2024 Fermat Contest
Solutions
(Grade 11)
Wednesday, February 28, 2024
(in North America and South America)
Thursday, February 29, 2024
(outside of North American and South America)
Wednesday, February 28, 2024
(in North America and South America)
Thursday, February 29, 2024
(outside of North American and South America)
©2023 University of Waterloo
Calculating, \(3
\left(\frac{5}{3}-\frac{1}{3}\right) = 3 \cdot \frac{5}{3} - 3 \cdot
\frac{1}{3} = 5 - 1 = 4\).
Alternatively, \(3
\left(\frac{5}{3}-\frac{1}{3}\right) = 3 \cdot \frac{4}{3} =
4\).
Answer: (D)
Simplifying, \(4x^2 - 3x^2 =
x^2\). When \(x=2\), this
expression equals \(4\).
Alternatively, when \(x = 2\), we have
\(4x^2 - 3x^2 = 4 \cdot 2^2 - 3 \cdot 2^2 = 16
- 12 = 4\).
Answer: (C)
The volume of a \(1 \times 1 \times
1\) cube is 1.
The volume of a \(2 \times 2 \times 2\)
cube is 8.
Thus, \(8\) of the smaller cubes are
needed to make the larger cube.
Answer: (E)
For there to be equal numbers of each colour of candy, there must
be at most \(3\) red candies and at
most \(3\) yellow candies, since there
are \(3\) blue candies to start.
Thus, Shuxin ate at least \(7\) red
candies and at least \(4\) yellow
candies.
This means that Shuxin ate at least \(7 + 4 =
11\) candies.
We note that if Shuxin eats \(7\) red
candies, \(4\) yellow candies, and
\(0\) blue candies, there will indeed
be equal numbers of each colour.
Answer: (A)
Square \(PQRS\) is made up of
\(16\) equal-sized small squares.
Of these, \(2\) are fully shaded and
\(8\) are half-shaded.
This shading is equivalent to fully shading \(2 + 8 \cdot \frac{1}{2} = 2 + 4 = 6\) of
the \(16\) small squares.
Thus, square \(PQRS\) is \(\frac{6}{16} = \frac{3}{8}\) shaded.
Answer: (E)
Using a calculator, \(\sqrt{15} \approx
3.87\) and \(\sqrt{50} \approx
7.07\).
The integers between these real numbers are \(4\), \(5\), \(6\), \(7\), of which there are \(4\).
Alternatively, we could note that integers between \(\sqrt{15}\) and \(\sqrt{50}\) correspond to values of \(\sqrt{n}\) where \(n\) is a perfect square and \(n\) is between \(15\) and \(50\). The perfect squares between \(15\) and \(50\) are \(16\), \(25\), \(36\), \(49\), of which there are \(4\).
Answer: (B)
Solution 1:
When a line is reflected in the \(y\)-axis, its \(y\)-intercept does not change (since it is
on the line of reflection) and its slope is multiplied by \(-1\).
Therefore, the new line has slope \(-3\) and \(y\)-intercept \(6\), which means that its equation is \(y = -3x + 6\).
The \(x\)-intercept of this new line is
found by setting \(y = 0\) and solving
for \(x\) which gives \(0 = -3x + 6\) or \(3x = 6\) or \(x =
2\).
Solution 2:
The \(x\)-intercept of the original
line is found by setting \(y = 0\) in
the equation of the line and solving for \(x\), which gives \(0 = 3x + 6\) or \(3x = -6\) or \(x
= -2\).
When the line is reflected in the \(y\)-axis, the \(x\)-intercept of the new line is the
reflection of the original line in the \(y\)-axis, and thus is \(x = 2\).
Answer: (A)
Using exponent laws, \(1000^{20} = (10^3)^{20} = 10^{60}\) and so \(n = 60\).
Answer: (B)
Since \(O\) is the centre of the
circle, then \(OA = OB = OC\).
