Advanced Math 6

Chapter at a Glance

• Chapter: 1
• Strand: Pure — Numbers
• Difficulty: Level 3 of 5
• Builds on: Multiples and factors (prime factorization), Writing and evaluating expressions

Try This First

Try this short activity before reading the lesson.

> Take $12$ coins. Put them in equal rows in every way you can: $1$ row of $12$, $2$ rows of $6$, $3$ rows of $4$. Write the row lengths down. Now do the same with $18$ coins. Which row lengths are in both lists? What is the biggest one?

The Big Idea

Every whole number bigger than $1$ is made from primes. $84 = 2^2 \cdot 3 \cdot 7$, and there is no other way to make it. This one fact does all the work in this unit.

Two numbers share some of their primes. The greatest common factor (GCF) is made from the shared primes only. Use the smallest power of each one. The least common multiple (LCM) must fit both numbers inside it. So it uses every prime that appears, with the biggest power of each one.

Lists of factors and multiples still work for small numbers. But a list is easy to get wrong. It is easy to miss one, and it does not work for a number like $840$.

Order of operations comes back in this unit too, with parentheses inside brackets. Every unit after this one uses it.

Words to Know

• prime: A whole number bigger than $1$ with only two factors, $1$ and itself. $2$ is the only even prime.
• index notation: A short way to write repeated primes: $2 \cdot 2 \cdot 2 = 2^3$.
• GCF: Greatest common factor. The biggest number that divides both. Made from the shared primes.
• LCM: Least common multiple. The smallest number both divide into. Made from every prime that appears.
• relatively prime: Two numbers with GCF $1$, like $8$ and $15$. Their LCM is just $8 \times 15 = 120$.

Formulas and Theorems

• GCF from primes: $\text{GCF} = \prod p^{\min}$
• Note: Shared primes, lowest power each.
• LCM from primes: $\text{LCM} = \prod p^{\max}$
• Note: Every prime that appears, highest power each.
• Product identity: $\text{GCF}(a,b) \times \text{LCM}(a,b) = ab$

Outcomes and Milestones

• Students can: place value and rounding; multi-digit × and ÷; primes, factor trees, ladder method; GCF and LCM; nested PEMDAS.
• Watch for: GCF by listing and missing 12; LCM of 8 and 12 as 96.
• Ready when: GCF(84, 120) and LCM(84, 120) from $84=2^2\cdot 3\cdot 7$, $120=2^3\cdot 3\cdot 5$.

Worked Examples and Your Turn

Study each example, then try the paired problem.

Example 1: Factor with a ladder

Problem: Write $360$ as a product of primes, in index notation.

Solution:

• Divide by the smallest prime that goes in. Keep going while it still goes in: $360 \div 2 = 180$, $180 \div 2 = 90$, $90 \div 2 = 45$. That is three twos.
• $45$ is odd, so try $3$: $45 \div 3 = 15$, $15 \div 3 = 5$. That is two threes.
• $5$ is prime, so stop. Read the primes down the side of the ladder: $2, 2, 2, 3, 3, 5$.

Answer: $360 = 2^3 \cdot 3^2 \cdot 5$

> Notice: The ladder never misses a prime. A factor tree gives the same primes, whichever way you split.

> Verification & Check: Multiply back: $8 \times 9 = 72$, and $72 \times 5 = 360$.

Your Turn:

• Prompt: Write $252$ as a product of primes, in index notation.
• Answer: $252 \div 2 = 126$, $\div 2 = 63$, $\div 3 = 21$, $\div 3 = 7$. So $252 = 2^2 \cdot 3^2 \cdot 7$.

Example 2: GCF and LCM from the primes

Problem: You know $84 = 2^2 \cdot 3 \cdot 7$ and $120 = 2^3 \cdot 3 \cdot 5$. Find $\text{GCF}(84, 120)$ and $\text{LCM}(84, 120)$ without writing any lists.

Solution:

• Line up the primes. Both numbers have $2$ and $3$. Only $84$ has $7$. Only $120$ has $5$.
• For the GCF, take each shared prime with its smallest power: $2^2 \cdot 3 = 12$. Why: a common factor cannot have more twos than the number with fewer twos.
• For the LCM, take every prime with its biggest power: $2^3 \cdot 3 \cdot 5 \cdot 7 = 840$. Why: a common multiple must hold three twos for $120$, a $7$ for $84$ and a $5$ for $120$.

Answer: $\text{GCF} = 12$, $\text{LCM} = 840$

> Notice: Both answers came from the same two lines of primes. Nothing was listed, so nothing could be missed.

> Verification & Check: $12 \times 840 = 10080$ and $84 \times 120 = 10080$. The GCF times the LCM always equals the two numbers multiplied together.

Your Turn:

• Prompt: You know $60 = 2^2 \cdot 3 \cdot 5$ and $72 = 2^3 \cdot 3^2$. Find the GCF and the LCM.
• Answer: GCF $= 2^2 \cdot 3 = 12$. LCM $= 2^3 \cdot 3^2 \cdot 5 = 360$. Check: $12 \times 360 = 4320 = 60 \times 72$.

