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GCF and LCM: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 The shared twos

Two numbers have prime-power forms $2^4\times13$ and $2^5\times7$ . What is the greatest power of $2$ that divides both numbers?
1. Hint 1
  A common factor must fit inside both prime lists.
2. Hint 2
  Compare how many copies of $2$ each number supplies.
Answer

$2^4$ , or $16$ .

Full solution

The first number has four copies of $2$ , and the second has five.
Both supply four copies, but the first does not supply a fifth.
$2^4=16$
Therefore $16$ is the greatest power of $2$ that divides both.

Answer

$2^4$ , or $16$ .

Key idea

A shared prime can appear in a common factor no more often than in the number with fewer copies.
Problem 1: Mark complete
Problem 2 The required fives

Two numbers are $3\times5^3$ and $5\times31$ . What is the highest power of $5$ that divides every positive common multiple of the two numbers?
1. Hint 1
  A common multiple must contain enough prime copies for each number.
2. Hint 2
  Compare the demand for copies of $5$ in the two given numbers.
Answer

$5^3$ , or $125$ .

Full solution

The first number needs three copies of $5$ , and the second needs one, which three copies also cover.
A multiple of the first must contain all three copies, so every common multiple holds at least three copies of $5$ .
$5^3=125$
So $125$ divides every common multiple.
The LCM, $3\times5^3\times31$ , is itself a common multiple with only three copies of $5$ , so no higher power of $5$ divides every common multiple.

Answer

$5^3$ , or $125$ .

Key idea

A common multiple must hold each prime at least as often as the number with more copies of it.
Problem 2: Mark complete
Problem 3 The two prime lists

Two numbers have prime-power forms $2^3\times3$ and $3^2\times13$ . What is their least common multiple?
1. Hint 1
  A common multiple must contain every prime copy that each number contains.
2. Hint 2
  For each prime, take the larger count found in either number.
Answer

$2^3\times3^2\times13$ , or $936$ .

Full solution

Go prime by prime.
The prime $2$ appears three times in the first number and not at all in the second, so take three copies.
The prime $3$ appears once and twice, so take two copies.
The prime $13$ appears only in the second number, once, so take one copy.

Multiply the highest powers together.
$2^3\times3^2\times13=936$

Check that both numbers, $24$ and $117$ , divide it.
$936\div24=39$ $936\div117=8$

Answer

$2^3\times3^2\times13$ , or $936$ .

Key idea

The LCM takes every prime that appears in either number, at its higher count.
Problem 3: Mark complete
Problem 4 The shortcut in reverse

Use the listing method to find $\operatorname{LCM}(16, 20)$ . Then use the product rule to find $\operatorname{GCF}(16, 20)$ without listing any factors.
1. Hint 1
  The LCM is the first value shared by the two lists of multiples, and the product rule links it to the GCF.
2. Hint 2
  Once the LCM is known, divide the product $16\times20$ by it.
Answer

LCM $80$ ; GCF $4$ .

Full solution

List multiples of each number until a value appears in both lists.
The multiples of $16$ begin $16$ , $32$ , $48$ , $64$ , $80$ , and those of $20$ begin $20$ , $40$ , $60$ , $80$ .
The first shared value is $80$ , so the LCM is $80$ .

For two numbers, the GCF times the LCM equals their product, so the GCF is the product divided by the LCM.
$16\times20=320$ $320\div80=4$
Thus the GCF is $4$ .

The prime powers agree: $16=2^4$ and $20=2^2\times5$ share only the prime $2$ , and its lower power is $2^2=4$ .

Answer

LCM $80$ ; GCF $4$ .

