13 (a) 2a = \sqrt{b} where b is a positive integer from 8 to 17 - OCR - GCSE Maths - Question 13 - 2023 - Paper 3

We can rearrange the equation: [ a = \frac{1}{2} \sqrt{b} ] Now we calculate ( b ) from 8 to 17 and check which values of ( 2a ) are integers:

1. For b = 8: [ a = \frac{1}{2} \sqrt{8} = \frac{1}{2} \cdot 2\sqrt{2} \text{ (not an integer)} ]
2. For b = 9: [ a = \frac{1}{2} \sqrt{9} = \frac{1}{2} \cdot 3 = 1.5 \text{ (not an integer)} ]
3. For b = 10: [ a = \frac{1}{2} \sqrt{10} \text{ (not an integer)} ]
4. For b = 11: [ a = \frac{1}{2} \sqrt{11} \text{ (not an integer)} ]
5. For b = 12: [ a = \frac{1}{2} \sqrt{12} = \frac{1}{2} \cdot 2\sqrt{3} \text{ (not an integer)} ]
6. For b = 13: [ a = \frac{1}{2} \sqrt{13} \text{ (not an integer)} ]
7. For b = 14: [ a = \frac{1}{2} \sqrt{14} \text{ (not an integer)} ]
8. For b = 15: [ a = \frac{1}{2} \sqrt{15} \text{ (not an integer)} ]
9. For b = 16: [ a = \frac{1}{2} \sqrt{16} = \frac{1}{2} \cdot 4 = 2 \text{ (integer)} ]
10. For b = 17: [ a = \frac{1}{2} \sqrt{17} \text{ (not an integer)} ]

Thus, the only positive integer value for a satisfying the equation is a = 2.

If a is only described as an integer and not restricted to being a positive integer, then in addition to a = 2, we would also consider a = -2. However, for the equation given, the valid integer value of a still remains as 2 when focusing on positive integers.