Working backwards, since Jin had chocolates left after eating half of
them, she had
chocolates after giving chocolates to Brian.
Since Jin gave chocolates to
Brian and had chocolates after
doing this, Jin was given
chocolates by Ingrid.
Ingrid gave one third of her chocolates to Jin, which means Ingrid
started with
chocolates.
Answer:
Since of is , we want to find so that of equals .
This means or and so
.
Answer:
Note that , and since there are
minutes in an hour, this means that minutes is equal to hours and minutes.
For hours and minutes to pass, hours will pass and then hours and minutes will pass.
In hours it will be 1:00
a.m. again.
In hours and minutes after 1:00 a.m., it will be
10:43 a.m.
Answer: 10:43
a.m.
As soon as two lockers are painted blue in the top row, the other
two lockers in the top row must be painted red.
Once the top row is painted, the colours of the lockers in the bottom
row are determined.
If the lockers in the top row are numbered , , , and , then there are six possibilities for
the two blue lockers.
They are and ,
and , and ,
and , and , and and .
Answer:
Let be on such that is perpendicular to as shown.
Since and are parallel and and are right angles, and are right angles as well, and
so is a rectangle.
Using and the fact that is a rectangle, we get .
Using , , and , we get .
In , and , and so
.
This means is
isosceles, and so .
By the Pythagorean theorem applied to right-angled , we have , and so .
Since , .
Answer:
Let be the amount of time
in hours that the train will take to make the trip if it is on time and
let be the distance in kilometres
between City A and City B.
If the train is minutes late,
then it is hours
late, which means it takes hours to
reach City B.
Similarly, if the train is
minutes early, then it takes hours to
reach City B.
From the given assumptions, this means and .
Cross multiplying these equations, we get and .
Equating the values of and
solving for , we get Substituting into gives
.
The speed in km/h that the train should travel in order to arrive on
time is .
Answer:
We are given that , but and has a right angle at , so . Since , .
Using that , we get by
similar reasoning. Therefore, .
Applying the Pythagorean theorem to , we get that .
Applying the Pythagorean theorem to , we get that .
The perimeter of is
Since the perimeter of is given to be , we conclude
that .
Answer:
Suppose is a Tim
number. That is, , , , , and are the digits of the number and since
it has five digits, .
Since is a multiple of , it must be a multiple of both and .
Since is a multiple of , either or .
Since is a multiple of , is a multiple of .
We are also given that and
.
Combining and with the fact that is a multiple of , we have that is a multiple of
.
Since is a multiple of , this means is a multiple of .
We are looking for triples
with the property that or and is a multiple of . However, we must keep in mind that
is a digit, which means
or .
We will consider two cases: and
. We note that once we choose
values for and , the value of will be determined.
: In this case, we
need to be a multiple of
, so is a multiple of .
Combined with the requirement that , this means or
. We also need to remember
that , which means is one of , , , , , , , , and for a total of nine
possibilities.
: In this case, we
need to be a multiple of
. For the same reason as in the
previous case, .
If , then which is not a multiple of
.
If , then which is a multiple of .
If , then which is not a multiple of
.
If , then which is not a multiple of
.
If , then which is a multiple of .
If , then which is not a multiple of
.
Therefore, we must have that or . Keeping in mind that , we get that can be , , , , , , or for a total of seven
possibilities.
This gives a total of Tim
numbers, which are listed below. One can check that all of these integers are indeed Tim
numbers.
Answer:
Solution 1
Multiplying by gives and multiplying by gives .
Adding and gives , and after dividing
through by , we get .
The fraction is in
lowest terms, which means and
, so .
Solution 2
We will try to solve the system of equations for , , and in terms of and and see what happens. For this
particular system, it is impossible to express , , and in terms of only and . However, a relationship between and will be revealed by the process of
attempting to solve the system.
Subtracting the second equation from the first, we get , and subtracting from this equation
gives .
After dividing by , we get .
