Arithmetic: divisibility, remainders and multiplicative change
| English | Español |
|---|---|
| least common multiple/liːst ˈkɒmən ˈmʌltɪpl/ | least common multiple |
| remainder/rɪˈmeɪndə/ | remainder |
A decision before an answer
- A 20% decrease followed by a 20% increase does not restore the starting value: the second percentage uses a smaller base.
- Your goal: Use prime factorisation for divisibility, greatest common factors and least common multiples.
Read the relationship
- An integer divisible by d can be written dk; a remainder r gives dk+r with 0≤r<d for a positive divisor. Prime factorisation identifies divisibility and common multiples. The greatest common factor takes shared prime powers at their smaller exponents; the least common multiple takes every needed prime power at its larger exponent.
- Interpret remainders with the original divisor.
LCM of 12 and 18 is:
12=2²·3 and 18=2·3²; maximum powers give 36.
Use the defining rule
- Keep odd/even and positive/negative properties separate. A product of two odd integers is odd; a product with an even integer is even. A statement about integers need not hold for real numbers. For quantitative comparison, exploit a remainder form to expose possible cases rather than assume the smallest positive example is the only value.
- Combine ratios and percentage changes with the correct reference total.
A price 200 rises 10% then falls 10%. Final price:
200·1.1·0.9=198.
Check the conditions
- A ratio a:b allocates a whole into a+b equal parts when the groups exhaust it. A percentage change multiplies the original value by 1+r or 1−r with r expressed as a decimal. Successive changes multiply factors. A part-of-a-part calculation changes the reference group at each stage; label the denominator before multiplying.
- Combine ratios and percentage changes with the correct reference total.
Known: 18=2×3² and 24=2³×3. Shared smaller powers give GCF=2×3=6; needed larger powers give LCM=2³×3²=72. If n=5k+2 then 2n=5(2k)+4, so the remainder on division by 5 is 4. For a price 100 decreased then increased by 20%, P_final=P_initial(1−0.20)(1+0.20)=100×0.8×1.2=96. A 2:3 division of 250 gives 100 and 150 because each of five parts is 50.
If n has remainder 3 on division by 7, n+2 has remainder ____.
n=7k+3, so n+2=7k+5.
Apply the task format
- Estimate magnitude and check integer restrictions after solving. Fractions, roots and negative exponents follow their ordinary domain restrictions; even roots represent the nonnegative principal value. Do not cancel a term in a sum as if it were a factor. Exact arithmetic can be faster and safer than decimal calculator entries.
- Combine ratios and percentage changes with the correct reference total.
Equal percentage increases and decreases use different bases. A remainder must be reported relative to the named divisor.
Which answer fits this case?
Use prime factorisation for divisibility, greatest common factors and least common multiples
Every real number has integer parity.
Odd/even applies to integers, not arbitrary real numbers.
Keep the distinctions
- least common multiple 最小公倍数 — The smallest positive integer divisible by each given positive integer.
- remainder 余数 — The nonnegative amount left after division by a positive integer, smaller than that divisor.
- Use prime factorisation for divisibility, greatest common factors and least common multiples.
- Interpret remainders with the original divisor.
- Combine ratios and percentage changes with the correct reference total.
Match each term with its precise meaning in this lesson.
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
The order follows the stated process; check each stage before the next.