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What does this show about 6?","back":"6 is a factor of 42.","mcqOptions":[{"text":"6 is a factor of 42.","correct":true,"explanation":"A factor divides a number exactly with no remainder. Since $42 \\div 6 = 7$, the number 6 is therefore a factor of 42."},{"text":"6 is a multiple of 42.","correct":false,"explanation":"This reverses the factor-multiple relationship. Since $42 \\div 6$ is an integer, 6 is a factor of 42, while 42 is a multiple of 6."},{"text":"42 is a factor of 6.","correct":false,"explanation":"This reverses the direction of divisibility: 42 does not divide 6 exactly. The calculation shows that 6 divides 42 exactly, so 6 is a factor of 42."},{"text":"7 is a factor of 6.","correct":false,"explanation":"The quotient 7 is not automatically a factor of the divisor 6. The exact division $42 \\div 6 = 7$ shows that 6 is a factor of 42."}]},{"id":"ef6dd3cb-2323-4ce8-aaa1-1dc69ca0964f","cardType":"basic","points":2,"front":"Why is 1 not a prime number?","back":"1 has only one factor.","mcqOptions":[{"text":"1 has only one factor.","correct":true,"explanation":"A prime number must have exactly two distinct positive factors: 1 and itself. For the number 1 these are the same factor, so it has only one."},{"text":"1 has exactly two factors.","correct":false,"explanation":"Counting 1 twice as both 1 and itself is a common mistake. Distinct factors are counted only once, so 1 has one factor and is not prime."},{"text":"1 has no positive factors.","correct":false,"explanation":"The number 1 divides itself exactly, so it does have a positive factor. It is not prime because it has only one distinct factor, rather than exactly two."},{"text":"1 is an even number.","correct":false,"explanation":"Evenness is determined by divisibility by 2, and 1 is odd. Its actual reason for not being prime is that it has only one distinct positive factor."}]},{"id":"87917010-a798-2570-1c8c-c74ec6158c73","cardType":"cloze","points":null,"front":"A prime number has exactly [ ] factors: 1 and [ ].","back":"A prime number has exactly two factors: 1 and itself.","mcqOptions":[{"text":"A prime number has exactly two factors: 1 and itself.","correct":true,"explanation":"By definition, a prime number has exactly two distinct positive factors. These are always 1 and the prime number itself."},{"text":"A prime number has exactly one factors: 1 and itself.","correct":false,"explanation":"Treating 1 and the number itself as one factor confuses prime numbers with the number 1. A prime has two distinct factors: 1 and itself."},{"text":"A prime number has exactly two factors: 1 and 2.","correct":false,"explanation":"Only the prime number 2 has 2 as its second factor. In general, every prime has exactly two factors: 1 and itself."},{"text":"A prime number has exactly three factors: 1 and itself.","correct":false,"explanation":"Having three factors does not satisfy the definition of a prime number. A prime has exactly two distinct positive factors: 1 and itself."}]},{"id":"6ba05888-336e-3b5f-2ed5-5edd835eed30","cardType":"cloze","points":null,"front":"A prime factorisation should never include [ ].","back":"A prime factorisation should never include 1.","mcqOptions":[{"text":"1","correct":true,"explanation":"The number 1 is not prime, so it cannot appear as a prime factor. Including it also adds no useful information because multiplying by 1 changes nothing."},{"text":"0","correct":false,"explanation":"Although 0 should not appear in a prime factorisation, it is not the intended defining exclusion here. Prime factorisations never include 1 because 1 is not a prime number."},{"text":"2","correct":false,"explanation":"The number 2 is the smallest prime and is included whenever it divides the original number. It is 1 that must never appear because 1 is not prime."},{"text":"3","correct":false,"explanation":"The number 3 is a valid prime factor and should be included when it divides the number. A prime factorisation never includes 1 because 1 is not prime."}]},{"id":"4b865946-e6f9-4bf6-3c68-da29007ead71","cardType":"basic","points":1,"front":"What does $2^3$ mean in index notation?","back":"$2 \\times 2 \\times 2$","mcqOptions":[{"text":"$2 \\times 2 \\times 2$","correct":true,"explanation":"An exponent tells you how many times the base is used as a factor. Therefore, $2^3$ means multiplying three factors of 2 together."