Next, we subtract times the
second equation from times the
first equation to get
We now have two expressions for , and if we set them equal, we get an
equation involving and . We can manipulate this equation as
follows. and since is in lowest terms, and , so .
Answer:
Fix a positive integer and
let (we are interested in even
).
The two-element subsets of
consisting of two integers that have a sum of are , , , and so on up to . There are subsets in the list.
A subset of contains two integers with a sum
of exactly when it contains
both of the integers from at least one of the subsets above.
We will now count the number of subsets of with the property that no two
integers in the subset have a sum of .
We will do this by counting the number of subsets of that contain at most one of the
two integers from each of the
two-element subsets listed earlier.
Suppose is one of these
subsets. The set could contain neither nor , it could contain and not , and it could contain and not .
This gives possibilities for each
of the subsets.
Every integer in appears in
exactly one of these subsets, so
a choice of one of these three possibilities for each of the subsets completely determines .
These choices are independent, which means the number of subsets with the desired property is exactly
.
Since there are subsets
in total, we have
We want to find values of that
satisfy .
Using the formula above, this is equivalent to , which is equivalent to .
The table below has positive integer values of in the left column, the value of in the middle column, and in the right column.
|
|
|
1 |
3 |
1 |
2 |
9 |
4 |
3 |
27 |
16 |
4 |
81 |
64 |
5 |
243 |
256 |
We see that is the smallest
even positive integer for which , so the answer is .
Answer:
The cod accounts for of the total pieces
of fish sold.
Therefore, the number of pieces of cod sold was .
Answer:
Rearranging the given equation, we get which implies that .
Therefore, .
Answer:
The slope of is .
Since is neither horizontal
nor vertical, the slope of a line perpendicular to it is the
negative
of the reciprocal of , or .
Answer:
Factoring as a
difference of squares, we have .
Then
Answer:
Solution 1
Substituting into gives or .
Substituting into gives or which means .
Substituting into gives , and substituting into gives .
Solution 2
The equation is equivalent
to .
The equation is equivalent
to .
Substituting
into gives .
Substituting these values of
and into gives .
Therefore,
Answer:
If we let , then we
have , and since , we have . Thus, .
Since , , and so we can enumerate the
triples by considering possible values of starting at .
When , we get the triples , , , , , , , and for a total of triples.
When , we get the triples , , , , , and for a total of triples.
When , we get the triples , , , and for a total of triples. When , we get the triples and for a total of triples.
If , then since is at least , . This means cannot be any larger than , so we have found all triples.
In total, there are
triples.
Answer:
The positive integer is
even and the sum of its digits is which is a multiple of , so itself is divisible by both and . (Similar to the well-known
divisibility rule for , a positive
integer is divisible by exactly
when the sum of its digits is divisible by .)
Factoring, we have , and it can be checked that .
Therefore, , so the distinct prime factors of are , , , and , the sum of which is .
Answer:
Since every line segment must be used exactly once, the integer
will have seven digits with each digit from to used exactly once.
Observe that the path that goes from to to to to to to to creates the integer .
We will now argue that this is the largest integer that can be
created.
Since every digit from through
is used exactly once, if we start
with any digits other than and
, in that order, the integer
formed must be smaller than .
This means that the path forming the largest possible integer must start
with , then go to , and then .
From , the options are to go to
or . If the path continues to , then the integer would start with
, which is guaranteed to be
smaller than , regardless of
how the path is completed.
The path must start with , , , , so the first three digits of the
largest inter are .
The line segments connecting to
and to have already been used, which means
that the path must continue to
since otherwise we would have to reuse a line segment. We now have shown
that the first four digits of the largest integer are .
Since the path arrived at from
, there is nowhere to go from
but to , and then by similar reasoning, it must
continue to and finally to , so the largest integer must be .
Answer:
Solution 1
Let equal the volume of the
pool in litres and let be the
rate at which each hose outputs water in litres per hour.
With three hoses, the pool will fill in hours at a rate of litres per hour. This means or .
One of the hoses stops working after of the hours, which means that the hose stops
working when there are litres in the pool, or
more litres to be
added to the pool.
