Technique explanation
Teaching Explanation
The squares of all numbers ending in 5 follow a rigid, predictable structure. Every such square ends in the digits 25. To find the beginning of the square, take the digit(s) before the 5 (let's call this $a$) and multiply it by the next consecutive integer ($a + 1$). For example, for $35^2$, the prefix is $3 \times 4 = 12$, and the suffix is 25, giving 1225. This pattern is a specific case of the algebraic identity $(10a+5)^2 = 100a(a+1) + 25$.
Method Conditions
Squaring a number that ends in the digit 5.
Standard Fallback
Standard long multiplication ($n \times n$).
Misconception & Correction
Misconception: Squaring the first digit instead of multiplying by the next integer (e.g., $35^2 = 925$).
Correction: The rule requires $a \times (a+1)$, not $a \times a$. For 35, we use $3 \times 4$, not $3 \times 3$.
Module
Worked examples
Worked Examples
- Calculate $35^2$. Step 1: Identify the prefix $a = 3$. Step 2: Multiply $a$ by $a+1$: $3 \times 4 = 12$. Step 3: Append 25: 1225. Verification: $35 \times 35 = 1225$ (using standard multiplication).
- Calculate $75^2$. Step 1: Identify the prefix $a = 7$. Step 2: Multiply $a$ by $a+1$: $7 \times 8 = 56$. Step 3: Append 25: 5625. Verification: $75 \times 75 = 5625$.
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