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Archit3110


\(A^3 + B^3\) is divisible by 10
it means both a & b have unit digits as 0 ; i.e both are even integers


Hey Archit3110,

Considering the highlighted sentence above, can we say it is always true?
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Archit3110


\(A^3 + B^3\) is divisible by 10
it means both a & b have unit digits as 0 ; i.e both are even integers


Hey Archit3110,

Considering the highlighted sentence above, can we say it is always true?

EgmatQuantExpert
well the highlighted part wont be true always ; given that A^3 + B^3 divisible by 10 ; so either of A or B can be 0 and other can be a factor of 10
but since its mentioned in the question that A & B are natural numbers , so either of them cannot be '0' in this case ,and a natural no ending with 0 would be an even integer only in this case..
had it been given in question that a & b are whole number then #2 would have been insufficient....
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Solution



Given:

    • A and B are distinct natural numbers.

To Find:

    • Is A+B odd?

Approach and Working:

    • A+B is odd when one out of A or B is even and another is odd as Even +Odd = Odd

So, we have to find whether the even-odd nature of A and B is different or not.

Analyse Statement 1: Unit digit of A x B is 6.

Unit digit of A x B can be 6 when units digit of (A, B) = (1, 6) or (2,3) or (4,4) or (2,8), or (4,9) or (7,8), or (6,6) etc.
However, in some cases the even-odd nature of (A, B) is different and in some cases the even-odd nature of (A, B) is same.

Hence, we cannot find the answer from statement 1.

Analyse Statement 2: \(A^3 + B^3\) is divisible by 10.

For \(A^3 + B^3\) to be divisible by 10, the units digit of \(A^3 + B^3\) must be 0.

    • The units digit of \(A^3 + B^3\) can be 0 when units digit (A, B) is (1, 9), or (2, 8), or (3, 7), or (4, 6), or (5, 5), or (6, 4), or (7, 3), or (8, 2), or (9, 1), or (0,0).

In all the given cases, the nature of (A, B) is same.
Hence, \(A^3 + B^3\) is always an even number.

Therefore, statement 2 alone is sufficient to find the answer.

Correct answer: B
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Archit3110


\(A^3 + B^3\) is divisible by 10
it means both a & b have unit digits as 0 ; i.e both are even integers


Hey Archit3110,

Considering the highlighted sentence above, can we say it is always true?

EgmatQuantExpert
well the highlighted part wont be true always ; given that A^3 + B^3 divisible by 10 ; so either of A or B can be 0 and other can be a factor of 10
but since its mentioned in the question that A & B are natural numbers , so either of them cannot be '0' in this case ,and a natural no ending with 0 would be an even integer only in this case..
had it been given in question that a & b are whole number then #2 would have been insufficient....


Well this is a limited scenario you are talking of. When it is stated that A and B are distinct natural number and not integers and we know that A^3+B^3 is divisible by 10, we can never conclude that either of A or B can be 0.

The best solution is trying out the cubes of single digit natural numbers.
For example,
1^3+9^3 = 0 (mod 10)
2^3+8^3 = 0 (mod 10)
3^3+7^3 = 0 (mod 10)
4^3+6^3 = 0 (mod 10)
5^3+5^3 = 0 (mod 10)

As we can see in each case the summation of the 2 natural numbers will be only even and never odd.
Hence, statement 2 is sufficient but not statement 1.

Hence, answer option B should be correct.
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What will be the case, if A = 7^(1/3) and B = 3^(1/3)?. Because question is asking for natural number not integers.­­
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Krsumit
What will be the case, if A = 7^(1/3) and B = 3^(1/3)?. Because question is asking for natural number not integers.­­
­
Natural numbers mean positive integers, or by some definitions, non-negative integers. That's why the GMAT uses "positive integers" instead of "natural numbers."
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