How can i solve this LCM puzzle?

Question

I have this puzzle "A woman went to the market and a horse broke the eggs she had in her basket. The owner of the horse offered her to pay for the damage caused. He asked her how many eggs were broken by his horse. She said she didnt knew , but remebered that when she arrenged them in pairs, there was one left. Same thing happened when she arrenged them in groups of 3, 4, 5 and 6. But when she arrenged them in groups of 7 there was none left. what is the minimum quantity of eggs that were in the basket?" I tried showing the equations

• 2a+1=x
• 3b+1=x
• 4c+1=x
• 5d+1=x
• 6e+1=x
• 7f=x

But I really dont know what to do with them

Answer 1

You can rewrite your equations as

• $x - 1 = 2a$

• $x - 1 = 3b$

• $x - 1 = 4c$

• $x - 1 = 5d$

• $x - 1 = 6e$

• $x = 7f$

The first five equations imply that $x-1$ is a multiple of $2,3,4,5$, and $6$.

The least common multiple of $2,3,4,5$, and $6$ is $60$

So $x = 60n + 1$ for some integer, $n$.

The last equation implies that $x$ is a multiple of $7$.

Since $56$ is a multiple of $7$ we can say $x = 56n + (4n + 1)$.

So $4n + 1$ needs to be a multiple of $7$.

So just keep adding $4$ to $1$ until you get a multiple of $7$

$$1, 5, 9, 13, 17, \color{red}{21}$$

where $21 = 1 + 4(5)$ implies that $n = 5$

So $x = 60(5)+1 = 301$

CONFIRMATION

• $301 - 1 = 2 \cdot 150$

• $301 - 1 = 3 \cdot 100$

• $301 - 1 = 4\cdot 75$

• $301 - 1 = 5 \cdot 60$

• $301 - 1 = 6 \cdot 50$

• $301 = 7 \cdot 43$

Answer 2

Hint

Let $m$ be the least common multiple of $2,3,4,5,6$.

Find $x$ such that $x \equiv +1 \pmod{m}$ and $x \equiv 0\pmod{7}.$

If the Chinese Remainder Theorem is not your thing...

if $a \equiv +1 \pmod{m}$, then so is $\{a + m, a+2m, a + 3m, \cdots\}.$

Find the element in that set that is a multiple of $7$.

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Edit
If $b \equiv m\pmod{7}$, then $(1 + kb) \equiv (1 + km) \pmod{7}.$