MYP 3 · Maths
NUMBER
Highest common factor
QUESTION 1 [6 marks] — Criterion B
Medium
Investigate the relationship between the HCF of two numbers and the HCF of their sum with one of the numbers.
a.
Find HCF$(12, 18)$, then find HCF$(12, 12+18)$ = HCF$(12, 30)$. What do you notice?
[2] Show Solution
HCF$(12,18)=6$. HCF$(12,30)=6$. Both give the same HCF.
b.
Test this again with the pair 15 and 25: find HCF$(15,25)$ and HCF$(15, 15+25)$.
[2] Show Solution
HCF$(15,25)=5$. HCF$(15,40)=5$. Again the same HCF.
c.
Write a general statement (conjecture) describing this pattern, using $a$ and $b$ to represent the two numbers.
[2] Show Solution
Conjecture: HCF$(a,b)$ = HCF$(a, a+b)$ for any two positive integers $a$ and $b$. This works because any common factor of $a$ and $b$ must also divide $a+b$, and any common factor of $a$ and $a+b$ must also divide $(a+b)-a=b$.
QUESTION 2 [5 marks] — Criterion C
Medium
Two methods can be used to find the HCF of 84 and 126: listing factors, or prime factorization.
a.
Using prime factorization, find HCF$(84, 126)$, showing full working.
[3] Show Solution
$84 = 2^2 \times 3 \times 7$. $126 = 2 \times 3^2 \times 7$. Common factors (lowest power of each shared prime): $2^1 \times 3^1 \times 7^1 = 42$. So HCF $= 42$.
b.
Explain, in your own words, why the prime factorization method works — what does taking the 'lowest power of each common prime' actually represent?
[2] Show Solution
Each prime factor with its lowest shared power represents the largest amount of that prime that BOTH numbers can be divided by. Multiplying these together gives the largest number that divides both — by definition, the HCF.
QUESTION 3 [5 marks] — Criterion D
Medium
A school is arranging chairs for an assembly. There are 96 boys and 72 girls, and the principal wants to arrange them into identical rows, with boys and girls not mixed within a row, using the maximum possible number of chairs per row.
a.
Find the maximum number of chairs per row that allows both 96 boys and 72 girls to be arranged in complete rows of that size.
[3] Show Solution
This requires HCF$(96,72)$. $96=2^5\times3$, $72=2^3\times3^2$. HCF $=2^3\times3=24$. Maximum chairs per row $=24$.
b.
How many rows of boys and how many rows of girls will there be in total?
[2] Show Solution
Boy rows $=96\div24=4$. Girl rows $=72\div24=3$. Total rows $=7$.
QUESTION 4 [6 marks] — Criterion A
Hard
A school is preparing identical care packages for a charity drive, using 168 notebooks and 252 pencils, with NOTHING left over.
a.
Find the maximum number of identical packages that can be made, using prime factorization.
[3] Show Solution
$168=2^3\times3\times7$. $252=2^2\times3^2\times7$. HCF (lowest powers of shared primes): $2^2\times3\times7=84$. Maximum packages $=84$.
b.
Find the number of notebooks and pencils in EACH package, and verify your answer by checking $84\times(\text{notebooks per package})=168$.
[3] Show Solution
Notebooks per package $=168\div84=2$. Pencils per package $=252\div84=3$. Check: $84\times2=168$ ?.
QUESTION 5 [7 marks] — Criterion B
Hard
Investigate the relationship between HCF$(a,b)$ and HCF$(a,b,c)$ when a third number $c$ is introduced, using $a=24, b=36, c=60$.
a.
Find HCF$(24,36)$ first (ignoring $c$ for now).
[2] Show Solution
$24=2^3\times3$, $36=2^2\times3^2$. HCF$=2^2\times3=12$.
b.
Now find HCF$(12, 60)$ — i.e. the HCF of your previous answer and the third number $c=60$. Compare this to what you'd get by finding HCF$(24,36,60)$ directly (all three prime-factorized together).
[3] Show Solution
$60=2^2\times3\times5$. HCF$(12,60)=12$ (since $12=2^2\times3$ divides evenly into $60=2^2\times3\times5$). Direct 3-way HCF: comparing $2^3\times3$, $2^2\times3^2$, $2^2\times3\times5$ — lowest shared powers give $2^2\times3=12$. Both methods agree: 12.
c.
Based on this, state a general rule for finding the HCF of THREE numbers using the HCF of only two numbers at a time.
[2] Show Solution
HCF$(a,b,c) = $ HCF$($HCF$(a,b), c)$ — you can find the HCF of any two of the numbers first, then find the HCF of that result with the third number, and you'll always get the same answer as computing all three together directly.
QUESTION 6 [5 marks] — Criterion C
Hard
A student says: 'The HCF of two numbers can never be bigger than the SMALLER of the two numbers, but it also can't be bigger than half of either number, since factors are always small.'
a.
Test the first part of the claim (HCF can't exceed the smaller number) using HCF$(15,15)$ — two IDENTICAL numbers.
[2] Show Solution
HCF$(15,15)=15$ — this equals (not exceeds) the smaller number, confirming the first part isn't violated, though it shows the HCF CAN equal the smaller number exactly.
b.