},{"text":"$2 \\times 3$","correct":false,"explanation":"Multiplying the base by the exponent is a common index error. The exponent 3 means use 2 as a factor three times, giving $2 \\times 2 \\times 2$."},{"text":"$3 \\times 3 \\times 3$","correct":false,"explanation":"This incorrectly uses the exponent as the repeated factor. In $2^3$, the base 2 is repeated three times, so it means $2 \\times 2 \\times 2$."},{"text":"$2 + 2 + 2$","correct":false,"explanation":"An index represents repeated multiplication, not repeated addition. Since the base is 2 and the exponent is 3, the expression means $2 \\times 2 \\times 2$."}]},{"id":"8aaf6277-69cc-f48c-5312-a04bb054b9d4","cardType":"basic","points":2,"front":"Write 84 as a product of prime factors.","back":"$84 = 2^2 \\times 3 \\times 7$","mcqOptions":[{"text":"$84 = 2^2 \\times 3 \\times 7$","correct":true,"explanation":"Dividing 84 gives $84 = 2 \\times 42 = 2 \\times 2 \\times 21 = 2^2 \\times 3 \\times 7$, and every factor on the right is prime."},{"text":"$84 = 2 \\times 3 \\times 7$","correct":false,"explanation":"One factor of 2 has been omitted, so the right-hand side equals 42 rather than 84. The complete factorisation is $84 = 2^2 \\times 3 \\times 7$."},{"text":"$84 = 2^2 \\times 3^2 \\times 7$","correct":false,"explanation":"The exponent on 3 has been increased incorrectly, making the product three times too large. There is only one factor of 3, so the correct power is $3^1$."},{"text":"$84 = 2^3 \\times 3 \\times 7$","correct":false,"explanation":"This includes one extra factor of 2, so the product is 168 rather than 84. There are exactly two factors of 2, giving $2^2 \\times 3 \\times 7$."}]},{"id":"7eb2f12b-cd69-a312-86a7-27fd293726d2","cardType":"basic","points":1,"front":"For HCF using prime factors, which primes do you keep?","back":"Primes that appear in every number.","mcqOptions":[{"text":"Primes that appear in every number.","correct":true,"explanation":"The HCF must divide every number, so only prime factors shared by all the numbers can be included. Any unshared prime would prevent the result dividing at least one number."},{"text":"Primes that appear in any number.","correct":false,"explanation":"Keeping every prime that appears in at least one number is the method associated with finding the LCM. For the HCF, retain only primes appearing in every number."},{"text":"Primes that appear in one number.","correct":false,"explanation":"A prime found in only one number cannot be part of a common factor. For the HCF, keep primes that occur in every number being considered."},{"text":"Primes that are below every number.","correct":false,"explanation":"A prime being smaller than each number does not mean it divides them. The HCF uses only prime factors that appear in every number’s prime factorisation."}]},{"id":"7d561f66-3b1f-6992-63ff-7381247cf92f","cardType":"cloze","points":null,"front":"For HCF, choose the [ ] power of each shared prime.","back":"For HCF, choose the smallest power of each shared prime.","mcqOptions":[{"text":"smallest","correct":true,"explanation":"A common factor cannot contain a prime more times than any number being compared. Therefore, the HCF uses the smallest exponent of each shared prime."},{"text":"largest","correct":false,"explanation":"Choosing the largest power is the rule used for the LCM, not the HCF. The HCF must divide every number, so it uses the smallest shared power."},{"text":"total","correct":false,"explanation":"Adding the powers would introduce too many copies of the prime and produce a result that may not divide every number. The HCF uses the smallest shared power."},{"text":"average","correct":false,"explanation":"Averaging exponents is not part of finding an HCF and could even produce a non-integer exponent. The HCF takes the smallest power of each shared prime."}]},{"id":"f3315558-3beb-b9e0-4d92-d47fd95f836b","cardType":"basic","points":2,"front":"Given $54 = 2 \\times 3^3$ and $126 = 2 \\times 3^2 \\times 7$, find the HCF.","back":"$18$","mcqOptions":[{"text":"$18$","correct":true,"explanation":"For the HCF, use each common prime at its smallest power: $2^1 \\times 3^2 = 2 \\times 9 = 18$."},{"text":"$6$","correct":false,"explanation":"This uses only one factor of $3$ instead of the shared $3^2$. Taking the smallest common powers gives $2 \\times 3^2 = 18$."