Starting at 11:00 a.m., the new rate at which water is entering the
pool is , and litres need to be added to
the pool.
If we let be the amount of time
in hours remaining to fill the pool with two hoses, then we have and so
.
Substituting , we
get Therefore, another hours are required, so the pool will
be full hours after 11:00
a.m., or at 9:30 p.m.
Solution 2
With three hoses working, the pool will fill in hours, which means that one hose could
fill the pool in hours. We will
say that the pool takes
hose-hours to fill.
At the time when one of the hoses stops working, three hoses have
been working for hours. In other
words, which is the same as if one hose had been working for hours. In other words, hose hours have been spent putting
water in the pool.
This means hose hours are
still required. There are two hoses still working, so it will take and
additional hours
to fill the pool.
Since hours after 11:00
a.m. is 9:30 p.m., the pool will be filled at 9:30 pm.
Answer: 9:30 p.m.
Substituting and into the two equations gives which can be simplified to
get
Dividing the first equation by
gives from which we can subtract to get .
Rearranging gives which
can be factored to get .
Therefore, there are two possible values of , which are and . Substituting these values into
and solving for gives and , respectively.
The ordered pairs and
are the two
possibilities. It can be checked that the two systems of equations
both have the
unique solution .
Answer: and
If we set , then Expanding, we have .
This means , so
the answer is .
Answer:
For the square root in the numerator to be defined, we need or , and for the square root in the
denominator to be defined, we need or . Therefore,
for the expression to be an integer, we must have .
Suppose is equal
to some integer . Then .
Squaring both sides, we get or . This equation can be rearranged to get and then solved for
to get .
Rearranging, we have Observe that for all , so the cancellation above is
justified.
Since and are both integers, must also be an
integer, so is a multiple of
.
Since is a divisor of and it must be positive, we have that
must be one of , , , , , or .
Setting equal to each of
these values and solving gives Note
that is the ratio of two square
roots, which means it must be nonnegative. As well, is not an integer, so this
means , , , , or .
Substituting these values of into
, we get
that the possible values of are
, , , , and , of which there are .
Answer:
The two-digit perfect squares are , , , , , and .
Notice that no two of these perfect squares have a common first digit,
which means that in a mystical integer, either a digit is the rightmost
digit or there is only one possibility for the digit to its right.
This means, for each possible first digit, the largest mystical integer
with that first digit is the mystical integer with as many digits as
possible.
To make a mystical integer starting with have as many digits as possible, we
must follow the by a , which must be followed by and then . No two-digit perfect square starts
with , so is the longest, and hence, the
largest mystical integer with its first digit equal to .
The only mystical integer with its first digit equal to is since no two-digit perfect square has
a first digit of .
The longest mystical integer starting with is .
The longest mystical integer starting with is .
There are no mystical integers starting with , , or .
The longest mystical integer with a first digit of is .
The longest mystical integer starting with is . This is the only five-digit
mystical integer, so it must be the largest.
Answer:
Suppose , , , and are the digits through in some order and are placed in the
top-left subgrid as shown below.
In order for the bottom-left subgrid to contain the integers from
through , the leftmost two cells in the bottom
row must contain and in some order. Similarly, in order for
the top-right subgrid to contain the digits from through , the third cells in the first two rows
must contain and in some order.
Therefore, there are four possible ways to fill in the rest of the
cells, excluding the bottom-right cell. They are shown below.
In each of these four grids, the top-left, top-right, and bottom-left
subgrids all satisfy the condition that each digit occurs exactly once
in it.
In the first three of four, the three digits already in the bottom-right
subgrid are different, so there is exactly one way to place a digit in
the bottom right cell to make the grid satisfy the condition.
For the first of the four grids above, we need to place in the bottom-right cell. For the
second, we need to place . For the
third, we need to place .
The fourth of these grids lacks
and in the bottom-right subgrid.
With only one cell left to fill, there is no way to place a digit in it
so that every digit occurs in the subgrid.