Key idea

For two numbers, dividing their product by the LCM gives the GCF, just as dividing it by the GCF gives the LCM.
Problem 4: Mark complete
Problem 5 The prize bags

A teacher has $72$ stickers and $54$ pencils. She splits all of them into identical prize bags, with none left over, so that each bag holds the same number of stickers and the same number of pencils. Every bag must hold at least $4$ pencils. What is the greatest number of bags she can make, and what does each bag contain?
1. Hint 1
  The number of bags must divide both totals, so list every number that divides both $72$ and $54$ .
2. Hint 2
  For each candidate, work out the pencils per bag and drop any count that gives fewer than $4$ .
Answer

$9$ bags, each with $8$ stickers and $6$ pencils.

Full solution

Each bag gets an equal share of both totals, so the number of bags must be a common factor of $72$ and $54$ .
The factors of $54$ are $1$ , $2$ , $3$ , $6$ , $9$ , $18$ , $27$ and $54$ , and all of them except $27$ and $54$ divide $72$ .
So the common factors are $1$ , $2$ , $3$ , $6$ , $9$ and $18$ .

The largest, $18$ , is the GCF, but it gives too few pencils.
$54\div18=3$
Three pencils is fewer than the required $4$ .

The next common factor down is $9$ .
$54\div9=6$ $72\div9=8$
Each of the $9$ bags holds $6$ pencils, which meets the condition, and $8$ stickers, with nothing left over.
No number between $9$ and $18$ divides both totals, so $9$ bags is the greatest possible.

Answer

$9$ bags, each with $8$ stickers and $6$ pencils.

Key idea

An extra condition on each group can rule out the GCF, so check the smaller common factors in turn.
Problem 5: Mark complete
Problem 6 The observation window

Two screens change at the same instant. After that, one changes every $6$ seconds and the other every $14$ seconds. At which times strictly between $50$ and $150$ seconds after the starting instant do they change together?
1. Hint 1
  Find the spacing between changes that happen together, then locate those times inside the stated window.
2. Hint 2
  Compare the prime factors of the two cycle lengths and take enough copies for both.
Answer

$84$ seconds and $126$ seconds after the starting instant.

Full solution

The cycle lengths have prime-power forms $6=2\times3$ and
$14=2\times7.$
The screens change together after a common multiple of the lengths.
The smallest positive spacing, in seconds, is
$2\times3\times7=42$

The listing method agrees: positive multiples of $6$ begin $6$ , $12$ , $18$ , $24$ , $30$ , $36$ , $42$ , while those of $14$ begin $14$ , $28$ , $42$ .
The first shared entry is $42$ .

Each common change time must contain a $2$ , a $3$ and a $7$ , so it is a multiple of $42$ , and every multiple of $42$ is a multiple of both $6$ and $14$ .
Common changes occur at multiples of $42$ .
$42\times2=84$ $42\times3=126$
The previous multiple is $42$ , below the window, and the next is $168$ , above it.
Thus precisely the changes at $84$ and $126$ seconds qualify.

Answer

$84$ seconds and $126$ seconds after the starting instant.

Key idea

An LCM gives the spacing between coincidences, and the stated time window determines which coincidences count.
Problem 6: Mark complete
Problem 7 The hidden pair

Two positive whole numbers have GCF $9$ and LCM $216$ , and neither number is $9$ or $216$ . Find the two numbers.
1. Hint 1
  Each number is a multiple of the GCF and a factor of the LCM.
2. Hint 2
  Use the product rule to find the product, then test each candidate pair's GCF.
Answer

$27$ and $72$ .

Full solution

The GCF divides both numbers, so each is a multiple of $9$ .
Both numbers divide the LCM, so each is also a factor of $216$ .
$216\div9=24$
Write each number as $9$ times a whole number.
If $9$ times the whole number goes into $9\times24$ some number of times, then the whole number goes into $24$ that same number of times, so it divides $24$ .
So each number is $9$ times a factor of $24$ .
Leaving out $9$ and $216$ , the candidates are $18$ , $27$ , $36$ , $54$ , $72$ and $108$ .

For two numbers, the GCF times the LCM equals their product.
$9\times216=1{,}944$
The candidates that pair up to this product are $18$ and $108$ , $27$ and $72$ , and $36$ and $54$ .

