IGCSE数学 Number 质因数 HCF 标准形式 上下限
Number 是 0580 的第一块内容，也是最容易被低估的一部分。很多学生以为这一块只是小学算术，翻开真题才发现质因数、标准形式、上限下限、复利与比例分配全挤在这里。Core 与 Extended 的差别在这一模块格外明显：同样是标准形式，Core 考生只在 Paper 3 上做运算，Extended 考生还要在非计算器卷上手算。下面按 11 个具体考点拆开，每个考点给出运算规则、一个算例与阅卷时真正扣分的地方。
Number is the first topic of the Cambridge IGCSE Mathematics 0580 syllabus for 2025 to 2027, and it is the part candidates underestimate most. Prime factors, standard form, bounds, compound interest and ratio all sit inside this one topic. The split between Core and Extended shows up clearly here, because Core candidates only calculate with standard form on Paper 3 while Extended candidates meet it on the non-calculator paper as well. The eleven sections below follow syllabus order, and each one ends where marks are actually lost.
数集与数的分类
0580 考纲的 C1.1 把数集列得很具体：自然数、整数（正整数、零、负整数）、质数、平方数、立方数、公因数、公倍数、有理数与无理数，还包括倒数。很多学生以为带根号的数都是无理数，其实根号九等于三，就是一个整数，只有开不尽方的根号才是无理数。1 既不是质数也不是合数，这一点在数质数个数时最容易丢分。倒数与相反数是两回事：3 的倒数是三分之一，相反数是负三。有理数的判据是能写成两个整数之比，0.4 与 0.333 循环都能化成分数，π 与根号二不能。把这几个定义压缩成一句话记住，选择题几乎不会错。
Syllabus point C1.1 lists the number types precisely: natural numbers, integers, primes, squares, cubes, common factors, common multiples, rational and irrational numbers, and reciprocals. A number sitting under a root sign is not automatically irrational, since the square root of 9 is the integer 3, and only roots that cannot be evaluated exactly are irrational. One is neither prime nor composite, a single fact that decides several multiple-choice marks. A reciprocal and a negative are entirely different ideas, and the practical test for rationality is whether the number can be written as one integer divided by another; 0.4 and 0.333 recurring can, while pi and root two cannot.
质因数分解与 HCF、LCM
把一个数写成质因数的乘积，是 HCF 与 LCM 的地基。72 拆成 2 的三次方乘 3 的二次方，写成指数形式比写一长串乘号省时间。求最大公因数取共有质因数的最低次幂，求最小公倍数取所有出现过的质因数的最高次幂。24 与 36 的 HCF 是 12，LCM 是 72，两个结果相乘刚好等于 24 乘 36。学生最常把这两个取法弄反，考场上可以用一对小数字验算。画维恩图是同一套逻辑，交集取低次幂，并集取高次幂。题目里出现把某数写成质因数乘积这几个字，答案就必须全部是质数。
Writing a number as a product of primes is the base for both the highest common factor and the lowest common multiple. For 72 that product is 2 cubed times 3 squared, and the index form saves real time under exam pressure. The HCF takes the lowest power of each shared prime, the LCM takes the highest power of every prime that appears, and for 24 and 36 the two answers are 12 and 72. Those answers multiply to the product of the original pair, which is a quick self-check worth remembering. Swapping the two rules is the commonest slip of all, and a Venn diagram with shared primes in the middle makes the logic visible.
分数、小数、百分数的互换
同一个数换三种写法，是 Number 里最高频的操作。八分之三等于 0.375，等于 37.5%，这三个结果要能瞬间互换。常见的错误是换算时忘了百分号代表分母一百，于是 37.5% 被写成 0.375% 或者 3.75。循环小数化分数有固定套路：把 0.444 循环设成 x，十倍减原式得 9x 等于 4，再解出九分之四。比较大小时统一化成小数最稳，负数那一侧要记得不等号方向反转。题目要求从小到大排列就先分清正负再动手。分数运算写到最后一步再化简，中间步骤化简容易把数字搞乱。
Three forms of the same number, and the fastest one depends on the question in front of you. Three eighths is 0.375 and also 37.5 percent, and all three should arrive without thinking. A recurring decimal converts by a fixed routine: call it x, multiply by ten, subtract the original, then solve the small equation. The percentage sign means a denominator of one hundred, which is exactly why 37.5 percent becomes 0.375 and never 3.75. When a list must be ordered, decimals give the cleanest comparison, and negative values flip the inequality.