There are ways to order the
digits from through as , , , and . We have shown that there are three
ways to fill the grid for each of these orderings, so the answer is .
Answer:
Two numbers are reciprocals if their product is . Therefore, we are looking for real
numbers for which .
Multiplying through by gives
. Expanding gives
which can be
rearranged to get .
Given that the problem states that there are exactly three real numbers
with the given condition, these three real numbers must be the solutions
to the cubic equation above. A cubic equation has at most three real
roots, which means this cubic must have exactly three real roots. The
sum of the roots of a cubic is the negative of the coefficient of , which means the answer is .
However, in the interest of presenting a solution that does not assume
the question “works”, we will factor this cubic to find the three real
numbers.
A bit of guessing and checking reveals that , which means is a factor of the cubic. We can
factor to get
By the quadratic formula, the other two roots of the cubic are The three real numbers are , , and which have a sum of
Answer:
From the similarity of and , we
get .
Substituting, we get which can be rearranged to get .
From the similarity of and , we
get .
Substituting, we get which can be rearranged to get .
We now know the lengths of ,
, and . Since , we get
Answer:
Let be the -coordinate of and note that . The parabola has reflective
symmetry across the -axis, which
implies has reflective
symmetry across the -axis. Thus,
the -coordinate of is .
Therefore, , and so the area
of is .
The point lies on the parabola
with equation and has the
same -coordinate as . Therefore, the coordinates of are .
Since is vertical, its length is
the difference between the -coordinates of and .
It is given that is below the
-axis, so its -coordinate is negative. The -coordinate of is , and so we have that .
Since is a square, , and so .
Rearranging this equation gives , and using the quadratic
formula, we get The quantity
is negative, and we are
assuming is positive, which means
.
We know that the area of is
, so the area is .
Answer:
In this solution, “”
(without explicitly indicating the base) means “”.
Using logarithm rules, we have Since and , we have .
Answer:
The multiples of from
through are , , , , , , , and .
The only multiple of in this list
is .
Jolene chooses the red ball numbered with probability since there are exactly five
red balls.
If Jolene chooses the red ball numbered with , then there will be green balls remaining with a multiple
of on them. The probability that
Tia chooses a ball numbered with a multiple of in this situation is .
The probability that Jolene and Tia win the game if Jolene chooses
the ball numbered with is .
Jolene chooses a red ball other than the one numbered with with probability .
In this situation, there are
green balls numbered with a multiple of , so the probability that Tia chooses a
green ball numbered with a multiple of is .
The probability that Jolene and Tia win the game if Jolene does not
choose the red ball numbered by
is .
Since exactly one of the events “Jolene chooses the ball numbered
with ” and “Jolene does not
choose the ball numbered with ”
must occur, the probability that Jolene and Tia win the game is equal to
the sum of the probabilities that they win in each of the two
situations.
Therefore, the probability that they win the game is
Answer:
The remainder when is
divided by is . This means that and there is a unique
integer such that .
Similarly, there are unique integers and such that and .
This gives three equations Subtracting the first equation
from the second, the first from the third, and the second from the third
gives the three equations Since , , and are integers, so are , , and . Therefore, must be a divisor of , , and .
Since is a divisor of and , it is a factor of .
Since is a divisor of and , it is a factor of .
Since is a divisor of and , it is a factor of .
Now note that , , and , so is a divisor of all three
integers.
We have shown that must be a
divisor of and that is a divisor of all three integers.
This implies that the possible values of are the positive divisors of , which are , , , and .
If , then the remainder is when each of , , and is divided by . In this case, .
If , then the remainder is when each of , , and is divided by . In this case, .
If , then the remainder is when each of , , and is divided by . In this case, .
If , then the remainder is
when each of , , and is divided by . In this case, .
Answer:
Solution 1
Let be the foot of the
altitude of from
to , and let .
It is given that
is isosceles and since , we must have that . As well, is equilateral so .
Points , , and are on a line, so .
By very similar reasoning, , which means is isosceles with .
Since , we have
and so .
Similarly, ,
and since is
isosceles, bisects , which means .
