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For the holiday, Isabella is putting Easter eggs into as many baskets as needed until she runs out of eggs. If she places 3 eggs into each basket except one, she'll have to place 2 eggs in the final basket. If she places 5 eggs into each basket except one, she'll also have to place 2 eggs into the final basket. If Isabella has fewer than 20 eggs total, how many eggs will she have to put in the final basket if she places 7 in all other baskets?
A.0
B.1
C.2
D.3
E.4

Why is knowing the total number of baskets unnecessary to solve this problem?

InanimateBeing
  • 3,074
  • The total number of baskets depends on the strategy for distributing eggs, and is not a constant. If she has 17 eggs, she can put 3 eggs in each of 5 baskets and 2 eggs in number 6. Or she can put 5 eggs in each of 3 baskets and 2 eggs in number 4. – aschepler Aug 17 '22 at 18:59
  • There are only two positive integers $(n)$, less than $(20)$, such that $~n \equiv 2\pmod{3}~$ and $~n \equiv 2\pmod{5}.~$ That is, $n$ must be an element in ${2,17}.$ $(n=2)~$ is eliminated by the problem's description. – user2661923 Aug 17 '22 at 23:53
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2 Answers2

1

Let $n$ be the total number of eggs.
Then, we have:
$n=3x+2$
$n=5y+2$
$\Rightarrow 3x=5y$

Since $n\in N\Rightarrow x,y\in N$. So we for each $y=3,6,9\dots$ we'll get $x\in N$.
$\Rightarrow n=17,32,47\dots$

But, $n<20$ (given)
$\Rightarrow n=17$

So, remainder when $17$ is divided by $7$ is $3$.
Thus, $3$ is the answer.

Since, we are given that:

putting Easter eggs into as many baskets as needed until she runs out of eggs

So, this means that $x,y$ may take any (natural number) value.
Even if she had $1000$ eggs and she put $1$ in each, she would be able to do so as she has no problem in arranging $1000$ baskets for the task.
Image she has hired a basket providing agency whose motto is "YOU ASK, WE DELIVER".

InanimateBeing
  • 3,074
0

Let $b_1$ be the number of baskets the first time and $b_2$ be the number of baskets the second time and $e$ the number of eggs.

We have:

$3(b_1 - 1) + 2 = e \implies 3b_1 - 1 = e$

and

$5(b_2 - 1) + 2 = e \implies 5b_2 - 3 = e$

So we get $5b_2 - 3b_1 = 2 \implies 3b_1 + 2 = 0 \mod 5 \implies b_1 = 1 \mod 5$

Trying $b_1 = 6$, we would get $e = 17$ and hence the answer is $3$.

sku
  • 2,643