Goldman Sachs

Interview Question

Technology Analyst Summer Intern Interview

An analytical question about: Say you have ten stacks of

  coins with ten coins in each stack, each coin weighing 1g. However, one of the stacks has 2g coins instead of 1g coins. If you have a weighing scale and can only make one measurement, how do you determine which stack of coins weighs 2g each coin?

Interview Answer

6 Answers


1-If all the coins are made from the same material and same diameter, then the heavier coin stack should be slightly taller.

2- Weight all the stacks at once, and remove each stack until you see a dip in the tota weight .

Just 'cause on Mar 28, 2013

I didn't quite understand what you meant. So there are 10 coins, 9 weigh 1 gram, and one weighs 2 grams? The second sentence seems contradictory.

If you can only make one measurement, then you would divide the stacks into two. Place each stack on either scale, and whichever one is heavier contains the 2g coin.

Anonymous on May 1, 2013

No, there are 10 stacks of coins with 10 coins in each stack. 9 of the stacks have 10 coins that weigh 1g in them and 1 stack has 10 coins that weigh 2g. Your goal is to find which stack out of the 10 total stacks contains coins that weigh 2g and not choose either of the 9 stacks that contain coins that weigh 1g when you can only do one measurement

Anonymous on May 2, 2013

And replying to Just'cause; can't use that method, as observing the change in total weight as you remove coins constitutes taking/observing another measurement. I asked the interviewer if I could do that :)

Anonymous on May 2, 2013

gauss' trick- more or less. assemble as 1+10, 2+9, 3+8 etc. weigh that. Number over 55 is the number of the stack with 2g coins.

Anonymous on May 22, 2013

It's simple. Take 1 coin from the first stack, 2 coins from the second, 3 from the thirds - you see where this is going - finally 10 from tenth stack. put them on weighing scale.

If in result you get 56 - it was first stack, 57 - second.... 65 - tenth stack.

kuba on Sep 26, 2013

Add Answers or Comments

To comment on this, Sign In or Sign Up.