This means that \(\triangle AOB\) and
\(\triangle COB\) are both isosceles
with \(\angle ABO = \angle BAO = \angle BAC =
25\degree\).
Thus, \(\angle AOB = 180\degree - \angle ABO -
\angle BAO = 130\degree\).
Since \(\angle AOC\) is a straight
angle, then \(\angle BOC = 180\degree - \angle
AOB = 180\degree - 130\degree = 50\degree\).
Answer: (D)
After David is seated, there are \(4\) seats in which Pedro can be seated, of
which \(2\) are next to David.
Thus, the probability that Pedro is next to David is \(\frac{2}{4}\) or \(\frac{1}{2}\).
Answer: (C)
Each of the \(4\) lines can
intersect each of the other \(3\) lines
at most once.
This might appear to create \(4 \times 3 =
12\) points of intersection, but each point of intersection is
counted twice – one for each of the \(2\) lines.
Thus, the maximum number of intersection points is \(\dfrac{4 \times 3}{2} = 6\).
The diagram below demonstrates that \(6\) intersection points are indeed
possible:
Answer: (D)
When a list of 5 numbers \(a\),
\(b\), \(c\), \(d\), \(e\)
has the property that \(a+b+c =
c+d+e\), it is also true that \(a+b=d+e\).
With the given list of \(5\) numbers, it is likely easier to find two pairs
with no overlap and with equal sum than to find two triples with one
overlap and equal sum.
After some trial and error, we can see that \(6+21 = 10 + 17\), and so the list \(6\), \(21\), \(5\), \(10\), \(17\) has the given property, which means that \(5\) is in the
middle.
(We note that these two pairs are the only such pairs, after allowing
for switching the numbers in each pair and/or switching the pairs.)
Answer: (A)
Expanding, \((x+m)(x+n) = x^2 + nx + mx
+ mn = x^2 + (m+n)x + mn\).
The constant term of this quadratic expression is \(mn\), and so \(mn
= -12\).
Since \(m\) and \(n\) are integers, they are each divisors of
\(-12\) and thus of \(12\).
Of the given possibilities, only \(5\) is not a divisor of \(12\), and so \(m\) cannot equal \(5\).
We can check that each of the other four choices is a possible value of
\(m\).
Answer: (E)
We note first that \(\triangle
ACB\) has a right angle and a \(60\degree\) angle and so it is a \(30\degree\)-\(\,60\degree\)-\(\,90\degree\) triangle.
Since \(AB = \sqrt{3}\), then using the
known ratios of side lengths, we can see that \(BC = 1\) and \(AC
= 2\).
Next, we note that \(\triangle ACE\)
has two \(45\degree\) angles and so is
an isosceles right-angled triangle.
This means that \(CE = AC = 2\) and
\(\angle ACE = 90\degree\).
Also, \(AE = \sqrt{2}AC =
2\sqrt{2}\).
Further, since \(\angle BCD\) is a
straight angle, then \[\angle ECD =
180\degree - \angle ACB - \angle ACE = 180\degree - 60\degree -
90\degree = 30\degree\] Since \(\triangle CED\) has a \(30\degree\) angle and a right-angle, it is
also a \(30\degree\)-\(\,60\degree\)-\(\,90\degree\) triangle.
Using the known ratios of sides, since \(CE =
2\), we have \(DE = 1\) and
\(CD = \sqrt{3}\).
Therefore, the perimeter of \(ABDE\) is
\[AB + BC + CD + DE + AE = \sqrt{3} + 1 +
\sqrt{3} + 1 + 2\sqrt{2} = 2 + 2\sqrt{2} + 2\sqrt{3}\]
Answer: (E)
We first note that \(197 = 8 \cdot 24 +
5\).
This tells us that the time that is \(197\) hours from now is \(8\) days and \(5\) hours.
