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Question of the Week  38 (A and B are distinct single digit integers)
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01 Mar 2019, 02:22
Question Stats:
34% (02:17) correct 66% (02:14) wrong based on 47 sessions
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Re: Question of the Week  38 (A and B are distinct single digit integers)
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01 Mar 2019, 03:22
EgmatQuantExpert wrote: eGMAT Question of the Week #38A and B are distinct natural numbers. Is A + B odd? Statement 1: Unit digit of A x B is 6. Statement 2: \(A^3 + B^3\) is divisible by 10. Statement 1: A*B = x6......Unit digit is 6. A = 23 B = 12 Unit digit is 23*12 =xx6 23 + 12 = 35=odd. Again, A=6 B=16 6*16 =x6 A + B = 6 + 16 = 22=even NOT sufficient. Statement 2: A^3 + B^3 is divisible by 10. It means unit digit of A^3 + B^3 is 0. 0 represent even integers. A + B must be even. statement 2 is sufficient.



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Re: Question of the Week  38 (A and B are distinct single digit integers)
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01 Mar 2019, 04:22
sum or subtraction of an even and odd integer will only result in odd integer.... now #1 unit digit of a*b= 6 a =1, b=6 ; sum odd a=6,b=6; sum even in sufficient #2[/b] \(A^3 + B^3\) is divisible by 10 it means both a & b have unit digits as 0 ; i.e both are even integers so sum of a+b ; will not be odd integer IMO B sufficient EgmatQuantExpert wrote: eGMAT Question of the Week #38A and B are distinct natural numbers. Is A + B odd? Statement 1: Unit digit of A x B is 6. Statement 2: \(A^3 + B^3\) is divisible by 10.
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Re: Question of the Week  38 (A and B are distinct single digit integers)
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01 Mar 2019, 05:17



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Question of the Week  38 (A and B are distinct single digit integers)
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01 Mar 2019, 05:25
EgmatQuantExpert wrote: Archit3110 wrote: \(A^3 + B^3\) 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? EgmatQuantExpertwell 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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Question of the Week  38 (A and B are distinct single digit integers)
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06 Mar 2019, 01:42
Solution Given:• A and B are distinct natural numbers. To Find: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 evenodd 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 evenodd nature of (A, B) is different and in some cases the evenodd 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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Question of the Week  38 (A and B are distinct single digit integers)
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06 Mar 2019, 01:42






