Question

Seven thieves robbed a kings treasury and with great difficulty escaped the palace and into the jungle. After a long walk thru the woods in the middle of night, they decided to stop and nap for a while.
Two thieves woke up early and started counting the heist. They found that all the jewels were identical diamonds. They tried to figure out how to split . If they divided the total number of jewels between 2 of them they had 1 jewel left. If they divided among 3, they had 1 left. If they divided among 4, they still had 1 left. If they divided it among 5, again they had 1 left. If they divided it among 6, they still had 1 jewel left. Finally they decided to wake the seventh thief and divided it among the 7 of them , now they didn’t have any left.

How many jewels did they steal?​

Answer 1

Let the number of jewels stolen be J.

• When divided between two thieves, 1 jewel was left:
• $\frac{J}{2}\to$ remainder = 1

• When divided among three thieves, 1 jewel was left:
• $\frac{J}{3}\to$ remainder = 1

• When divided among four thieves, 1 jewel was left:
• $\frac{J}{4}\to$ remainder = 1

• When divided among five thieves, 1 jewel was left:
• $\frac{J}{5}\to$ remainder = 1

• When divided among six thieves, 1 jewel was left:
• $\frac{J}{6}\to$ remainder = 1

• When divided among seven thieves, no jewel was left:
• $\frac{J}{7}\to$ remainder = 0

From the last point, we conclude that J is a multiple of 7.

• Here, LCM (2, 3, 4, 5, 6) = 60
• Since 1 jewel was left each time except for the last one, we find the value of J by putting r = 1, 2, 3, 4, 5, 6, ... in J = 60r + 1.

• When r = 1, J = 61, not a multiple of 7.
• When r = 2, J = 121, not a multiple of 7.
• When r = 3, J = 181, not a multiple of 7.
• When r = 4, J = 241, not a multiple of 7.
• When r = 5, J = 301, a multiple of 7.

∴ they stole 301 jewels in total.