Since Anila’s grandmother’s activities are the same every day, then in
\(197\) hours and \(5\) minutes she will be doing the same
thing as she is doing in \(5\) hours
and \(5\) minutes, at which point she
is doing yoga.
Answer: (C)
Of the row and column products, only \(135\) and \(160\) are divisible by \(5\). This means that \(5\) must go in the square in the 2nd row,
3rd column.
Of the row and column products, only \(21\) and \(56\) are divisible by \(7\). This means that \(7\) must go in the square in the 1st row,
1st column.
Of the row and column products, only \(108\) and \(135\) are divisible by \(9\). This means that \(9\) must go in the square in the 2nd row,
2nd column.
So far, this gives the following grid:
In the 2nd row, the product is \(135\) which means that the missing entry
is \(\frac{135}{5 \cdot 9} = 3\).
In the 1st column, the product is \(21\) which means that the missing entry
is \(\frac{21}{7 \cdot 3} = 1\).
The 3rd row, whose product is \(48\),
thus includes \(1\) and two more
integers between \(1\) and \(9\). The only divisor pair of \(48\) with both divisors less than \(10\) is \(48 = 6
\cdot 8\).
Since \(8\) is not a divisor of \(108\), then \(N\) must be \(6\).
We can complete the square as follows:
Answer: (C)
Since \(b + d > a + d\), then
\(b > a\). This means that \(a\) does not have the greatest value.
Since \(c + e > b + e\), then \(c > b\). This means that \(b\) does not have the greatest value.
Since \(b + d = c\) and each of \(b\), \(c\), \(d\)
is positive, then \(d < c\), which
means that \(d\) does not have the
greatest value.
Consider the last equation \(a + c = b +
e\) along with the fact that \(a < b
< c\).
From this, we see that \(e = c +
(a-b)\).
Since \(a < b\), then \(a-b\) is negative and so \(e < c\).
This means that \(c\) has the greatest
value.
Answer: (C)
Since \(3x + 2y = 6\), then
\((3x + 2y)^2 = 6^2\) or \(9x^2 + 12xy + 4y^2 = 36\).
Since \(9x^2 + 4y^2 = 468\), then \[12xy = (9x^2 + 12xy + 4y^2) - (9x^2 + 4y^2) = 36
- 468 = -432\] and so \(xy =
\frac{-432}{12} = -36\).
(With some additional work, we can find that the solutions to the system
of equations are \((x,y) = (-4,9)\) and
\((x,y) = (6,-6)\).)
Answer: (A)
Suppose that when the three dice are rolled, the numbers rolled
are \(x\), \(y\) and \(z\).
Since there are 6 possibilities for each of \(x\), \(y\)
and \(z\), there are \(6 \cdot 6 \cdot 6 = 216\) possible
outcomes.
Also, the sum, \(S\), of the three
rolls is at least \(3 \cdot 1 = 3\) and
at most \(3 \cdot 6 = 18\).
The outcome "\(S > 5\)" is the
complement of the outcome "\(S \leq
5\)".
Thus, the probability that \(S>5\)
is \(1\) minus the probability that
\(S \leq 5\).
It is easier to compute the probability that \(S \leq 5\) directly by listing the rolls
that give this.
If \(S = 3\), then \(x+y+z=3\) and so \((x,y,z)=(1,1,1)\).
If \(S = 4\), then \(x+y+z=4\) and so \(x\), \(y\)
and \(z\) must be \(1\), \(1\)
and \(2\) in some order. Thus, \((x,y,z) = (2,1,1)\) or \((1,2,1)\) or \((1,1,2)\).
If \(S = 5\), then \(x+y+z=5\) and so \(x\), \(y\)
and \(z\) must be \(1\), \(1\)
and \(3\), or \(1\), \(2\)
and \(2\) in some order. There are
\(3\) arrangements in each case and so
\(6\) triples in total.
