Question

1. dig deeper! a store sells cashews in $\frac{2}{3}$-pound bags. you buy some bags and repackage the cashews into 1-pound bags. what is the least number of bags you should buy so that you do not have any cashews left over?

Explanation:

Step1: Define the problem

We need to find the least number of $\frac{2}{3}$-pound bags of cashews to buy so that when repackaged into 1-pound bags, there is no cashew left over. Let the number of $\frac{2}{3}$-pound bags be $n$. The total weight of cashews is $\frac{2}{3}n$ pounds, and this should be a multiple of 1 pound (i.e., an integer). So we need to find the smallest positive integer $n$ such that $\frac{2}{3}n$ is an integer.

Step2: Solve for n

We want $\frac{2}{3}n = k$, where $k$ is a positive integer. To make $k$ an integer, $n$ must be a multiple of 3 (because the denominator is 3). Let's start with the smallest positive multiple of 3, which is $n = 3$. Then $\frac{2}{3} \times 3 = 2$, which is an integer. Wait, but let's check $n = 3$: total weight is 2 pounds, which can be repackaged into 2 one - pound bags. But wait, maybe we made a mistake. Wait, actually, we want the total weight to be a multiple of 1 pound, so $\frac{2}{3}n=m$, where $m\in\mathbb{Z}^+$. We can rewrite this as $n=\frac{3m}{2}$. For $n$ to be an integer, $m$ must be an even integer. The smallest positive even integer is $m = 2$. Then $n=\frac{3\times2}{2}=3$. Wait, but if $m = 1$, $n=\frac{3}{2}$, not an integer. If $m = 2$, $n = 3$. Let's verify: 3 bags of $\frac{2}{3}$ pounds each: $3\times\frac{2}{3}=2$ pounds, which can be repackaged into 2 one - pound bags with no remainder. But wait, is there a smaller $n$? Let's check $n = 1$: $\frac{2}{3}$ pound, can't make a 1 - pound bag. $n = 2$: $2\times\frac{2}{3}=\frac{4}{3}$ pounds, which is not an integer. $n = 3$: $3\times\frac{2}{3}=2$ pounds, which is an integer. Wait, but maybe we need the total weight to be a multiple of 1, so the least common multiple of the numerator and denominator? Wait, the weight per bag is $\frac{2}{3}$ pounds. We need the total weight $W=\frac{2}{3}n$ to be an integer. So we need $\frac{2n}{3}\in\mathbb{Z}$. This implies that $3$ divides $2n$. Since 2 and 3 are coprime, 3 divides $n$. So the smallest positive integer $n$ divisible by 3 is $n = 3$. Wait, but let's think again. Suppose we want to have a total weight that is a whole number of pounds. So we need $\frac{2}{3}n=1,2,3,\cdots$. The smallest $n$ for which $\frac{2}{3}n$ is an integer is when $n = 3$ (since $\frac{2}{3}\times3 = 2$), or wait, if we want the total weight to be at least 1 pound, the first time when $\frac{2}{3}n$ is an integer is when $n = 3$ (gives 2 pounds) or wait, maybe I messed up. Wait, no, let's take $n = 3$: 3 bags of $\frac{2}{3}$ pounds: total weight is 2 pounds, which can be repackaged into 2 one - pound bags. If we take $n = 3$, we have no cashews left over. But wait, is there a smaller $n$? Let's check $n = 3$ is the first $n$ where $\frac{2}{3}n$ is an integer. Wait, actually, the problem is to find the least number of bags so that when repackaged into 1 - pound bags, there is no remainder. So the total weight must be a positive integer. So we need $\frac{2}{3}n = k$, $k\in\mathbb{N}$. So $n=\frac{3k}{2}$. The smallest positive integer $k$ such that $n$ is an integer is $k = 2$, so $n = 3$.

Wait, but maybe another approach: we need to find the least common multiple of the denominators? No, the weight per bag is $\frac{2}{3}$, so to get a whole number of pounds, the number of bags times $\frac{2}{3}$ must be a whole number. So we need to find the smallest $n$ where $\frac{2n}{3}$ is a whole number. So $2n$ must be divisible by 3. Since 2 and 3 are coprime, $n$ must be divisible by 3. The smallest positive integer divisible by 3 is 3. So $n = 3$.

Answer:

The least number of bags is 3.