Test the second part of the claim ('can't be bigger than half of either number') using HCF$(20,20)$. Does this disprove the second part of the claim?
[3] Show Solution
HCF$(20,20)=20$, which is NOT less than or equal to half of 20 (which is 10) — in fact it equals the FULL number, not half. This disproves the second part of the student's claim: the HCF of two equal numbers is the number itself, which is clearly larger than half of it (for any number greater than 0).
QUESTION 7 [6 marks] — Criterion D
Hard
A tile installer has two types of square tiles to cover sections of a floor with NO cutting: one section requires tiles that fit evenly into a $180$cm length, and another section requires tiles fitting evenly into a $252$cm length. The installer wants to use the SAME tile size for both sections, as large as possible.
a.
Find the largest tile side length (in cm) that fits evenly into both 180cm and 252cm.
[3] Show Solution
$180=2^2\times3^2\times5$. $252=2^2\times3^2\times7$. HCF$=2^2\times3^2=36$. Largest tile size: 36 cm.
b.
Find how many tiles fit along each section's length, and calculate the TOTAL number of tiles needed if both sections are also 36cm wide (i.e. each section is a single row of square tiles).
[3] Show Solution
Section 1: $180\div36=5$ tiles. Section 2: $252\div36=7$ tiles. Total tiles needed $=5+7=12$.
QUESTION 8 [6 marks] — Criterion A
Hard
A school is preparing identical care packages for a charity drive, using 168 notebooks and 252 pencils, with NOTHING left over.
a.
Find the maximum number of identical packages that can be made, using prime factorization.
[3] Show Solution
$168=2^3\times3\times7$. $252=2^2\times3^2\times7$. HCF (lowest powers of shared primes): $2^2\times3\times7=84$. Maximum packages $=84$.
b.
Find the number of notebooks and pencils in EACH package, and verify your answer by checking $84\times(\text{notebooks per package})=168$.
[3] Show Solution
Notebooks per package $=168\div84=2$. Pencils per package $=252\div84=3$. Check: $84\times2=168$ ?.
QUESTION 9 [7 marks] — Criterion B
Hard
Investigate the relationship between HCF$(a,b)$ and HCF$(a,b,c)$ when a third number $c$ is introduced, using $a=24, b=36, c=60$.
a.
Find HCF$(24,36)$ first (ignoring $c$ for now).
[2] Show Solution
$24=2^3\times3$, $36=2^2\times3^2$. HCF$=2^2\times3=12$.
b.
Now find HCF$(12, 60)$ — i.e. the HCF of your previous answer and the third number $c=60$. Compare this to what you'd get by finding HCF$(24,36,60)$ directly (all three prime-factorized together).
[3] Show Solution
$60=2^2\times3\times5$. HCF$(12,60)=12$ (since $12=2^2\times3$ divides evenly into $60=2^2\times3\times5$). Direct 3-way HCF: comparing $2^3\times3$, $2^2\times3^2$, $2^2\times3\times5$ — lowest shared powers give $2^2\times3=12$. Both methods agree: 12.
c.
Based on this, state a general rule for finding the HCF of THREE numbers using the HCF of only two numbers at a time.
[2] Show Solution
HCF$(a,b,c) = $ HCF$($HCF$(a,b), c)$ — you can find the HCF of any two of the numbers first, then find the HCF of that result with the third number, and you'll always get the same answer as computing all three together directly.
QUESTION 10 [5 marks] — Criterion C
Hard
A student says: 'The HCF of two numbers can never be bigger than the SMALLER of the two numbers, but it also can't be bigger than half of either number, since factors are always small.'
a.
Test the first part of the claim (HCF can't exceed the smaller number) using HCF$(15,15)$ — two IDENTICAL numbers.
[2] Show Solution
HCF$(15,15)=15$ — this equals (not exceeds) the smaller number, confirming the first part isn't violated, though it shows the HCF CAN equal the smaller number exactly.
b.
Test the second part of the claim ('can't be bigger than half of either number') using HCF$(20,20)$. Does this disprove the second part of the claim?
[3] Show Solution
HCF$(20,20)=20$, which is NOT less than or equal to half of 20 (which is 10) — in fact it equals the FULL number, not half. This disproves the second part of the student's claim: the HCF of two equal numbers is the number itself, which is clearly larger than half of it (for any number greater than 0).
QUESTION 11 [6 marks] — Criterion D
Hard
A tile installer has two types of square tiles to cover sections of a floor with NO cutting: one section requires tiles that fit evenly into a $180$cm length, and another section requires tiles fitting evenly into a $252$cm length. The installer wants to use the SAME tile size for both sections, as large as possible.
a.
Find the largest tile side length (in cm) that fits evenly into both 180cm and 252cm.
[3] Show Solution
$180=2^2\times3^2\times5$. $252=2^2\times3^2\times7$. HCF$=2^2\times3^2=36$. Largest tile size: 36 cm.
b.
Find how many tiles fit along each section's length, and calculate the TOTAL number of tiles needed if both sections are also 36cm wide (i.e. each section is a single row of square tiles).
[3] Show Solution
Section 1: $180\div36=5$ tiles. Section 2: $252\div36=7$ tiles. Total tiles needed $=5+7=12$.