Therefore, there are \(1 + 3 + 6 = 10\)
triples with \(S \leq 5\), and so the
probability that \(S > 5\) is equal
to \(1 - \frac{10}{216} \approx
0.954\).
Of the given choices, this is closest to \(0.95\).
Answer: (B)
Suppose that the radius of the cylinder is \(r\) and the height of the cylinder is \(h\).
This means that the volume of the cylinder is \(\pi r^2 h\); the volume of half of the
cylinder is \(\frac{1}{2}\pi r^2
h\).
Also, the radius of the cone is \(\frac{1}{2}r\) and the height of the cone
is \(h\).
This means that the volume of the cone is \(\frac{1}{3}\pi \left(\frac{1}{2}r\right)^2
h\) or \(\frac{1}{12}\pi r^2
h\).
When the cone is divided into two pieces by a horizontal plane at half
of its height, the top portion of the cone is a cone with the same
proportions, but with dimensions \(\frac{1}{2}\) of those of the larger
cone.
This means that the volume of the top portion is \(\left(\frac{1}{2}\right)^3 = \frac{1}{8}\)
of that of the cone, which equals \(\frac{1}{8} \cdot \frac{1}{12}\pi r^2 h\)
or \(\frac{1}{96}\pi r^2 h\).
To see this in another way, we note that this top portion of the cone
has height \(\frac{1}{2}h\) and should
have radius \(\frac{1}{2} \cdot
\frac{1}{2}r\) (because the radius decreases proportionally to
the height). This means that the volume of this portion is \(\frac{1}{3}\pi \left(\frac{1}{4}r\right)^2 \cdot
\frac{1}{2}h\) which is again \(\frac{1}{96}\pi r^2 h\).
Using this information, the bottom portion of the cone has volume \(\frac{7}{8} \cdot \frac{1}{12} \pi r^2 h =
\frac{7}{96} \pi r^2 h\).
Now, when the cone is in the cylinder and the cylinder is filled with
water to half of its height, the volume of the bottom half of the
cylinder is filled with the bottom portion of the cone and with the
water.
Therefore, the volume of water is the difference between half of the
volume of the cylinder and the volume of the bottom portion of the cone,
or \(\frac{1}{2}\pi r^2 h - \frac{7}{96}\pi
r^2 h = \frac{48}{96}\pi r^2 h - \frac{7}{96}\pi r^2 h =
\frac{41}{96}\pi r^2 h\).
When the cone is removed, the water then occupies a cylinder with radius
\(r\) and volume \(\frac{41}{96}\pi r^2 h\).
If the depth of the water in this configuration is \(d\), then \(\pi
r^2 d = \frac{41}{96}\pi r^2 h\) and so \(d = \frac{41}{96}h\), which means that the
depth of the water is \(\frac{41}{96}\)
of the height of the cylinder.
Answer: (B)
Since the second column includes the number \(1\), then step (ii) was never used on the
second column, otherwise each entry would be at least \(2\).
To generate the \(1\), \(3\) and \(2\) in the second column, we thus need to
have used step (i) \(1\) time on row
\(1\), \(3\) times on row \(2\), and \(2\) times on row \(3\).
This gives:
| \(1\) | \(1\) | \(1\) |
| \(3\) | \(3\) | \(3\) |
| \(2\) | \(2\) | \(2\) |
We cannot use step (i) any more times, otherwise the entries in
column \(2\) will increase. Thus, \(a = 1 + 3 + 2 = 6\).
To obtain the final grid from this current grid using only step (ii), we
must increase each entry in column \(1\) by \(6\) (which means using step (ii) \(3\) times) and increase each entry in
column \(3\) by \(4\) (which means using step (ii) \(2\) times). Thus, \(b = 3 + 2 = 5\).
Therefore, \(a + b = 11\).
