If a and b are positive integers, is 10^a + b divisible by 3

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

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