指数法则与根式
C1.7 的指数法则要能顺着用也要能倒着用。同底相乘指数相加，同底相除指数相减，幂的幂指数相乘，任何非零数的零次幂等于一。负指数写成倒数，2025 年启用的新考纲还把二分之一与负二分之一这两档指数写进要求，实际考的就是开平方与倒数的平方根。Extended 的 C1.18 新增根式：根号七十二化简成六倍根号二，分母带根号要上下同乘把它消掉。这一步最容易在没化到最简就交卷，阅卷时答案留着根号七十二不给满分。根号里能提出完全平方数就一定先提出来。
The index laws in C1.7 have to work in both directions, forwards when simplifying an expression and backwards when solving for a power. Same-base powers add their indices under multiplication and subtract them under division, a power of a power multiplies them, and any non-zero number to the power zero is one. The syllabus in use from 2025 also requires indices of one half and minus one half, which is really square roots and their reciprocals in disguise. Extended candidates meet surds in C1.18, simplify the root of 72 to six root two, and rationalise any denominator still carrying a root.
标准形式与三件套运算
标准形式只认一种写法：一个介于 1 与 10 之间的数乘 10 的整数次幂，前面的数不能写成 12 或者 0.4。0.00042 写成 4.2 乘 10 的负四次方，小数点移动几位，指数就是几，往右移指数取负。乘法把前面的数相乘、指数相加，除法把前面的数相除、指数相减。考纲写明 Core 考生只在 Paper 3 上做标准形式的运算，Paper 1 与 Paper 2 不考这一类计算，这个差别决定了你该练哪一份卷。很多学生以为写成 42 乘 10 的负五次方也算对，其实前面的数超出范围就不得分。
Standard form allows exactly one shape, a number between 1 and 10 multiplied by an integer power of ten, so 12 times ten to the fifth is not standard form. Moving the decimal point three places to the right turns 0.00042 into 4.2 times ten to the minus four, and the exponent simply counts those places. Multiplication multiplies the leading numbers then adds the indices, while division divides them then subtracts. The syllabus states that Core candidates only calculate with standard form on Paper 3, which tells you exactly which paper to rehearse this on.
有效数字、小数位与估算
估算的写法是先修约再算，不是先算再修约。0.765 乘 41.3 除以 9.79 这样一道题，三个数各取一位有效数字，变成 0.8 乘 40 除以 10，答案是 3.2，与精确值有差距但数量级对。有效数字从第一个非零数字开始数，0.0305 只有三位。考纲写明 Paper 3 与 Paper 4 的非精确答案默认给三位有效数字，题目另说才改。答案恰好是四五位有效数字时不要自行修约，考纲专门写了这一条。学生最常把 5764 修约到最近的千位答成 5000，按一位有效数字应该是 6000。小数位与有效数字两个要求同时出现时，以题目点明的那个为准。
Estimation means rounding first and calculating second, never the reverse. Take 0.765 times 41.3 divided by 9.79: at one significant figure the three numbers become 0.8, 40 and 10, and the estimate comes out at 3.2. Significant figures are counted from the first non-zero digit, so 0.0305 carries only three of them. The syllabus sets three significant figures as the default for non-exact answers and adds a warning against rounding anything already exact to four or five. Rounding 5764 to the nearest thousand gives 6000, not 5000, because the leading digit alone decides it.
上限与下限
上限下限考的是修约的反向推理。一条边长 24 厘米、精确到厘米，真实长度落在 23.5 与 24.5 之间，上限写作 24.5，下限写作 23.5。考纲的 C1.10 只要求 Core 给出数据本身的上下限，E1.10 才要求 Extended 用上下限去算结果，比如用 23.5 乘 24.5 求矩形面积的上限。求速度的下限要用距离的下限除以时间的上限，这一步最容易把分子分母选反。面积与体积的上限一律拿最大值相乘，减法的下限是大数减大数。写答案时把不等式一并写上，只写一个数通常只得一半分。
Bounds reverse the rounding that produced the measurement. A length given as 24 centimetres to the nearest centimetre really lies between 23.5 and 24.5, so those two numbers are the lower and upper bounds. Core candidates only give bounds for the data itself under C1.10, while Extended candidates under E1.10 must push bounds through a calculation, multiplying 23.5 by 24.5 for the upper bound of a rectangle area. A lower bound for speed comes from dividing the lowest distance by the highest time, and choosing those two the wrong way round is where most marks disappear.