Answer: \(11\)
We note that \[\begin{aligned}
ac + bd - ad - bc &= ac - ad - bc + bd\\
& = a(c-d) - b(c-d) \\
& = (a-b)(c-d)\end{aligned}\] Since each of \(a\), \(b\), \(c\), \(d\)
is taken from the set \(\{1,2,3,4,5,6,7,8,9,10\}\), then \(a - b \leq 9\) since the greatest possible
difference between two numbers in the set is \(9\).
Similarly, \(c-d \leq 9\).
Now, if \(a-b = 9\), we must have \(a = 10\) and \(b
= 1\).
In this case, \(c\) and \(d\) come from the set \(\{2, 3, 4, 5, 6, 7, 8, 9\}\) and so \(c-d \leq 7\).
Therefore, if \(a-b = 9\), we have
\((a-b)(c-d) \leq 9 \cdot 7 =
63\).
If \(a - b = 8\), then either \(a = 9\) and \(b =
1\), or \(a = 10\) and \(b = 2\).
In both cases, we cannot have \(c - d =
9\) but we could have \(c - d =
8\) by taking the other of these two pairs with a difference of
\(8\).
Thus, if \(a - b = 8\), we have \((a-b)(c-d) \leq 8 \cdot 8 = 64\).
Finally, if \(a - b \leq 7\), the
original restriction \(c - d \leq 9\)
tells us that \((a-b)(c-d) \leq 7 \cdot 9 =
63\).
In summary, the greatest possible value for \(ac + bd - ad - bc\) is 64 which occurs, for
example, when \(a = 9\), \(b = 1\), \(c =
10\), and \(d = 2\).
Answer: \(64\)
Solution 1:
Suppose that \(BE = AC = x\) and
\(DE = y\).
Extend \(BC\) to point \(F\) so that \(BC
= DE = y\).
Since \(BC + DE = 288\), then \(BF = BC + CF = BC + DE = 288\).
Also, \(\triangle BED\) is congruent to
\(\triangle ACF\) by
side-angle-side.
Therefore, \[\angle BAF = \angle BAC + \angle
ACF = \angle BDE + \angle DBE = 90\degree\] since \(DE\) and \(AC\) are parallel.
Next, \(\triangle BED\) is similar to
\(\triangle BAF\) since both are
right-angled and they share an angle at \(B\).
Therefore, \(\dfrac{DE}{BD} =
\dfrac{FA}{BF}\) and so \(\dfrac{DE}{120} = \dfrac{120}{288}\), which
gives \(DE = \dfrac{120 \cdot 120}{288} =
50\), as required.
Solution 2:
Suppose that \(BE = AC = x\) and \(DE = y\).
Since \(DE + BC = 288\), then \(BC = 288 - y\).
We note that \(\triangle BED\) is
similar to \(\triangle BCA\) because
each is right-angled and their angles at \(B\) are common.
Therefore, \(\dfrac{BE}{DE} =
\dfrac{BC}{AC}\) and so \(\dfrac{x}{y}
= \dfrac{288 - y}{x}\).
Manipulating, we obtain \(x^2 =
y(288-y)\) and so \(x^2 = 288y -
y^2\) or \(x^2 + y^2 =
288y\).
Also, using the Pythagorean Theorem in \(\triangle BED\) gives \(x^2 + y^2 = 120^2\).
Since \(x^2 + y^2 = 288y\) and \(x^2 + y^2 = 120^2\), then \(288y = 120^2\) which gives \(2 \cdot 12 \cdot 12 \cdot y = 120 \cdot
120\) and so \(2y = 10 \cdot
10\) or \(y = 50\).
Therefore, \(DE = 50\).
Answer: \(50\)
Throughout this solution, we use the fact that if \(N\) is a positive integer with \(N > 1\) and \(N\) has prime factorization \(p_1^{a_1} p_2^{a_2} \cdots p_m^{a_m}\) for
some distinct prime numbers \(p_1, p_2,
\ldots, p_m\) and positive integers \(a_1, a_2, \ldots, a_m\), then the number of
positive divisors of \(N\) including 1
and \(N\) is equal to \((1+a_1)(1+a_2) \cdots (1+a_m)\).