From the double-angle formula, . Using this fact and
the fact that , we can compute the area of as
Since is
equilateral, , and since
, .
Rearranging and using that , we get .
Since , we get as well.
Therefore, the area of is
The areas of and
are equal, so the
ratio is .
Solution 2
Let , which implies as well since is equilateral.
Since has a right
angle at and is isosceles, the
Pythagorean theorem implies that or , so .
Therefore, the area of is .
Since is isosceles
and right angled at , . As well, is equilateral, which means
. Using that
, , and lie on the same line, we can compute
.
By similar reasoning, , which implies that is isosceles with .
Using the two known angles in , we get as
.
This means the area of is .
The areas of and
are equal, so the
ratio is .
Answer:
Since has
and a common difference of
, we have that for all . By similar reasoning, we get
that for all .
For fixed , the term appears in both sequences exactly
when there is some such that
.
Using the closed forms established earlier, the term appears in both sequences exactly
when there exists such that , which is equivalent
to .
In other words, is in both
sequences exactly when the expression is a positive integer,
which happens exactly when is
a positive multiple of .
Note that is a positive
integer, so is at least , which means is positive for all .
Since is a multiple of , is a multiple of exactly when is a multiple of , but and have no prime factors in common, is a multiple of exactly when is a multiple of .
We have now shown that is in
both sequences exactly when is a
multiple of .
Let for some . Then and we wish to
find the smallest possible for
which .
This inequality is equivalent to which is equivalent to . The smallest positive
integer with this property is
. Therefore, is the smallest integer that is larger than
and is in both sequences.
Answer:
For every real number ,
, which means
that and are both at most .
Therefore, has a maximum
possible value of , so .
The only way that
is if and .
Since corresponds to an angle in
a triangle, .
This and together imply
that . We conclude that
has a right angle at
.
Since is an angle in a
triangle, we have that . However, we have already shown that this triangle
has a right angle not at , which
means .
Multiplying this inequality through by gives .
This and together imply
that , which means
.
It follows that is a
- - triangle with hypotenuse .
The side is opposite the angle, so its length is half
that of the hypotenuse, or .
The side is opposite the angle, so its length is .
The area of is .
Suppose the length of the altitude from to is . Then the area of is also equal
to .
Therefore, , and so .
Answer:
Squaring both sides of the equation gives .
Substituting into the
above equation gives .
From , we get , and from , we get .
Substituting into , we get .
Regardless of the values of
and , . Using this observation as
well as and , we get the following
equivalent inequalities.
If , then
both and are negative, so their product is
positive.
If ,
then and , so their product is
non-positive.
If , then and are both positive, so their product
is positive.
Therefore, in order for , we need that .
We have shown that is at least
and at most , but to be sure these are the actual
minimum and maximum possible values of , we must show that there are solutions
to the system of equations that have taking each of these two values.
If , the equations become
or and which can be factored as . Substituting gives , so or .
Since and at least one of
and is , it must be the case that they are both
. This gives the solution .
If , then the
first equation gvies or . The second equation
gives .
Substituting into
this equation and rearranging gives .
Multiplying
through by gives , and after
substituting gives
which is equivalent to .
Factoring, we get so
, which means . Substituting this into
and solving leads
to .
Therefore,
is a soluton.
We have shown that in any solution, we must have and that there
are solutions with
and . Thus, and , so .
Answer:
With , , and so the given
identity implies .
Since we need to find , it remains to find the
value of .
With , . Applying the identity gives
.
With , . Applying the identity gives .
With , .
Applying the identity gives .
Let , , and . We have
shown that the following three equations hold: Adding the first and third
equations gives .
Since , this
means , and so .
We already know that , so .
Remark: It can be shown that the given
condition implies that
for all except and and that and are cannot be
uniquely determined. However, the values of , , and are uniquely determined, which is
all that was needed for this problem.
Answer:
(Note: Where possible, the solutions to parts (b) and (c) of each
Relay are written as if the value of is not initially known, and then is substituted at the end.)