| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2007 |
| Session | January |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof |
| Type | Proof by exhaustion with cases |
| Difficulty | Moderate -0.8 This is a straightforward proof by exhaustion requiring students to list all 1- and 2-digit perfect squares (1, 4, 9, 16, 25, 36, 49, 64, 81) and observe their final digits. The method is explicitly stated, no algebraic manipulation is needed, and the generalization (that no perfect square ends in 2, 3, 7, or 8) follows immediately from the pattern observed. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps |
Use the method of exhaustion to prove the following result.
No 1- or 2-digit perfect square ends in 2, 3, 7 or 8.
State a generalisation of this result. [3]
M1: Check all perfect squares from $1^2$ to $9^2$ for single digits and $10^2$ to $99^2$ for two digits
M1: Verify that the final digit of each perfect square is only 0, 1, 4, 5, 6, or 9
A1: Generalisation: No perfect square ends in 2, 3, 7 or 8 (for any positive integer)
---4 Use the method of exhaustion to prove the following result.\\
No 1 - or 2 -digit perfect square ends in $2,3,7$ or 8\\
State a generalisation of this result.
\hfill \mbox{\textit{OCR MEI C3 2007 Q4 [3]}}