We are told that \(N\) is a positive
multiple of 2024.
Now, \(2024 = 8 \cdot 253 = 2^3 \cdot 11 \cdot
23\).
This means that \(N\) has at least
\(3\) prime factors (namely \(2\), \(11\) and \(23\)) and that at least one of these prime
factors has an exponent of at least \(3\).
Let \(D\) be the number of positive
divisors that \(N\) has. We are told
that \(100 < D < 110\).
Since \(D = (1+a_1)(1+a_2) \cdots
(1+a_m)\) and \(N\) has at least
\(3\) prime factors, then \(D\) is a positive integer that can be
written as the product of at least \(3\) positive integers each greater than
\(2\).
\(D\) cannot equal \(101\), \(103\), \(107\), or \(109\), since each of these is prime (and so
cannot be written as the product of \(3\) integers each at least \(2\)).
\(D\) also cannot equal \(106\) because \(106 = 2 \cdot 53\) (both \(2\) and \(53\) are prime), which means that \(106\) cannot be written as the product of
three integers each greater than \(1\).
The possible values of \(D\) that
remain are \(102\), \(104\), \(105\), \(108\).
We note that \(102 = 2 \cdot 3 \cdot
17\) and \(104 = 2^3 \cdot 13\)
and \(105 = 3 \cdot 5 \cdot 7\) and
\(108 = 2^2 \cdot 3^3\).
Case 1: \(D = 102\)
Since the prime factors of \(N\)
include at least \(2\), \(11\) and \(23\), then the prime factorization of \(N\) includes factors of \(2^a\), \(11^b\) and \(23^c\) for some positive integers \(a\), \(b\)
and \(c\) with \(a \geq 3\).
If a fourth prime power \(p^e\) was
also a factor of \(N\), then \(D\) would be divisible by \((1+a)(1+b)(1+c)(1+e)\). (\(D\) could have more factors if \(N\) had more prime factors.)
Since \(D = 102 = 2 \cdot 3 \cdot 17\)
has only \(3\) prime factors, it cannot
be written as the product of \(4\)
integers each greater than \(1\).
Thus, \(N\) cannot have a fourth prime
factor.
This means that \(N = 2^a 11^b 23^c\),
which gives \(D = (1+a)(1+b)(1+c) = 2 \cdot 3
\cdot 17\).
This means that \(1+a\), \(1+b\) and \(1+c\) are equal to \(2\), \(3\)
and \(17\), in some order, and so \(a\), \(b\)
and \(c\) are equal to \(1\), \(2\)
and \(16\), in some order.
For \(D\) to be as small as possible,
the largest exponent goes with the smallest prime, the next largest
exponent with the next smallest prime, and so on. (Can you see why this
makes \(N\) as small as
possible?)
Therefore, the smallest possible value of \(N\) in this case is \(N = 2^{16} 11^2 23^1 = 182\,386\,688\).
Case 2: \(D = 105\)
Using a similar argument, we can determine that \(N = 2^a 11^b 23^c\) with \(1+a\), \(1+b\) and \(1+c\) equal to \(3\), \(5\)
and \(7\) in some order, meaning that
\(a\), \(b\) and \(c\) equal \(2\), \(4\), \(6\)
in some order.
Therefore, the minimum value of \(N\)
is this case is \(N = 2^6 11^4 23^2 =
495\,685\,696\).
Case 3: \(D = 104\)
Since \(D = 2^3 \cdot 13\) has \(4\) prime factors, then \(N\) cannot have more than \(4\) prime factors. (If \(N\) had 5 or more prime factors, then the
product equal to \(D\) would include at
least \(5\) integers, each at least
\(2\).)
