Math Revolution and GMAT Club Contest! For positive integers a and b,

QUESTION #1:

For positive integers a and b, when a^4 - b^4 is divided by 3, what is the remainder?

(1) When a + b is divided by 3, the remainder is 0
(2) When a^2 + b^2 is divided by 3, the remainder is 2

ANSWER:

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

(1) \((a+b)\) is divisible by 3 --> (*) is always divisible by 3 --> sufficient

(2) \((a^2 + b^2) = 3m +2\) --> \(a^2 = 3q + 1\) and \(b^2 =3p +1\)--> \(a^2 - b^2 = 3 (q-p) +1 -1 = 3(q-p)\) --> (*) is divisible by 3 --> sufficient

Answer: D

MATH REVOLUTION OFFICIAL SOLUTION:

If we alter the original condition and question, it becomes \(a^4-b^4=3t\)? (t is an integer). If we factor the question, the result is (a-b)(a+b)(a^2+b^2)=3t?, which is question. Then, if we look at 1), a-b=3n (n is an integer) becomes \(a^4-b^4=3n(a+b)(a^2+b^2)\), which is surely divisible by 3. This is a “yes” and sufficient.

If we look at 2), it looks like it will be \(a^2+b^2=3p+2\) (p is a positive integer) and not sufficient. Mistake Type 4(B) states “If answer choice A and B seem too attractive and obvious, consider answer choice D”. This Mistake Type commonly applies to a case similar to 1) and also to integer question, which is one of key questions. So, if we directly substitute 1 and 1 to a and b in 2), it yields (a,b)=(1,1) ← a-b=multiples of 3. This is a “yes”. If we substitute 1 and 2, it yields (a,b)=(1,2) ← a+b=multiples of 3. This is also a “yes”. Thus, in any case, a4-b4 is always a multiple of 3. This is a “yes” and sufficient. The correct answer choice is D.

In case of GMAT math DS problems, students have to be extra cautious about Mistake Type 4(B). Remember, “If answer choice A and B seem too attractive and obvious, consider answer choice D”. It is nearly impossible to solve a problem in 2 to 3 minutes during the exam. So students have to approach problems with a complete sense of logic based on Mistake Types. Also, please remember that if one condition seems easy when the other seems rather challenging, the answer will be D. This type of question decides a score of 50 or a perfect 51.

Thanks Bunuel. For option 2, I have an alternate approach to this. Please let me know, if this can be solved in this manner?
\((a^2+b^2)=3p+2\). It can also be written as \(((a-b)^2+2ab)=3p+2\). So this means (a-b) is multiple of 3. Hence \(a^4-b^4\) is divisible to 3.
Please suggest if this solution is ok or it can be solved something similar to this?

I don't completely understand the solution posted by Math Revolution. Could Bunuel or chetan2u please post their explanation/solution? Thanks!


For positive integers a and b, when a^4 - b^4 is divided by 3, what is the remainder?

(1) When a + b is divided by 3, the remainder is 0
(2) When a^2 + b^2 is divided by 3, the remainder is 2

Hi Victor,

I find plugging in works best with these questions.

I got D

1) a+b / 3 = Rem0

This basically means that when you add the numbers up they add up to 3 or a multiple of 3.
Take 3 for easy calculation. Plug in a few ridiculous numbers in there.

Try a = -1, b = -2
Try a = +7, b = -4
Try a = 1, b = 2

Looks sufficient to me.
PS If you plug those into the original equation they work as well. Try 7^4 - (-4)^4 / 3 = 2401 - 254 / 3 = 2145 / 3 (just add the individual numbers up and you'll see that its divisible).

2) a^2 + b^2 / 3 = Rem2

Same idea. Plug in the same set of numbers or different ones. Doesn't matter really. You'll find this to be true as well.

Let's try with our ridiculous example.

7^2 + (-4)^2 / 3 = 49+16/3 = 65/3 = Rem 2 (63 closest)

Try same idea with bunch of numbers if you have the time. Looks good to me.

Answer = D



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(2) When a^2 + b^2 is divided by 3, the remainder is 2

Was able to know statement (2) is sufficient by trial and error. However, is there a more intuitive way to figure it out without trial and error? Thank you!

Statement 1- (a-b)= 0 mod 3
\(a^4-b^4= a-b)(a+b)(a^2+b^2)\)
hence, \(a^4-b^4\)= 0 mod 3

Sufficient

Statement 2- \(a^2+b^2\)= 2 mod 3

\(a^2\)= 1 mod 3
\(b^2\)= 1 mod 3

\(a^4\)= 1 mod 3
\(b^4\)= 1 mod 3

\(a^4-b^4\)= (1-1) mod 3= 0 mod 3
Sufficient

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