比与比例、正反比例
比例分配按份数走，不按平均数走。240 按 3 比 5 分，先算总份数 8，一份 30，两部分是 90 与 150。正比例写成 y 等于 k 乘 x，反比例写成 y 等于 k 除以 x，判断依据是两者同增还是此增彼减。常见的错误是把反比例题直接按正比例套，答案会差出一大截。题目给三个量求第四个量时，先写出单位比再交叉相乘。单位不同要先统一，小时与分钟混着算最容易出错。比化简到最后要互质，三个数的比也一样。
Ratio questions divide by shares, never by an average. Splitting 240 in the ratio 3 to 5 means eight shares of 30, giving 90 and 150. Direct proportion takes the form y equals k times x, inverse proportion takes y equals k divided by x, and the test is whether both quantities rise together or one rises as the other falls. Reading an inverse case as direct is the error that costs most, because the answer still looks plausible. Convert units before forming the ratio, and reduce every ratio to coprime numbers at the end.
百分比变化、复利与折旧
百分比变化只做一件事：找到乘数。涨 15% 就乘 1.15，降 15% 就乘 0.85，连续两年各涨 15% 是乘 1.15 的平方，不是加 30%。单利每年只按本金计息，复利每年按新余额计息，5000 元存四年、年利率 3%，复利终值是 5000 乘 1.03 的四次方。折旧按乘数递减，每年折旧 20% 就是每年乘 0.8。很多学生以为先涨后降能回到原价，涨 20% 再降 20% 其实只剩原价的 96%。求原价要用除法，用减法算回去一定错。
Every percentage change reduces to a single multiplier. A rise of 15 percent multiplies by 1.15, a fall of 15 percent multiplies by 0.85, and two successive rises of 15 percent multiply by 1.15 squared rather than adding 30 percent. Simple interest charges the original amount every year, compound interest charges the new balance, so 5000 at 3 percent for four years ends at 5000 times 1.03 to the fourth power. Depreciation works the same way with a multiplier below one, and a 20 percent rise followed by a 20 percent fall leaves only 96 percent of the starting value.
非计算器卷的手算策略
Paper 1 是 1 小时 30 分钟 80 分且不许用计算器，考纲写明这份卷不考 C1.14 使用计算器，考的全是基本功。八分之一、三分之一、五分之一这几个分数的三套写法要背到条件反射，乘法先约分再乘能省一半步骤。平方数记到 15，立方数记到 5，开方时先看末位数字。时间上 80 分给 90 分钟，平均一分多钟一题，卡住就先跳过再回头。阅卷时看到中间步骤保留分数、只在最后一步化小数，说明这个学生清楚哪里该精确。草稿纸分区写，改错时不用重抄整题。
Paper 1 lasts 1 hour 30 minutes for 80 marks with no calculator allowed, and the syllabus excludes C1.14 Using a calculator from that paper entirely. Fluent recall of eighths, thirds and fifths in all three forms is worth more here than any technique, and cancelling before multiplying halves the written work. Knowing squares to 15 and cubes to 5 turns most root questions into a one-line check. With roughly a minute per mark, leaving a stubborn question and returning later protects the rest of the paper, and keeping working in fractions until the final line keeps the answer exact.
六周复习排期
六周够把 Number 过两遍。第 1 周把数集与质因数分解过一遍，第 2 周专攻分数、小数、百分数的互换与比较大小，第 3 周做指数法则与根式化简，第 4 周集中练标准形式与上限下限，第 5 周处理比例、速率与利息，第 6 周限时做两套真题并把错题回填到前面几周。每周留一次 25 分钟小测，只测当周内容，错题第二天重做。Extended 考生在根式与结果上下限上多留一周余量，Core 考生把标准形式的计算压在 Paper 3 的题型上。每周结束时把错题按考点归类，比重做整套卷更省时间。
Six weeks is enough to cover Number twice. Weeks one and two take number types, prime factorisation and the three-way conversion between fractions, decimals and percentages. Week three covers index laws and surds, week four concentrates on standard form and bounds, week five handles ratio, rates and interest, and week six runs two full past papers under timed conditions with every mistake traced back to its week. A 25-minute test at the end of each week keeps the material active, and sorting mistakes by syllabus point beats simply redoing whole papers.
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