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Is the product of the three nonzero numbers a, b, and c divisible by 8 06 Sep 2019, 10:55
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65% (01:33) correct 35% (01:28) wrong based on 100 sessions

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Is the product of the three nonzero numbers a, b, and c divisible by 81?

(1) The product of the three numbers a, b, and c is a multiple of 27.
(2) None of the three numbers a, b, and c is divisible by 9.
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Is the product of the three nonzero numbers a, b, and c divisible by 8 09 Sep 2019, 00:56
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(1) The product of the three numbers a, b, and c is a multiple of 27.

If the numbers are 1,2,27 - abc is not divisible by 81
If the numbers are 1,3,27 - abc is divisible by 81

1 is insufficient

(2) None of the three numbers a, b, and c is divisible by 9.

If none of the three numbers is divisible by 9, abc is not divisible by

If abc is divisible by 81, then abc must be divisible by 9

Therefore we can conclude that abc is not divisible by 81

2 is sufficient

Answer is (B)
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Is the product of the three nonzero numbers a, b, and c divisible by 8 03 Mar 2020, 03:42
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Bunuel
Is the product of the three nonzero numbers a, b, and c divisible by 81?

(1) The product of the three numbers a, b, and c is a multiple of 27.
(2) None of the three numbers a, b, and c is divisible by 9.

Solution


Step 1: Analyse Question Stem


• a, b, and c are non-zero numbers.
We need to find, whether the product of a, b, and c is divisible by 81 or not.

Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE


Statement 1: The product of the three numbers a, b, and c is a multiple of 27.
    • \(a*b*c = 27*n\), where n is an integer.
    • If n is a multiple of 3, then \(27*n\) is divisible by 81.
    • If n is not a multiple of 3, then \(27*n\) is not divisible by 81.
Hence, statement 1 is not sufficient, we can eliminate answer options A and D.
Statement 2: None of the three numbers a, b, and c is divisible by 9.
    • Since none of the numbers are divisible by 9, we need to check if one or two or all integers are divisible by 3, will that make the product divisible by 81 or not.
      o Best case, if all these are divisible by 3, then we can say that \(abc = 3*3*3*x\) , where x is an integer, which is not a multiple of 3.
         The maximum power of 3 which the product of number a, b, and c can contain is 3.
    • \(abc = 27 * x\) and we can clearly see that it is not divisible by 81.
Hence, statement 2 is sufficient, the correct answer is Option B.
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Is the product of the three nonzero numbers a, b, and c divisible by 8 02 Jun 2022, 02:08
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1. The product of the three numbers a, b, and c is a multiple of 27.

For a number to be divisible by 81, it should be in the form of 3^4Y(since 81 = 3^4), where Y can be any integer.
Since 27 = 3^3, the multiplication of a, b, and c will be the form of 3^3X, where X can be any product of three numbers. Since X can have any values, we cannot say whether the product of a, b, and c will be a multiple of 81.

2. None of the three numbers a, b, and c is divisible by 9.

For a number to be divisible by 81, it should be in the form of 3^4Y(since 81 = 3^4), where Y can be any integer. Thus, the highest power of 3 in the number should be atleast 4.
Every number in the form of 3^2A is divisible by 9, where A can be any integer.
Since none of the three numbers is divisible by 9, the highest power of 3 in any number can be at most 1, and for the product of three numbers, it will be at most 3.

Thus the number will not be divisible by 81.

Therefore, by statement 2 alone, we will get our answer.
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