Sky-High I.Q.

[Gary K]

The Bye-Bye Sky-High I.Q. Murder Case

Columbo must solve a murder case involving a genius accountant. The accountant, Oliver Brandt, gives Columbo a puzzle that some genius' are asked.

In a room there's several sacks of gold . . . as many sacks as you want. One sack contains artificial gold. Oliver says, the real gold coins weigh 1 pound, and the artificial gold weighs whatever you want. Columbo says; let's say 1 lb., and 1 ounce, and Oliver says, okay. Now Oliver says, there's a penny scale in the room, but you only have one penny. Columbo says; a scale that gives you a card with your weight on it, and Oliver says . . . correct. Now Oliver says . . . which sack has the artificial gold in it?

Later in the show, Columbo says to Oliver . . . one more question sir? If I place all those sacks of gold on the scale, and remove them, one by one? Oliver says, no, you use your penny, you get a card with the weight on it. Columbo said, I'll have to work on that tonight with Mrs. Columbo.

Hint: When Columbo gives him the answer, he used 3 bags in his calculation.

Whats your answer?

[Steve Holbrook]

Soooo........What's the answer?

[Gary K]

Talk about timing Steve Holbrook . . . I just copied the answer off of my files, as I was going to post it! You must of read my mind?

Here's Lieutenant Columbo's answer to the gold sack puzzle:

Answer: Columbo say's to Oliver Brandt . . . I know exactly what you did, and how you did it . . . "The gold sack puzzle sir." Mrs. Columbo figured it out. You said there was these sacks of gold in the room, and while Columbo was saying that, he picks up 3 little bags of chocolate coins wrapped in golden colored wrappers. Columbo went on to say . . . let's suppose there was 3 sacks of gold, as I have them right here, then holds up those 3 bags. Now Columbo goes over to a small kitchen scale, and lays the 3 bags down. He takes one coin out of bag one, two coins out of bag two, and three coins out of bag three. Now Columbo says, we have 6 coins, and we place all of them on the scale. If all 6 pieces were the real gold, it'd weigh 6 lbs. Now, if the first bag had the phony pieces, it'd weigh 6 lbs., and 1 oz. If the second bag had the phony gold, it'd weigh 6 lbs., and 2 oz., and if it were the third bag, it'd weigh 6 lbs., and 3 oz.

[Gary K]

FWD: Nine balls and a balance scale

There's 9 identical balls . . . all of the same size, and color, except one is slightly heavier than the rest. There's a balance scale, and you can only use it two times.

How would you weigh them, so you can find the heavy one?

[Gary K]

FWD: Math Competition

May 13, 2000

An eighth grader from Auroa, Ill., won the national MathCounts competition Friday, solving an algebra problem to beat out 227 other “mathletes.”

Ruozhou “Joe” Jia, 14, a pupil at Washington Middle School, solved the equation in 15 seconds. Contestants in the finals have 45 seconds to solve each problem.

The final question was: A lecture hall has 40 rows of seats. There are 10 seats in the first row, 12 seats in the second row, and so on, with two more seats in each row than in the previous row. How many seats are in the lecture hall?

Ruozhou said he computed his answer on pencil and paper, using the following algebraic formula to solve his problem: (a1 + an)n/2.

Needless to say, it took me longer than 15 seconds!

What's your answer?

[BobRR]

I hate algabra!!! Long hand, I have 476 seats. BobRR

[Gary K]

Thanks BobRR for trying to answer this math quiz. To be truthful, I couldn't make heads or tails of Ruozhou's formula either, so I drew out the question, and seen it had the shape of a trapezoid . . . see answers below:


Ruozhou's answer: 1960 seats.

My answer: The lecture hall is in the shape of a trapezoid, so I used the known to find the unknown. There's 10 seats in the front row, with 2 additional seats than in the previous one, and there's 40 rows.

Then 2 x 39 = 78 + 10 = 88 seats in the back row.

In a trapezoid you add the two bases, and multiply times the height altitude, and divide by 2 to find the area. Then; (10 + 88) x (40) / 2 = 1960 seats.

Gary K

[BobRR]

Sure missed that! BobRR

[Gary K]

FWD: Nine balls and a balance scale

There's 9 identical balls . . . all of the same size, and color, except one is slightly heavier than the rest. There's a balance scale, and you can only use it two times.

How would you weigh them, so you can find the heavy one?

Answer: Place 3 balls on each side of the balance scale. If it balances, then it's one of the three remaining balls. In which case, you'd place one ball on each side of the scale to expose the heavy one. If it balances, it's the remaining ball. The same process is used if the scale indicates it's one of the three when weighed the first time!