One
Puzzler entrant gets answers
correct, but can't be winner
One entrant in Young Saint Louis.com's December Math
Puzzler contest got all the answers right. But, she can't
be a gift certificate winner because her entry came in late.
Thirteen-year-old
Kelsey Provance answered all six of the Puzzlers correctly.
But, the Green Park Lutheran School student couldn't win
a $10 Borders' gift certificate. She didn't get her entry
in the mail before the Dec. 15 deadline.
But,
YSL.com wanted to mention Kelsey's good work because
the December Puzzlers proved to be difficult. She was the
only entrant who got all the answers correct.
Puzzlers
No. 1 and No. 6 were particularly difficult. In No. 1, there
were a lot of unknowns. In No. 6, a lot of entrants figured
out an answer but it wasn't the "cheapest" answer.
For that, you had to think "out of the box."
The
answers to all of the Puzzlers are listed below. In the
case of Puzzler No. 2, several entrants gave an answer that
was correct although the numbers were in different order
than Mr. Math Puzzler's answer.
Each
month, YSL.com publishes a series of six Math Puzzlers.
These are primarily to let our viewers have fun with math.
These questions aren't designed as a class assignment.
YSL.com
believes math can be fun. And the more fun you have, the
more likely you are to learn this important subject.
Answers
to December, 2001, Math Puzzlers
1. You have seven
coins. Five of them weigh the same and two are slightly heavier.
(The heavier coins are equal in weight to each other.) Using
only the balance scale, which is the fewest number of weightings
needed to determine which coins are the heavier ones?
Answer: Up to 3 weighings
Explanation:
This answer is fairly easy to understand with illustrations.
But, we'll be using words so it'll take a little longer.
The first weighing
involves putting three coins on each side of the balance,
leaving one coin off the bar. If the balance is even, that
would mean there is one heavier coin on each side of the balance
and the left-out coin was a light one. If the balance goes
down on one side or the other, it means you have one or two
heavy coins on the side that dips. (Remember, the balance
could go down with only one heavy coin, which would then mean
the seventh coin you left off might be a heavy one.) You won't
know for sure until the second weighing.
The second
weighing involves splitting up the three coins on one side.
In the case of the even-balance, you're looking to find the
"heavier" coin on each side. If this weighing finds
the balance with one coin on each side still even, the "heavier"
coin is the one left off. Then a third weighing with the three
coins on the other side will locate other "heavier"
coin.
In the case
of the uneven-balance, your second weighing focuses on the
"heavy" side by putting one coin on each side and
leaving one off. If the balance is even, both of the coins
on the bar are "heavier." If the balance dips, it
means the left-off coin and the coin on the side that dips
are the "heavier" ones.
2. Place the digits
1 through 7 in the boxes below so that each three-box-line,
including the diagonals, add up to 12.
Explanation:
This is a matter of trying different combinations until they
add up. Entrants in the contest found a number of different
combinations that worked. The above answer is just one of
the acceptable ones.
3. What is the
smallest number that can be divided by 6, leaving a remainder
of 5; divided by 5 and leaving 4, and divided by 4 and leaving
3?
Answer: 59
Explanation:
This involves finding the least common multiplier (LCM) and
then subtract one to leave each original number with the right
remainder. In this case, the LCM of 6, 5 and 4 is 60. Then,
to get the remainder, you take one off, or 59.
4. Ron and Jane
compete in a car rally, going several times around a closed
circuit. Ron can drive the circuit in 25 minutes, but Jane
takes 30 minutes. If the two drivers start at the same time,
how long will it take Ron to lap Jane?
Answer: 2 1/2 hours
Explanation:
This also is an LCM problem. The LCM of both 30 minutes and
25 minutes is 150 minutes, which reduces to 2 1/2 hours.
5. Pymm has many
dragons. A few years ago, one of these dragons, Alaranthus,
though not fully grown, weighed one thousand pounds plus two-thirds
of his own weight. How much did Alaranthus weigh?
Answer:
3,000 pounds
Explanation:
This is an algebra problem. Alaranthus' weight is W, therefore
the formula is
W = 1,000 +
2/3 W. When you subtract 2/3 W from both sides of the equation,
you have
1/3W = 1,000.
To find out the value of W, you multiply both sides by 3 so
that W = 3,000 pounds.
6. Betty has 13
chains with three links in each chain. She would like to connect
all 13 chains together to form one continuous circular chain.
If a jeweler charges $4 to cut a link open and $10 to weld
it back together again, what is the cheapest cost by which
Betty can have this done?
Answer:
$140
Explanation:
Sometimes you need to "think outside the box." A
common way to find this answer would be to cut a link in each
of the 13 three-link segments and reweld all. That answer
would be $182. However, you were to find the cheapest
way. To do that, you keep 10 of the segments intact and cut
each link (9 in total) of the other three segments. Using
the nine links, you can cut and weld together nine of the
segments ($14 each or $126). Then to make the final link,
you need to cut and reweld only one link of the other 10 segments
to put the whole thing together; that's another $14. The total
is then $140.