Now test each pair's GCF.
The pair $18$ and $108$ has the common factor $18$ , and so does the pair $36$ and $54$ , since $36=18\times2$ and
$54=18\times3.$
Both of those GCFs are larger than $9$ .

The remaining pair fits.
$27=3^3$ $72=2^3\times3^2$
The lower powers give a GCF of $3^2=9$ , and the higher powers give an LCM of
$2^3\times3^3=216.$
So the numbers are $27$ and $72$ .

Answer

$27$ and $72$ .

Key idea

Both numbers are multiples of the GCF and factors of the LCM, which limits the search to a short list.
Problem 7: Mark complete
Problem 8 The two answers

For $A=2^3\times3^2\times7$ and $B=2^2\times3^3\times5$ , a student reports GCF $108$ and LCM $7{,}560$ . Decide whether each answer is correct and explain any correction.
1. Hint 1
  A common factor must fit inside both numbers, while a common multiple must contain each number.
2. Hint 2
  Compare the counts of $2$ and $3$ , and notice which primes occur in just one number.
Answer

GCF $108$ is incorrect; the GCF is $36$ . LCM $7{,}560$ is correct.

Full solution

For a common factor, the two numbers supply at most two shared copies of $2$ and two shared copies of $3$ .
Neither $5$ nor $7$ is shared.
$2^2\times3^2=36$
The proposed $108$ contains three copies of $3$ , but $A$ has only two, so $108$ does not divide $A$ .

A common multiple needs three copies of $2$ , three copies of $3$ , and both of the primes appearing in just one number.
$2^3\times3^3=216$ $216\times5\times7=7{,}560$
Taking exactly these required copies gives the least common multiple, so the proposed LCM is correct.

Answer

GCF $108$ is incorrect; the GCF is $36$ . LCM $7{,}560$ is correct.

Key idea

The GCF respects the smaller supply of each prime, while the LCM meets the larger demand.
Problem 8: Mark complete
Problem 9 Three matching numbers

Three positive whole numbers are all equal to the same number greater than $1$ . Can their GCF times their LCM equal the product of all three numbers? Explain.
1. Hint 1
  Try it on three copies of one number, such as $5$ , $5$ and $5$ .
2. Hint 2
  Count how many copies of the common number each side multiplies together.
Answer

No.

Full solution

Call the common number $n$ , with $n>1$ .
Its largest shared factor is $n$ , and its smallest positive shared multiple is also $n$ .
Thus
$\operatorname{GCF}=n$ $\operatorname{LCM}=n$
Their product contains two copies of $n$ , while the product of the three original numbers contains three.

Multiplying the positive quantity $n\times n$ by the extra factor $n>1$ increases it.
The products are therefore unequal.
The rule for two numbers does not extend to this three-number setting.

Answer

No.

Key idea

For three equal numbers greater than one, multiplying the GCF and LCM leaves out one copy of the common number.
Problem 9: Mark complete
Problem 10 The added prime

A student takes two positive whole numbers greater than $1$ , multiplies each by $13$ , and claims that their GCF and their LCM are each multiplied by $13$ . Is the claim correct? Explain.
1. Hint 1
  Compare the prime lists before and after multiplying each number by $13$ .
2. Hint 2
  Track the count of $13$ in each number before and after, and compare the lower and the higher count.
Answer

Yes; both the GCF and the LCM are multiplied by $13$ .

Full solution

Multiplying each number by $13$ adds one copy of $13$ to each prime list and changes no other prime count.

Each number may already hold some copies of $13$ , possibly none.
Both counts rise by one, so the lower of the two counts rises by one, and the new GCF is the old GCF times $13$ .
The higher count also rises by one, so the new LCM is the old LCM times $13$ .
The rules select the same counts as before for every other prime.

Answer

Yes; both the GCF and the LCM are multiplied by $13$ .

Key idea

Multiplying both numbers by the same prime multiplies both their GCF and their LCM by that prime.
Problem 10: Mark complete

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