You’re a pharmacist and you’ve just taken delivery of ten bottles of 1,000 pills each. All pills are of the same type with the same dosage. But before you have a chance to put them away, your supplier calls to inform you that, due to a glitch at the factory, one of the ten bottles is tainted. The pills are supposed to contain 10 milligrams of medication, but all of the pills in the bad bottle contain one extra milligram.

Obviously, you can’t allow your customers to buy the overdosed pills, but this medication is very expensive so you can’t afford to throw away the whole lot. Fortunately, you have a smart assistant, who suggests weighing the pills. “Brilliant!”, you exclaim, “All we have to do is weigh each bottle – nine will register the same weight and one bottle will weigh an extra 1,000 mg”. “You could do it that way”, adds your assistant with a sly grin, “but that could take up to ten weighings. I can think of a plan that’s guaranteed to find the tainted bottle in only one weighing”.

What was your assistant’s plan? Hint: you may open the bottles and weigh any combination of pills and/or bottles you like. Leave your guess in a comment below. Today’s puzzle is adapted from the book *Aha!* by the late, great puzzlemaster Martin Gardner.

**Solution:** The key insight comes from the observation that if you weigh a different number of pills from each bottle then the excess weight can be used to identify the bad bottle. More specifically…

- Mark each bottle with a unique number from 1 to 10.
- Take one randomly selected pill from bottle one, two pills from bottle two, etc. Weigh the resulting 55 pills together (1+2+3+4+5+6+7+8+9+10 = 55) and note the result.
- If all 55 pills were legitimate, the expected result would be 550 mg (10 mg per pill times 55 pills) but the actual result is going to exceed the expected weight because you’ve included some number of overdosed pills in your sample. Subtract the actual weight from 550 mg to find the number of extra milligrams and, hence, the number of bad pills in your sample.
- Because you included a different number of pills from each bottle, you can trace the number of bad pills directly to the bad bottle. One bad pill implicates bottle one, two bad pills implicate bottle two, etc.

Hat’s off to Simon Banks, Doug Needham and Kimberly Cohen for submitting correct answers.

Morning Marc,

Label each bottle from 1-10, and then take twice that number of tablets from each of the bottles. e.g. #1 (take two tablets), #5 (table 10 tablets), #10 (take 20 tablets).

The perfectionist in me also says put the tablets from each bottle in paper bags so once you identify them you can put them back in the right jar. ;-)

Weigh the tablets together and if all the tablets equaled 10mg they would total:-

20+40+60+80+100+120+140+160+180+200 = 1100

So we can identify which bottle has the most tablets by the discrepancy of 1mg for each of the tablets.

22+40+60+80+100+120+140+160+180+200 = 1102 (bottle 1)

20+44+60+80+100+120+140+160+180+200 = 1104 (bottle 2)

20+40+66+80+100+120+140+160+180+200 = 1106 (bottle 3)

20+40+60+88+100+120+140+160+180+200 = 1108 (bottle 4)

20+40+60+80+110+120+140+160+180+200 = 1110 (bottle 5)

20+40+60+80+100+133+140+160+180+200 = 1113 (bottle 6)

20+40+60+80+100+120+154+160+180+200 = 1114 (bottle 7)

20+40+60+80+100+120+140+176+180+200 = 1116 (bottle 8)

20+40+60+80+100+120+140+160+198+200 = 1118 (bottle 9)

20+40+60+80+100+120+140+160+180+220 = 1120 (bottle 10)

Let’s suppose the bottles are numbered 1 – 10. Take 1 pill from bottle #1, 2 pills from bottle #2, and so on up through the last bottle. You will have a total of 1+2+3+4+5+…+10 = 55 pills. So, let’s weigh the 55 pills (all at once). If all the pills weighed 10 milligrams, the total weight would be 550 mg. However, we know that some bottle contains pills that weigh 11 mg each. So, when we weigh the 55 pills, if the result is

551 mg, we know bottle #1 is the one with the “tainted” pills. If the weight is 552 mg, bottle #2 is the culprit. The pattern continues… and if the weight is 560 mg, then bottle #10 would be the bad one.

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