What is the minimum number of packs of each ingredient needed without left over? Case 1

Ingredients A (3 kg per pack), B (4 kg per pack), and C (5 kg per
pack) must purchased in whole packs.
Given that 1 kg of A, 1 kg of B and 1 kg of C are needed to make a giant bread. We want to make some breads.


What is the minimum number of pack of each ingredient we have to purchase such that none is left over?

My attempt

*

*First, find the least common multiple of 3 kg, 4 kg, and 5 kg. The LCM is 60 kg.


*Second, find the number of packs of each ingredients.

*

*$A=60/3=20$ packs

*$B=60/4=15$ packs

*$C=60/5=12$ packs



*Third, additionally we can find the number of breads which is 60.
Case 2
Let's change to another recipe.

Ingredients A (3 kg per pack), B (4 kg per pack), and C (5 kg per
pack) must purchased in whole packs.
Given that 2 kg of A, 3 kg of B and 7 kg of C are needed to make a giant bread. We want to make some breads.


What is the minimum number of pack of each ingredient we have to purchase such that none is left over?

Here 60 cannot be divided by 7. What should I do next?
 A: Suppose the minimum number of packs of ingredients $A$, $B$, $C$ that we need to buy are $a$, $b$, $c$ respectively.  Then the total amounts of each ingredient are $3a$, $4b$, $5c$.  Let $n$ represent the number of breads that are made.  Then $2n$, $3n$, $7n$ are the total amounts of each ingredient that are used.  Therefore, we require
$$\begin{align}
3a &= 2n \\ 4b &= 3n \\ 5c &= 7n. \end{align}$$
Since $a, b, c, n$ are integers, we must have $n$ be multiples of $3$, $4$, and $5$ as you had observed in the first question; the least such common multiple is $60$, therefore let $n = 60k$ for some integer $k$.  Then we get
$$\begin{align}
a &= 40k \\ b &= 45k \\ c &= 84k. \end{align}$$
So long as $k$ is a positive integer, $a, b, c$ are also positive integers; thus $k = 1$ gives the minimum solution, $(a, b, c) = (40, 45, 84)$ packs.  This makes $60$ breads as in the first question.
It is worth mentioning that we need to be careful:  if, for instance, the ingredient ratio had been $2 : 4 : 7$ instead of $2 : 3 : 7$, then we only need $n$ to be a common multiple of $3$ and $5$, since the second equation in the system reduces:
$$\begin{align}
3a &= 2n \\ 4b &= 4n \color{red}{\iff b = n} \\ 5c &= 7n \end{align}$$
This gives us $n = 15k$ and for $k = 1$ we get $$(a, b, c) = (10, 15, 21).$$
A: In short, focus on minimum number of breads we need to make. Everything else follows from there.
To make a number of giant breads with given ingredients and quantities and no leftover ingredient, you need to ensure two things -
i) Considering each ingredient individually - you are purchasing quantity of each ingredient in multiple of what one bread requires
For example, if I can purchase ingredient $C$ only in multiple of $5$ kg packs and a bread requires $7$ kg, I need to purchase $35$ kg of the ingredient (making $5$ breads) or in multiple of that otherwise I would have leftover.
$5$ kg packs and $7$ kg required for a bread - so you need to purchase $7$ packs and you need to make $5$ breads.
ii) Now combining all ingredients - you are also purchasing quantity of each ingredient such that they can all make same number of breads.
Taking the above example, ingredient $C$ will have no leftover if I make $5$ breads.
But now if I take $B$, ingredient $B$ is available in $4$ kg pack and each bread requires $3$ kg. So ingredient $B$ will have no leftover if make $4$ breads and so purchase $3$ packs of $4$ kg each.
Ingredient $C$ requires me to make breads in multiple of $5$. Ingredient $B$ requires me to make breads in multiple of $4$. What does that mean? I of course need to make $5 \times 4 = 20$ (LCM) breads so I have no leftover of either the ingredient $B$ or of $C$.
Now if I combine $A$ into the mix, $A$ requires me to make breads in multiple of $3$ ($3$ kg packs and each bread requires $2$ kg).
That means I need to make $5 \times 4 \times 3 = 60 \,$ breads (LCM of $5, 4, 3$).
Now that we know the number of breads, we can find number of packs I need to purchase for each ingredient.
So to end it, for each ingredient, I find the minimum number of whole breads I need to make given how much is required for each bread and the package size it is available in. Then LCM of minimum number of whole breads I need to make for each ingredient gives me the number of breads I need to make. Rest follows from there.
