Exact arithmetic and estimation · Higher
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| prime factor/praɪm ˈfæktə/ | 质因数 | zhì yīn shù |
Can we pack without leftovers?
- A supplier packs 72 pencils and 90 pens into identical gift bags. How can we avoid leftovers?
- This lesson studies prime factor 质因数: A prime number that divides the integer exactly.
我们能做到打包无剩余吗？
- 供应商将72支铅笔和90支钢笔装入相同的礼品袋中。如何避免剩余？
- 本课学习质因数：能整除该整数的素数。
Choose the mathematical structure
- Prime factors reveal shared structure. Use the smallest common prime powers for the HCF and the largest for the LCM. Estimate before calculating; use brackets to preserve the order of operations.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
选择数学结构
- 质因数揭示了共同的结构。取最小公倍素幂计算最大公约数（HCF），取最大公倍素幂计算最小公倍数（LCM）。计算前先估算；使用括号保持运算顺序。
- 计算前请先明确允许的输入项和单位。方程应表达关系本身，而不仅仅是记录计算器按键过程。
Which description correctly defines prime factor?
A prime number that divides the integer exactly.
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
72=2^3×3^2 and 90=2×3^2×5. Their HCF is 2×9=18. Make 18 bags with 4 pencils and 5 pens each. Their LCM is 2^3×3^2×5=360.
通过验证案例进行推导
- 将结果与初始量进行比对。代入原始关系式，或在适当情况下对比图表与数值答案。
72=2^3×3^2 且 90=2×3^2×5。它们的最大公约数是2×9=18。制作18个袋子，每个袋子装4支铅笔和5支钢笔。它们的最小公倍数是2^3×3^2×5=360。
Exact arithmetic and estimation
Prime factors reveal shared structure
Combine the common powers 2 and 9 to justify 18 identical bags.
Find the HCF of 48 and 60.
48=2⁴×3 and 60=2²×3×5. The shared smallest powers give HCF=2²×3=12.
Test a tempting shortcut
- The HCF divides both numbers; the LCM is a multiple of both. They answer different questions. A decimal estimate is not an exact fraction.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
The HCF of two positive integers is always larger than either integer. This claim is false. Explain which definition or assumption it violates.
检验一个诱人的捷径
- 最大公约数能整除两个数；最小公倍数是两个数的倍数。它们解答不同的问题。小数估算不等于精确分数。
- 当捷径失效时，找出其违背的假设。保留精确值直到题目要求的最终舍入步骤。
两个正整数的最大公约数总是大于其中任何一个整数。此说法是错误的。请解释它违反了哪个定义或假设。
Find the LCM of 8 and 12.
8=2³ and 12=2²×3. The largest prime powers give LCM=2³×3=24.
The HCF of two positive integers is always larger than either integer.
The HCF divides both numbers; the LCM is a multiple of both. They answer different questions. A decimal estimate is not an exact fraction.
Interpret a new situation
- For a non-calculator paper, keep fractions exact and show cancellation. For a calculator paper, enter the full expression and compare with your estimate.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
解读新情境
- 对于无计算器试卷，保持分数精确并展示约分过程。对于有计算器试卷，输入完整表达式并与估算结果进行比较。
- 完整解答需给出数学结果并解释其含义。检查其在所述背景下是否可行。
Work out 3/4 + 5/6 as a decimal.
Use denominator 12: 3/4+5/6=9/12+10/12=19/12.
Match each part of a complete solution to its purpose.
An assumption justifies the model; a check tests the result; interpretation connects it to the question.
Use this in your course
- 4MA1 · Higher · 1. Match the target tier and specification before assigning extensions.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
A prime number that divides the integer exactly. Choose the relationship, show the method, check its assumptions and interpret the result.
将其应用于你的课程
- 4MA1 · 高级 · 1. 在分配拓展内容前，请匹配目标层级和教学大纲要求。
- 在给出最终答案前先展示解题方法，并遵循试卷的计算器及公式使用规则。通过定位第一个无效步骤来复盘错误答案。
能恰好整除该整数的质数。选择正确的关系式，展示解题方法，检验其适用条件并解读结果。