a

(1) + (2) a = (c – 1)^2 AND b = c^2 – 1

After combining (1) and (2) -->
a + b = (c-1)^2 + c^2 - 1 = c^2 - 2c + 1 + c^2 - 1 = 2c(c-1)
a - b = (c-1)^2 - c^2 + 1 = c^2 - 2c + 1 - c^2 +1 = -2c + 2 = 2(1-c)

a^2 - b^2 = (a+b)(a-b) = 4c(c-1)(1-c) = -4c (c-1)^2

Hence , a^2 - b^2 = 4 * Integer , So the answer is Yes --> (1) + (2) SUFFICIENT

Answer : C

If a, b, and c are integers and abc ≠ 0, is a^2 – b^2 a multiple of 4?

(1) a = (c – 1)^2

(2) b = c^2 – 1

Question can be written as a^2 - b^2 or (a-b)*(a+b) is a multiple of 4
STAT1 alone is not sufficient as we don't know anything about b
STAT2 alone is not sufficient as we don't know anything about a
Taking both together

a = (c-1)^2 and b = c^2 - 1 or b = (c-1)*(c+1)

a+b = (c-1)^2 + (c-1)*(c+1) = (c-1)* (c-1 + c+1) = c*(c-1)
a-b = (c-1)^2 - (c-1)*(c+1) = (c-1)*(c-1 - (c+1)) = (c-1)*(-2) = -2*c*(c-1)
So, a^2 - b^2 = (a-b)*(a+b) = c*(c-1) * (-2*c*(c-1))
= -2*(c^2)*((c-1)^2)

If c is odd then c-1 will be even and c*c-1 will be a multiple of 2
else if c i even then again c*c-1) will be a multiple of 2

So, in both the cases c*(c-1) is a multiple of 2. So, -2*(c^2)*((c-1)^2) will be a multiple of 4 (Actually will be a multiple of 8 too)
So, Answer will be C

Hope it helps!



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st.1 alone is not sufficient as nothing is mentioned about b
st.2 alone is not sufficient as nothing is mentioned about a

st.1 and st.2

\(a= (c-1)^2\) and \(b= (c^2-1)\)

\(a^2-b^2= (a-b)(a+b)\)
put the value of a and b in the above expression, we have

\(a^2-b^2 = ((c-1)^2 - (c^2-1))((c-1)^2 + (c^2 - 1))\)

= \(((c^2+1-2c) - c^2 +1) ((c^2+1-2c) +c^2 - 1)\)

= \((2-2C)(2C^2-2c)\)

=\(4c(1-c)(c-1)\)

as can be seen, the above expression is a multiple of 4. hence answer is C.

Merging topics. Please ask if anything remains unclear.
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Ans : C
a^2-b^2=-4c(c-1)^2

Sent from my SM-J210F using GMAT Club Forum mobile app

Bunuel niks18 chetan2u amanvermagmat




How about this approach?
If I simplify question stem to best possible, what am I asked?
Is a / b even?

Why?
a. A multiple pf 4 is always even
b. Squaring of odd / even is always odd /even
c. Even - Even = Even.
or Odd - Odd = Even

I did above in my head.
Essentially, either of statements is clearly insuff
since I need to know both a and b.

What happens when I combine and simplify \(a^2\) - \(b^2\)
I get -2c + 1.

Even if I do not know value of c, I know -2 will always be even
Even + Odd = Odd. So, we do get an unique ans: NO
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Hello

Its NOT necessary for a & b to be even for this to be true. a & b could be both odd also. Eg, a=5, b=3, here a^2 - b^2 = 25-9 = 16, which is a multiple of 4.

The math on the previous posts is clear, so I can't add anything to that. That's a surefire way of getting the problem correct.

I took a different approach in plugging in numbers to the prompts. It was obvious right away that 1 and 2 alone are insufficient because they don't give full details about the variables.

After combining, start picking smart numbers for C and work out what A and B would be. I made C = 2, 3, and 4, and then worked out what that would make A and B equal to. This gave me a good enough sample size to test divisibility of 4 and to feel comfortable enough in picking answer choice C.

Is this a perfect approach? No, but I got to the right answer in well under 2 minutes with a high probability of being correct.

Given: a, b, and c are integers and abc ≠ 0

Target question: Is a² – b² a multiple of 4?

Statement 1: a = (c – 1)²
No information about b. NOT SUFFICIENT

If you're not convinced, let's TEST some values.
There are several values of a, b and c that satisfy statement 1. Here are two:
Case a: a = 4, b = 2 and c = 3. In this case, a² – b² = 4² – 2² = 12. So, the answer to the target question is YES, a² – b² IS divisible by 4
Case b: a = 4, b = 3 and c = 3. In this case, a² – b² = 4² – 3² = 7. So, the answer to the target question is NO, a² – b² is NOT divisible by 4
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: b = c² – 1
No information about a. NOT SUFFICIENT

If you're not convinced, let's TEST some values.
There are several values of a, b and c that satisfy statement 2. Here are two:
Case a: a = 10, b = 8 and c = 3. In this case, a² – b² = 10² – 8² = 36. So, the answer to the target question is YES, a² – b² IS divisible by 4
Case b: a = 9, b = 8 and c = 3. In this case, a² – b² = 9² – 8² = 17. So, the answer to the target question is NO, a² – b² is NOT divisible by 4
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that a = (c – 1)², so a² = (c – 1)⁴
Statement 2 tells us that b = c² – 1, so b² = (c² – 1)²

This means that: a² - b² = (c – 1)⁴ - (c² – 1)²
KEY: the right side is a DIFFERENCE OF SQUARE, which means we can factor it
We get: a² - b² = [(c – 1)² + (c² – 1)][(c – 1)² - (c² – 1)]
Expand and simplify to get: a² - b² = [(c² - 2x + 1) + (c² – 1)][(c² - 2x + 1) - (c² – 1)]
Simplify to get: a² - b² = [2c² - 2x][-2x + 2]
Factor out some 2's to get: a² - b² = [2(c² - x)][2(-x + 1)]
Simplify to get: a² - b² = 4(c² - x)(-x + 1)
We can see that a² - b² is clearly divisible by 4
So, the answer to the target question is YES, a² – b² IS divisible by 4

Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

Cheers,
Brent
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VeritasKarishma

can you please solve this \((2c(c-1))^2 - (2(1-c))^2\) in a way i can understand how to arrive at \(-4c (c-1)^2\)


i did this \(4c^2(c^2-2c+1)-(4(1-2c+c^2)\) now what ?

\(a = (c – 1)^2 = c^2 + 1 - 2c\)
\(b = c^2 – 1\)

\(a^2 – b^2 = (a + b)(a - b) \)
\(= (c^2 + 1 - 2c + c^2 - 1)(c^2 + 1 - 2c - c^2 + 1) \)
\(= (2c^2 - 2c)(2 - 2c)\)
\(= 2c (c - 1) * 2(1 - c)\)
\(= 2c (c - 1) * (-2)(c - 1)\)
\(= -4c(c - 1)^2\)
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