# Math egg puzzle Help

• January 24th 2007, 06:23 AM
Dragon
Math egg puzzle Help
a peddler is taking edge to the market to sell. The eggs are in a cart that can hold 500 eggs. If the eggs are removed fom the cart 2 3 4 5 or 6 at a time 1 egg s alwas left over If the eggs are removed 7 at a time no eggs are left over how many eggs are their in the cart?
• January 24th 2007, 07:20 AM
Soroban
Hello, Dragon!

Quote:

A peddler is taking eggs to the market to sell.
The eggs are in a cart that can hold 500 eggs.
If the eggs are removed fom the cart 2, 3, 4, 5 or 6 at a time,
. . one egg is always left over.
If the eggs are removed 7 at a time, no eggs are left over.
How many eggs are their in the cart?

Let $N$ be the number of eggs.

The LCM of 2, 3, 4, 5 and 6 is $60$.
. . Hence, $60$ (and any multiple of 60) is divisible by 2, 3, 4, 5 and 6.
To have a remainder of 1, $N$ must be of the form: . $60k + 1$
. . for some positive integer $k$.
That is: $N \:=\:60k + 1$ [1]

Since $N$ is divisible by $7\!:\;\;60k + 1 \:=\:7a$

If you're familiar with Modulo Arithmetic: . $60k \,\equiv \,-1\!\! \pmod{7}$

Reduce: . $4k \,\equiv \,6\!\! \pmod{7}$

Multiply by 2: . $8k\,\equiv\,12\!\!\pmod{7}\quad\Rightarrow\quad k\,\equiv\,5\!\!\pmod{7}$

Hence, $k$ is of the form: $7b + 5$ for some integer $b$.
That is: $k \:=\:7b + 5$ [2]

Substitute [2] into [1]: . $N \:=\:60(7b + 5) + 1\:=\:420b + 301$

Since $0 < N \leq 500$, then $b = 0$

Therefore: . $\boxed{N \,= \,301}$

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If you're not familiar with Modulo Arithmetic,
. . it can be solved with "ordinary" algebra.
It just takes longer.