VtProblem 1 – Give Up – Score: 0 / 7 (1632)
Problem:
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Suppose
and
common fraction.
Solution:
Solution 1:
From the given equations,
. What is the value of
and
? Express your answer as a
, so
Solution 2:
We have
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
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VtProblem 2 – Give Up – Score: 0 / 7 (1628)
Problem:
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The MATHCOUNTS Sprint Round competition consists of 30 problems with a time limit of
40 minutes. Suppose you complete 20 Sprint Round problems in 25 minutes. On average,
how many times as long will you be able to spend on each remaining problem than you did
on each of the first 20? Express your answer as a common fraction.
Solution:
The average time you spent per problem for the first 20 problems is
minutes.
There are
minutes remaining, and
problems remaining, so
the average time per remaining problem is
minutes. Hence, you can spend
times longer on the remaining problems.
Your Response(s):

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VtProblem 3 – Give Up – Score: 0 / 7 (8981)
Problem:
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Richard needs to fly from San Diego to Halifax, Nova Scotia and back in order to give an
important talk about mathematics. On the way to Halifax, he will get a speed boost from the
wind which blows at 50 miles per hour (mph). On the way back, the plane must,
unfortunately, fight this wind speed. If the talk lasts 2 hours, and if the distance between the
two cities is 3000 miles, how fast must the plane fly in mph if the entire trip is to take 13
hours?
Solution:
Let the plane's speed be . With the wind helping, the plane travels at
to the ground. So, it takes
way back takes
hours to make the trip to Halifax. Using similar logic, the
hours. Combining these with the 2 hours needed to give the talk
yields the total time of 13 hours:
both sides by
This simplifies to
, so since
mph relative
Subtracting 2 and multiplying
gives:
. This factors as
must be positive we have the solution
.
You might also simply use a little trial and error on the main time equation above. The two
fractions will be roughly equal and must contribute 11 total hours to the left hand side.
When is near
, the fractions are about each; so, trying a few values of quickly gets
to the answer with no messy algebra.
Your Response(s):
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VtProblem 4 – Give Up – Score: 0 / 7 (8979)
Problem:
Suppose and
value of
Solution:
are the solutions to the quadratic equation
.
By Vieta's formulas, we know that
and
.
Thus,
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. Find the
.
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VtProblem 5 – Give Up – Score: 0 / 7 (8980)
Problem:
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. Find the value
Let and be the roots of the quadratic equation
of
.
Solution:
Solution 1:
With a bit of cleverness, we may
write
.
But
quadratic.
, because this is what it means for
This leaves
to be a root of the given
. By Vieta's formula, we know
, so
.
Solution 2:
You don't necessarily need to be clever: we can also substitute
our expression to the equivalent
to change
Notice, interestingly, that we can calculate this expression without knowing the values
of and , and moreover, without knowing which root is being called , and which is !
Hint(s):
Can you substitute for ?
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VtProblem 6 – Give Up – Score: 0 / 7 (8978)
Problem:
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How many three-digit palindromes (numbers that read the same forward and backward)
satisfy the following property: the sum of three things -- the hundreds digit, the units digit,
and the product of the units and tens digits -- is eight more than the tens digit?
Solution:
The first step in this problem is decoding the property described. We let our number
be
where is non-zero. With this algebraic setup, we
need:
.
Simplifying, we have
, or
factoring trick, we subtract 2 from both sides to get:
gives
, or
Factoring
:
,
as
,
,
,
, and
only
answers:
,
Your Response(s):

. To use Simon's
. This
.
, and
gives use four possibilities for the pair
. Since
cannot be a digit, we have
, and
.
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VtProblem 7 – Give Up – Score: 0 / 7 (1637)
Problem:
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For what positive value of
will the following equation be true if
?
Solution:
Expanding
, we get
which simplifies to
Then
. Isolating
satisfies this equation is
Your Response(s):
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, we find
. The positive value of
that
.
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VtProblem 8 – Give Up – Score: 0 / 7 (1624)
Problem:
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Solve:
. Write your answers as a list of numbers, separated by
commas, e.g. "23, 19" (but without the quotes).
Solution:
Solution 1:
If
, then either
or
. Hence, if
, then
either
or
.
If
the solutions are
, then
. If
, then
. Therefore,
.
Solution 2:
Another approach is to square both sides. Squaring both sides, we
get
(because
for all ). This simplifies
to
, which factors as
, so the solutions
are
and
, as before. You must be careful when using this approach
because squaring both sides of an equation can introduce false solutions. Thus, we need to
check that both of these "solutions" work in the original expression before declaring them
correct.
Your Response(s):
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VtProblem 9 – Give Up – Score: 0 / 7 (1629)
Problem:
The product of four consecutive positive integers is 1 less than
these four numbers?
Solution:
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. What is the least of
By difference of squares,
factors of 460 and 462 that are close to each other, and we
find
and
, so
. We look for
. The least number
is
.
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
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VtProblem 10 – Give Up – Score: 0 / 7 (1630)
Problem:
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Two trains travel directly toward each other. One of the trains travels at a rate of 12 km/h
while the other travels at a rate of 20 km/h. When the trains are 72 km apart, a conductor at
the front of one of the trains releases the insane pigeon Hyde. Hyde flies first from the
slower of the two trains to the faster train at which point Hyde doubles back toward the
slower train. Hyde continues to fly back and forth between the trains as they approach,
always at a constant speed of 48 km/h. Assuming the trains never change speed until they
meet and magically stop, how many kilometers has Hyde flown when the trains meet?
Solution:
From the point of view of one train, the other train is travelling at a rate
of
km/h. Therefore, it takes
hours for the trains to meet. Hyde
flies at a rate of 48 km/h, so he covers a total distance of
km.
Hint(s):
You could try to plot Hyde's path, but this approach is very complicated. You know Hyde's
speed, so you can compute how far Hyde has flown simply by computing the time he is in
the air.
Your Response(s):
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VtProblem 11 – Give Up – Score: 0 / 7 (1633)
Problem:
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The terms of a particular sequence are determined according to the following rules:
* If the value of a given term is an odd positive integer , then the value of the following term
is
* If the value of a given term is an even positive integer , then the value of the following
term is
.
Suppose that the terms of the sequence alternate between two positive
integers
. What is the sum of the two positive integers?
Solution:
Suppose that is odd. Then the next term is
, which is even. Then the term
after that is
. Thus, we must solve the system of equations
Substituting the first equation into the second equation, we get
. Solving
for , we find
, so
, and
.
If is even, then with the same calculations, we find that the solution is
Your Response(s):

and
.
& Give Up
VtProblem 12 – Give Up – Score: 0 / 7 (1636)
Problem:
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Given positive integers and
possible positive value for
Solution:
We are given that
Multiplying both sides by
such that
?
, we get
and
, what is the smallest
, which we can write as
We can then apply Simon's Favorite Factoring Trick, by adding
sides:
to both
The left-hand side factors:
We want to minimize
factors
and
, which is equivalent to minimizing the sum of the two
.
Since the product of
and
is a constant, we can minimize their sum by
making them as close to each other as possible. Normally, we could
set
, but the problem states that
. The next two factors of 144
(whose product is 144) that are close to 12 as possible are 9 and 16. Hence, we can
set
and
, to get
and
, so the minimum value
of
is
.