Solution Step 0/0
A woman was carrying a basket of eggs to market when a passerby bumped her. She dropped the basket and all the eggs broke. The passerby, wishing to pay for her loss, asked:
“How many eggs were in your basket?“
“I don't remember exactly,“ the woman replied, “but I do recall that whether I divided the eggs by 2, 3, 4, 5, or 6, there was always 1 egg left over. When I took the eggs out in groups of 7, I emptied the basket.“
What is the least number of eggs that broke?
“How many eggs were in your basket?“
“I don't remember exactly,“ the woman replied, “but I do recall that whether I divided the eggs by 2, 3, 4, 5, or 6, there was always 1 egg left over. When I took the eggs out in groups of 7, I emptied the basket.“
What is the least number of eggs that broke?
Explanations
The lowest common multiple of 2, 3, 4, 5, and 6 is 60. We have to find a multiple of 7 that is larger by 1 than a multiple of 60. Now:60n + 1 = (7 x 8n) + 4n + 1.
The number (60n + 1) is divisible by 7 if (4n + 1) is divisible by 7. The lowest value of n that satisfies this condition is 5.
Therefore, there were 301 eggs in the basket.
[After Boris A. Kordemsky. The Moscow Puzzles: 359 mathematical recreations. 1972]
Check 101 eggs
Check 201 eggs
Check 301 eggs
Check 401 eggs