},{"text":"$9$","correct":false,"explanation":"This omits the factor $2$, even though both numbers contain it. The HCF includes $2^1$ and $3^2$, giving $18$."},{"text":"$54$","correct":false,"explanation":"This copies the first number, but $3^3$ is not a factor of $126$. The smallest shared powers are $2^1$ and $3^2$, giving $18$."}]},{"id":"c40acec0-40c9-c2b2-ddf0-a25fefec0107","cardType":"basic","points":1,"front":"For LCM using prime factors, which primes do you include?","back":"Every prime that appears.","mcqOptions":[{"text":"Every prime that appears.","correct":true,"explanation":"The LCM must be divisible by every original number, so it includes every prime appearing in any factorisation, using the greatest required power of each."},{"text":"Every prime shared by all.","correct":false,"explanation":"Including only shared primes is the method for finding an HCF. An LCM includes every prime that appears in any of the numbers."},{"text":"Every prime in the first number.","correct":false,"explanation":"Restricting the selection to the first number can omit factors needed for divisibility by the others. The LCM includes every prime that appears."},{"text":"Only the largest prime present.","correct":false,"explanation":"The numerical size of a prime does not determine whether it is included. The LCM requires every prime appearing in any of the factorisations."}]},{"id":"8a74c720-f927-a75b-738c-a880d44688eb","cardType":"cloze","points":null,"front":"For LCM, choose the [ ] power of each prime.","back":"For LCM, choose the largest power of each prime.","mcqOptions":[{"text":"largest","correct":true,"explanation":"The LCM must contain enough of each prime to be divisible by every number, so the largest power of each prime is selected."},{"text":"smallest","correct":false,"explanation":"Choosing the smallest power is the HCF method because it keeps only factors common to every number. For an LCM, choose the largest power."},{"text":"first","correct":false,"explanation":"The first power would include each prime only once, which may not provide enough factors. The LCM uses the largest power shown."},{"text":"common","correct":false,"explanation":"Restricting the choice to a common power confuses LCM with HCF. For the LCM, select the largest power of every prime that appears."}]},{"id":"612ae405-699b-1883-6fdb-cc2f26bd38a2","cardType":"basic","points":2,"front":"Given $45 = 3^2 \\times 5$ and $60 = 2^2 \\times 3 \\times 5$, find the LCM.","back":"$180$","mcqOptions":[{"text":"$180$","correct":true,"explanation":"The LCM uses the largest powers present: $2^2 \\times 3^2 \\times 5 = 4 \\times 9 \\times 5 = 180$."},{"text":"$15$","correct":false,"explanation":"This is the HCF, found from the shared factors $3 \\times 5$. The LCM uses the greatest powers of all primes, giving $180$."},{"text":"$60$","correct":false,"explanation":"This keeps only $3^1$, so the result is not divisible by $45$, which contains $3^2$. Using $3^2$ gives an LCM of $180$."},{"text":"$900$","correct":false,"explanation":"This is obtained by multiplying $45$ and $60$ and then dividing incorrectly. Their shared factors must not be duplicated, so the LCM is $180$."}]},{"id":"3967a4a5-22c5-1561-221f-1e920f2ce9d6","cardType":"basic","points":1,"front":"Two regular events repeat at fixed intervals. What calculation gives the next time together?","back":"Find the LCM of the intervals.","mcqOptions":[{"text":"Find the LCM of the intervals.","correct":true,"explanation":"Each event occurs at multiples of its interval. The first positive time that is a multiple of both intervals is their lowest common multiple."},{"text":"Find the HCF of the intervals.","correct":false,"explanation":"The HCF measures the greatest interval that divides both values; it does not give their next shared occurrence. That time is found using the LCM."},{"text":"Add the two intervals together.","correct":false,"explanation":"Adding the intervals does not generally produce a multiple of each one. The next time both events occur together is given by their LCM."},{"text":"Multiply the two intervals together.","correct":false,"explanation":"The product is a common multiple but may not be the first one, especially when the intervals share factors. The next meeting uses the LCM."