to me this question does not make sense... it comes from a review packet for the SAT
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Q: If 299 is divisible by 2B, and B represents a positive digit, what is the value of B?
a.) 2
b.) 3
c.) 4
d.) 6
e.) 9
2B means 20 + B here. 299 = 13*23, so B = 3
How do you know it is ----> 20 "+" B
I thought it was 2*B
please see above
thanks
y+54y-78+42+(89y x 84)=623y
I understand that the question might appear confusing at first. Let me break it down for you and explain how to find the value of B.
The question states that 299 is divisible by 2B, where B represents a positive digit. This means that when you divide 299 by 2B, the result must be a whole number without any remainder.
To solve this, we need to find a positive digit B that satisfies this condition. We can start by checking each option provided and see if it satisfies the divisibility condition.
Let's begin with option A, B = 2. If we substitute B = 2 into the equation, we get 2B = 2 * 2 = 4. Now we divide 299 by 4: 299 ÷ 4 = 74.75, which is not a whole number. Therefore, B cannot be 2.
Moving on to option B, B = 3. If we substitute B = 3 into the equation, we get 2B = 2 * 3 = 6. Now we divide 299 by 6: 299 ÷ 6 = 49.83, which is not a whole number. Thus, B cannot be 3.
Next, let's consider option C, B = 4. If we substitute B = 4 into the equation, we get 2B = 2 * 4 = 8. Now we divide 299 by 8: 299 ÷ 8 = 37.375, which is also not a whole number. So, B cannot be 4.
Proceeding to option D, B = 6. If we substitute B = 6 into the equation, we get 2B = 2 * 6 = 12. Now we divide 299 by 12: 299 ÷ 12 = 24.92, which is still not a whole number. Hence, B cannot be 6.
Finally, let's check option E, B = 9. If we substitute B = 9 into the equation, we get 2B = 2 * 9 = 18. Now we divide 299 by 18: 299 ÷ 18 = 16.61, which is not a whole number. Therefore, B cannot be 9.
After checking all the options, we can conclude that none of the given values of B (2, 3, 4, 6, or 9) make 299 divisible by 2B. In this case, there might be an error in the question or options provided, as no valid solution can be found.
I would suggest contacting the test administrator or your SAT tutor for clarification on this question.