Example 3: Cycles that line up

Problem: One lighthouse flashes every $15$ seconds. Another flashes every $20$ seconds. They flash together at exactly 9:00:00. When do they next flash together?

Solution:

• "Next time together" means the smallest number of seconds that is a whole number of both cycles. That is the least common multiple.
• Write the primes: $15 = 3 \cdot 5$ and $20 = 2^2 \cdot 5$. Take every prime with its biggest power: $2^2 \cdot 3 \cdot 5 = 60$.
• Add $60$ seconds to 9:00:00.

Answer: 9:01:00

> Verification & Check: $60 \div 15 = 4$ and $60 \div 20 = 3$. Both are whole numbers, so both lights flash at $60$ seconds. No smaller number works for both.

Your Turn:

• Prompt: One light blinks every $6$ seconds. Another blinks every $8$ seconds. They blink together now. In how many seconds do they next blink together?
• Answer: $6 = 2 \cdot 3$ and $8 = 2^3$. LCM $= 2^3 \cdot 3 = 24$ seconds.

Example 4: Lists have holes

Problem: Find $\text{GCF}(36, 60)$. Then find $\text{LCM}(8, 12)$. Both are traps for a list.

Solution:

• Common mistake: list the factors and pick the biggest one in both lists. Factors of $36$: $1, 2, 3, 4, 6, 9, 18, 36$. Factors of $60$: $1, 2, 3, 4, 5, 6, 10, 15, 20, 30, 60$. Biggest in both: $6$. But both lists skipped $12$. Lists have holes.
• Do this instead. Use primes: $36 = 2^2 \cdot 3^2$ and $60 = 2^2 \cdot 3 \cdot 5$. Shared primes, smallest powers: $2^2 \cdot 3 = 12$. A ladder cannot skip a factor.
• Now $\text{LCM}(8, 12)$. Common mistake: multiply them and get $96$. It is a common multiple, but not the smallest one. $24$ is smaller, and both $8$ and $12$ go into it.
• Do this instead. $8 = 2^3$ and $12 = 2^2 \cdot 3$. Biggest powers: $2^3 \cdot 3 = 24$. Multiplying counted the shared $2^2$ twice. That is why $96 \div 24 = 4$.

Answer: $\text{GCF}(36, 60) = 12$, not $6$. $\text{LCM}(8, 12) = 24$, not $96$.

> Notice: Multiplying two numbers always gives a common multiple. It is the least one only when the numbers share no prime.

Your Turn:

• Prompt: Find $\text{GCF}(24, 40)$ and $\text{LCM}(6, 9)$ using primes.
• Answer: $24 = 2^3 \cdot 3$ and $40 = 2^3 \cdot 5$, so GCF $= 8$. $6 = 2 \cdot 3$ and $9 = 3^2$, so LCM $= 2 \cdot 3^2 = 18$, not $54$.

Example 5: Parentheses inside brackets

Problem: Evaluate $2[3 + 4(10 - 7)^2] - 5$.

Solution:

• Start with the innermost parentheses: $10 - 7 = 3$.
• Do the power next: $3^2 = 9$. Why: powers come before multiplying.
• Now finish the square bracket. Multiply first: $4 \times 9 = 36$. Then add: $3 + 36 = 39$.
• The square bracket is done. Multiply by $2$: $78$. Then subtract: $78 - 5 = 73$.

Answer: $73$

> Verification & Check: Estimate: the square bracket is about $40$. Double it, about $80$. Take $5$ away. So $73$ makes sense. Working left to right gives a number in the hundreds, which is the wrong choice on a test.

Your Turn:

• Prompt: Evaluate $3[10 + 2(8 - 5)^2] - 4$.
• Answer: $8 - 5 = 3$, $3^2 = 9$, $2 \times 9 = 18$, $10 + 18 = 28$, $3 \times 28 = 84$, $84 - 4 = 80$.

Example 6: Equal pieces, as long as possible

Problem: A $90$ cm rope and a $126$ cm rope are cut into pieces. All the pieces are the same length, as long as possible, with nothing left over. How long is each piece? How many pieces are there altogether?

Solution:

• "Same length, as long as possible, nothing left over" means the biggest common factor.
• Write the primes: $90 = 2 \cdot 3^2 \cdot 5$ and $126 = 2 \cdot 3^2 \cdot 7$. Shared primes, smallest powers: $2 \cdot 3^2 = 18$.
• Count the pieces: $90 \div 18 = 5$ and $126 \div 18 = 7$. That is $12$ pieces.

Answer: $18$ cm; $12$ pieces

> Notice: The story tells you which one to use. "Biggest piece" or "most groups" means GCF. "Next time together" or "smallest amount that both" means LCM.

Your Turn:

• Prompt: Two ribbons are $48$ cm and $60$ cm long. Cut both into equal pieces, as long as possible. How long is each piece, and how many pieces are there?
• Answer: $\text{GCF}(48, 60) = 12$ cm. $48 \div 12 = 4$ and $60 \div 12 = 5$, so $9$ pieces.

Example 7: The remainder that stays the same

Problem: A number leaves a remainder of $2$ when you divide it by $3$, by $4$, and by $5$. What is the smallest such number with two digits?

The chapter continues in the book.