}]},{"id":"29e3dbb2-741b-c776-f451-fac508a8b370","cardType":"cloze","points":null,"front":"Before using LCM in a time question, put all intervals in [ ].","back":"Before using LCM in a time question, put all intervals in the same units.","mcqOptions":[{"text":"the same units","correct":true,"explanation":"LCM compares whole-number intervals directly, so all times must first be expressed in one unit, such as converting hours and minutes entirely into minutes."},{"text":"different units","correct":false,"explanation":"Applying LCM to mixed units compares unlike quantities and gives an invalid result. Convert every interval into the same units before calculating the LCM."},{"text":"the largest units","correct":false,"explanation":"Choosing the largest unit can introduce fractions and is not the requirement. All intervals must be converted into the same units before finding the LCM."},{"text":"the smallest units","correct":false,"explanation":"Small units are often convenient, but they are not always necessary. The essential requirement is that every interval is expressed in the same units."}]},{"id":"055cfd05-1e48-9423-bf39-8c33487ffc2a","cardType":"basic","points":2,"front":"A tram and coach both arrive at 10:00, then every 9 and 12 minutes. When next together?","back":"10:36","mcqOptions":[{"text":"10:36","correct":true,"explanation":"The LCM of $9$ and $12$ is $36$, so both services return together $36$ minutes after 10:00, which is 10:36."},{"text":"10:21","correct":false,"explanation":"This adds the two intervals, but $21$ is not a multiple of either $9$ or $12$. Their LCM is $36$, giving 10:36."},{"text":"10:12","correct":false,"explanation":"This uses only the coach's interval. After $12$ minutes the tram is not due, since $12$ is not a multiple of $9$; both return at 10:36."},{"text":"11:48","correct":false,"explanation":"This multiplies the intervals to get $108$ minutes, which is a shared return time but not the next one. The LCM is $36$ minutes, giving 10:36."}]},{"id":"fa4b5d79-f1c1-1848-e77e-b1fabb0a03fc","cardType":"basic","points":2,"front":"For $P = 2^3 \\times 3^2 \\times 5$ and $Q = 2^2 \\times 3 \\times 7$, find the HCF.","back":"$12$","mcqOptions":[{"text":"$12$","correct":true,"explanation":"The common primes are $2$ and $3$. Taking their smallest powers gives $2^2 \\times 3^1 = 4 \\times 3 = 12$."},{"text":"$6$","correct":false,"explanation":"This incorrectly reduces the shared power of $2$ from $2^2$ to $2^1$. The smallest powers actually present are $2^2$ and $3^1$, giving $12$."},{"text":"$36$","correct":false,"explanation":"This uses $3^2$ from $P$, but $Q$ contains only $3^1$, so $36$ cannot divide $Q$. Using the smaller powers gives $12$."},{"text":"$84$","correct":false,"explanation":"This includes the factor $7$, which appears only in $Q$ and is therefore not common. The shared prime powers produce an HCF of $12$."}]},{"id":"009ccbbd-ebdc-f0c8-7c1d-029928a5d0c0","cardType":"basic","points":2,"front":"For $P = 2^3 \\times 3^2 \\times 5$ and $Q = 2^2 \\times 3 \\times 7$, find the LCM.","back":"$2520$","mcqOptions":[{"text":"$2520$","correct":true,"explanation":"The LCM uses the highest power of each prime present: $2^3 \\times 3^2 \\times 5 \\times 7 = 2520$."},{"text":"$1260$","correct":false,"explanation":"Using $2^2$ instead of the higher power $2^3$ misses a factor of $2$. The LCM is $2^3 \\times 3^2 \\times 5 \\times 7 = 2520$."},{"text":"$7560$","correct":false,"explanation":"Multiplying by an unnecessary extra factor of $3$ makes the result three times too large. Using each prime’s highest power gives an LCM of $2520$."},{"text":"$504$","correct":false,"explanation":"Omitting the factor $5$ gives $504$, but the LCM must contain every prime factor needed by either number. Including $5$ gives $2520$."}]},{"id":"11e49fd7-76ca-c6e8-a64f-fd72347457a1","cardType":"basic","points":2,"front":"Given $12 = 2^2 \\times 3$, $18 = 2 \\times 3^2$, and $30 = 2 \\times 3 \\times 5$, find the HCF.","back":"$6$","mcqOptions":[{"text":"$6$","correct":true,"explanation":"The common prime factors with the lowest powers are $2^1$ and $3^1$. Therefore, the HCF is $2 \\times 3 = 6$."},{"text":"$2$","correct":false,"explanation":"This includes the common factor $2$ but overlooks that all three numbers are also divisible by $3$. Their HCF is therefore $2 \\times 3 = 6$."},{"text":"$3$","correct":false,"explanation":"This includes the common factor $3$ but overlooks that all three numbers are also divisible by $2$. Their HCF is therefore $2 \\times 3 = 6$."