Rob figures this out quickly.
Rob makes a new pile of coins containing one coin from the first pile, 2 coins from the 2nd pile, 3 from the 3rd, etc... all the way up to 10 from the 10th pile. His new pile has 55 coins (1+2+3+...+10). If his coins all weighed the correct weight (say the correct weight of one Loonie is X grams), then the total weight of the new pile should be 55X grams. But since the new pile contains some lighter coins, it actually weighs less, say Y grams. X-Y must be an even number of grams, as each "light Loonie" is exactly one gram lighter than each "correct-weight Loonie". Therefore:
 | If X-Y is 1 (gram), then one coin is too light, and it must be from pile 1. Pile 1 is therefore the bogus pile. |
| If X-Y is 2 (grams), then two coins are too light, and they must be from pile 2. Pile 2 is therefore the bogus pile |
| etc. |
Rob the takes all 90 correct-weight loonies, and takes his co-workers out for ice cream.
