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Hey, Banuel, how 10^a+ b will divisible by 3 when b=6.
eg 10+6 = 16/3 will not be divisible.
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Hey, Banuel, how 10^a+ b will divisible by 3 when b=6.
eg 10+6 = 16/3 will not be divisible.

(1)+(2) Statement (1) tells us that \(b\) is of the form \(2* \text{odd integer}\), while statement (2) tells us that \(b\) is less than 10. As a result, \(b\) could be 2 or 6. If \(b = 2\) (i.e., one less than a multiple of 3), then \(10^a + b\) will be divisible by 3. However, if \(b = 6\) (a multiple of 3), then \(10^a + b\) will not be divisible by 3. Therefore, even together, the statements are insufficient to determine whether \(10^a + b\) is divisible by 3.
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Bunuel @wizako egmat
How is it in any form divisible by 3?
if we just take the first statement, isn't that enough to conclude that the equation is not divisible by 3?

why is it E then?
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Bunuel @wizako

Bunuel
Official Solution:


If \(a\) and \(b\) are positive integers, is \(10^a + b\) divisible by 3?

Note that \(10^a\), where \(a\) is a positive integer, will always be 1 more than a multiple of 3: 10, 100, 1,000, and so on. Therefore, for \(10^a + b\) to be divisible by 3, \(b\) must be one less than a multiple of 3, i.e., 2, 5, 8, 11, 14, and so on. In this case, \(10^a + b = (\text{multiple of 3} + 1) + (\text{multiple of 3} - 1) = 2*\text{multiple of 3}\), making \(10^a + b\) divisible by 3.

(1) \(\frac{b}{2}\) is an odd integer.

The above statement implies that \(b=2* \text{odd integer}\). Therefore, \(b\) could be 2, 6, 10, 14, 18, and so on. Thus, \(b\) could be one less than a multiple of 3, for example 2 or 14. However, it could also be a multiple of 3, such as 6 or 18, as well as two less than a multiple of 3, such as 10 or 22. Not sufficient.

(2) The remainder of \(\frac{b}{10}\) is \(b\).

This statement implies that \(b\) is less than 10: i.e., 1, 2, 3, ..., or 9. However, this information is also insufficient to determine if \(10^a + b\) is divisible by 3.

(1)+(2) Statement (1) tells us that \(b\) is of the form \(2* \text{odd integer}\), while statement (2) tells us that \(b\) is less than 10. As a result, \(b\) could be 2 or 6. If \(b = 2\) (i.e., one less than a multiple of 3), then \(10^a + b\) will be divisible by 3. However, if \(b = 6\) (a multiple of 3), then \(10^a + b\) will not be divisible by 3. Therefore, even together, the statements are insufficient to determine whether \(10^a + b\) is divisible by 3.


Answer: E


egmat
How is it in any form divisible by 3?
if we just take the first statement, isn't that enough to conclude that the equation is not divisible by 3?

why is it E then?

Have you read the solution provided carefully enough?
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