Question bank — LCM — prime factorization method, relationship HCF × LCM = product

False — if one number divides the other (e.g. 4 and 12), the LCM equals the larger one (12), not something bigger. It is $\ge$ both, never strictly greater than both.

False — if one divides the other (e.g. 6 and 18), the HCF equals the smaller one (6). HCF is $\le$ both, and can equal the smaller.

True — coprime means every prime's $\min$ exponent is 0, so $\text{HCF}=1$, and $\text{LCM}=\frac{a\times b}{1}=a\times b$.

True — for every prime, $\min (\alpha ,\beta )\le \max (\alpha ,\beta )$ (recall $\alpha ,\beta$ are the exponents in the two recipes), so every prime power in the HCF is present (to at least that power) in the LCM. Hence HCF is a factor of LCM.

True — same reason as above stated the other way: since HCF divides LCM, LCM is a multiple of it. In fact $\text{LCM}=\text{HCF}\times \frac{a}{\text{HCF}}\times \frac{b}{\text{HCF}}$.

True — doubling adds one factor of 2 to each recipe, so the max exponent of 2 rises by exactly 1, multiplying the LCM by 2. (The other primes' maxima are unchanged.)

True — this holds for all pairs of two positive integers, because $\min +\max =\alpha +\beta$ for each prime (the exponents in the two recipes). It only fails when you try to stretch it to three or more numbers.

False — take HCF $=6$, LCM $=36$: both the pair $(12,18)$ and the pair $(6,36)$ have HCF $6$ and LCM $36$, yet are different pairs. HCF and LCM fix the product and which primes appear, not $a$ and $b$ individually. (Given HCF, LCM and one number, though, the other is forced.)

"Least" describes the final number being smallest, not the exponents. You must take the max power so the result is divisible by both; taking the min would give a number that is not even a multiple of the larger one.

The rule HCF $\times$ LCM $=$ product only works for exactly two numbers. For three or more it breaks, as this very example shows ($24\ne 48$). Use min/max per prime directly.

$9$ is not prime — it is $3^2$. You must break down to genuine primes: the recipe of $18$ is $2^1\times 3^2$. Using $9$ as if it were a prime block breaks the max/min-per-prime method.

They are coprime (HCF $=1$), but coprime numbers still have an LCM — it is simply their product, $12\times 35=420$. Every pair of positive integers has an LCM.

The answer is right but the reason is wrong. You don't "ignore" the 5; you include $5^{\min (1,0)}=5^0=1$. The 5 contributes a factor of 1 because 72 has none — that is the principled reason it drops out.

You must divide by the highest common factor, not just any common factor. Dividing $a\times b$ by a smaller common factor overcounts and gives a number larger than the true LCM.

Because the LCM must be a multiple of both. If a prime is in one number's recipe, any multiple of that number must contain that prime — so the LCM cannot leave it out.

The max is the minimum requirement to be divisible by both. Any exponent above the max would still be a common multiple but a strictly bigger number — so the max is exactly enough, and enough-but-no-more is smallest.

The HCF must divide both numbers, and a divisor can use a prime at most as many times as the number it divides. So its exponent is capped by the smaller of the two — that is why HCF uses the min.

Here $\alpha ,\beta$ are a prime's exponents in $a$ and $b$. HCF supplies exponent $\min$ and LCM supplies $\max$; multiplying them adds exponents to $\min +\max =\alpha +\beta$, which is exactly the exponent of that prime in $a\times b$. True for every prime means the whole products match.

It is a direct algebraic rearrangement of the proven identity HCF $\times$ LCM $=ab$. As long as HCF is known and the pair is two numbers, the shortcut is exact, not an approximation.

It is $a$ itself. Every prime's max exponent is $\max (\alpha ,\alpha )=\alpha$, so nothing changes. Likewise HCF($a,a$)$=a$, and indeed $a\times a=a\times a$. ✓

It is $1$ — every shared prime power is $p^0=1$. The number 1 is the "empty product" of primes (a recipe with no ingredients) and is the HCF of any coprime pair.

$1$ has an empty prime recipe, so max exponents come entirely from $n$: LCM $=n$. HCF takes min exponents, all $0$, so HCF $=1$. Check: $1\times n=1\times n$. ✓

Since $b$'s recipe already contains all of $a$'s primes to at least equal powers, mins give $a$ (HCF $=a$) and maxes give $b$ (LCM $=b$). No arithmetic needed.

$0$ is technically a multiple of every number, but "least positive" excludes it — otherwise 0 would always win and the concept would be useless. We want the smallest positive common multiple.

By the standard convention it is $n$ — every integer divides $0$, so the largest number dividing both $0$ and $n$ is the largest divisor of $n$, which is $n$ itself. (This is why $\gcd (0,n)=n$ in the Euclidean algorithm's base case.)

It is taken to be $0$ — the only common multiple of $0$ and $n$ is $0$ (since $0$'s only multiple is $0$). This is the one place the "smallest positive" phrasing breaks, so LCM with a zero is defined separately as $0$.

Both are conventionally $0$: every integer divides $0$ so there is no largest common divisor in the usual sense, and the convention $\gcd (0,0)=0$, $\text{lcm}(0,0)=0$ is chosen to keep formulas consistent. This is a degenerate corner — normal LCM/HCF assumes positive integers.

Only when the numbers are equal. LCM $=$ HCF forces $\max =\min$ for every prime, meaning $\alpha =\beta$ throughout (equal exponents in both recipes), so $a=b$. Otherwise LCM is strictly larger.