},{"text":"$12$","correct":false,"explanation":"Taking the highest powers resembles the LCM method and makes the result too large; $12$ does not divide $18$ or $30$. The HCF is $6$."}]},{"id":"86f3f0fc-3de7-b042-989d-4677f26e5636","cardType":"basic","points":2,"front":"Given $12 = 2^2 \\times 3$, $18 = 2 \\times 3^2$, and $30 = 2 \\times 3 \\times 5$, find the LCM.","back":"$180$","mcqOptions":[{"text":"$180$","correct":true,"explanation":"The LCM uses the highest power of every prime present: $2^2 \\times 3^2 \\times 5 = 180$, which is divisible by all three numbers."},{"text":"$60$","correct":false,"explanation":"Using only one factor of $3$ overlooks the $3^2$ in $18$, so $60$ is not divisible by $18$. The correct LCM is $180$."},{"text":"$90$","correct":false,"explanation":"Using only one factor of $2$ overlooks the $2^2$ in $12$, so $90$ is not divisible by $12$. The correct LCM is $180$."},{"text":"$360$","correct":false,"explanation":"Including an unnecessary extra factor of $2$ doubles the smallest common multiple. The highest required powers give $2^2 \\times 3^2 \\times 5 = 180$."}]},{"id":"afebf876-a72f-4c5d-0460-1358af2d6ca4","cardType":"cloze","points":null,"front":"Two numbers are coprime if their HCF is [ ].","back":"Two numbers are coprime if their HCF is 1.","mcqOptions":[{"text":"1","correct":true,"explanation":"Coprime numbers share no prime factors, so their only positive common factor is $1$. Consequently, their highest common factor is $1$."},{"text":"0","correct":false,"explanation":"An HCF is a positive factor shared by the numbers, and every whole number has $1$ as a factor. Coprime numbers therefore have HCF $1$, not $0$."},{"text":"2","correct":false,"explanation":"An HCF of $2$ means both numbers are even and share the prime factor $2$, so they are not coprime. Coprime numbers have HCF $1$."},{"text":"-1","correct":false,"explanation":"At GCSE, the HCF is defined as the greatest positive common factor, so it cannot be negative. For two coprime numbers, the HCF is $1$."}]},{"id":"0146379c-787b-85ce-45fb-0736602b7840","cardType":"basic","points":2,"front":"Two numbers have HCF 6 and LCM 72. After taking out the HCF, what do the leftover parts multiply to?","back":"$72 \\div 6 = 12$","mcqOptions":[{"text":"$72 \\div 6 = 12$","correct":true,"explanation":"Writing the numbers as $6a$ and $6b$, where $a$ and $b$ are coprime, gives an LCM of $6ab$. Therefore, $ab = 72 \\div 6 = 12$."},{"text":"$72 \\times 6 = 432$","correct":false,"explanation":"Multiplying by the HCF counts the common factor again instead of removing it. Dividing the LCM by the HCF gives the leftover product: $72 \\div 6 = 12$."},{"text":"$72 \\div 36 = 2$","correct":false,"explanation":"Dividing by $6^2$ removes the common factor twice. The LCM contains one shared factor of $6$, so the leftover parts multiply to $72 \\div 6 = 12$."},{"text":"$72 - 6 = 66$","correct":false,"explanation":"Taking out a common factor requires division, not subtraction. Dividing the LCM by the HCF gives the product of the coprime leftover parts: $72 \\div 6 = 12$."}]},{"id":"fbf5e9c5-5394-5c90-238b-085305a26d7d","cardType":"basic","points":3,"front":"Two whole numbers greater than 8 have HCF 6 and LCM 72. Find them.","back":"18 and 24","mcqOptions":[{"text":"18 and 24","correct":true,"explanation":"The numbers are $6 \\times 3$ and $6 \\times 4$. Since $3$ and $4$ are coprime and multiply to $12$, the HCF is $6$ and LCM is $72$."},{"text":"12 and 36","correct":false,"explanation":"These numbers have HCF $12$ and LCM $36$, so neither required value is obtained. The pair $18$ and $24$ has HCF $6$ and LCM $72$."},{"text":"12 and 72","correct":false,"explanation":"Although the LCM is $72$, both numbers are divisible by $12$, making their HCF $12$ rather than $6$. The required numbers are $18$ and $24$."},{"text":"24 and 36","correct":false,"explanation":"Although the LCM is $72$, these numbers share a factor of $12$, so their HCF is $12$. The pair with HCF $6$ is $18$ and $24$."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Component 1"},"heading":"Prime Factors, HCF and LCM","title":"Prime Factors, HCF and LCM Flashcards","topbar":{"backHref":"/gcse/maths/eduqas/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"22045d12-0f9a-4fa9-9a57-0ba0dfb1d2fc","walled":"$L38"}],"$L39","$L3a"]