Therefore, \(N = 2^a 11^b 23^c\) and
\(D = (1+a)(1+b)(1+c)\), or \(N = 2^a 11^b 23^c p^e\) for some prime
\(p \neq 2, 11, 23\) and \(D = (1+a)(1+b)(1+c)(1+e)\).
This means that \((1+a)(1+b)(1+c) = 2^3 \cdot
13\) or \((1+a)(1+b)(1+c)(1+e) = 2^3
\cdot 13\).
In the case that \(N\) has three prime
factors, we note that \(104 = 26 \cdot 2 \cdot
2 = 13 \cdot 4 \cdot 2\) are the only two ways of writing \(104\) as the product of \(3\) integers each of which is at least
\(2\).
These give corresponding minimum values of \(N\) of \(N =
2^{25} \cdot 11 \cdot 23 = 8\,489\,271\,296\) and \(N = 2^{12} \cdot 11^3 \cdot 23 = 125\,
390\,848\).
In the case that \(N\) has four prime
factors, then \((1+a)(1+b)(1+c)(1+e) = 2 \cdot
2 \cdot 2 \cdot 13\) means that \(a\), \(b\), \(c\)
and \(e\) are \(1\), \(1\), \(1\), \(12\) in some order.
This in turn means that the corresponding smallest possible value of
\(N\) is \[N
= 2^{12} \cdot 3 \cdot 11 \cdot 23 = 3\,108\,864\] We note here
that the prime power \(p^e\) has become
\(3^1\) in order to minimize both \(p\) (since \(p
> 2\)) and its exponent.
Case 4: \(D = 108\)
Since \(D = 2^2 \cdot 3^3\) has
\(5\) prime factors, then \(N\) cannot have more than \(5\) prime factors.
If \(N\) has \(5\) prime factors, then we need to use the
factorization \(D = 2 \cdot 2 \cdot 3 \cdot 3
\cdot 3\).
This is not possible, however, because the power \(2^a\) must have \(a \geq 3\) which would mean that one of the
five factors of \(D\) would have to be
at least \(4\).
If \(N\) has \(4\) prime factors, then \(D\) must be partitioned as \(9 \cdot 3 \cdot 2 \cdot 2\) or \(6 \cdot 3 \cdot 3 \cdot 2\) or \(4 \cdot 3 \cdot 3 \cdot 3\). (Since two of
the prime factors have to be combined, either two \(2\)s, two \(3\)s, or a \(2\) and a \(3\) are combined.)
These give minimum values of \(N = 2^8 \cdot
3^2 \cdot 11 \cdot 23 = 582\,912\) and \(N = 2^5 \cdot 3^2 \cdot 11^2 \cdot 23 =
801\,504\) and \(N = 2^3 \cdot 3^2
\cdot 11^2 \cdot 23^2 = 4\,408\,648\).
If \(N\) has \(3\) prime factors, then we must use one of
the factorizations \(D = 27 \cdot 2 \cdot
2\) or \(D = 18 \cdot 3 \cdot
2\) or \(D = 12 \cdot 3 \cdot
3\) or \(D = 9 \cdot 4 \cdot 3\)
or \(D = 6 \cdot 6 \cdot 3\).
These gives corresponding minimum values \[\begin{aligned}
N & = 2^{26} \cdot 11 \cdot 23 = 16\,978\,542\,592 \\
N & = 2^{17} \cdot 11^2 \cdot 23 = 364\,773\,376\\
N & = 2^{11} \cdot 11^2 \cdot 23^2 = 131\,090\,432\\
N & = 2^8 \cdot 11^3 \cdot 23^2 = 180\,249\,344\\
N & = 2^5 \cdot 11^5 \cdot 23^2 =
2\,726\,271\,328\end{aligned}\] Combining Cases \(1\) through \(4\), the minimum possible value of \(N\) is \(582\,912\).
The sum of the digits of \(582\,912\)
is \(5 + 8 + 2 + 9 + 1 + 2 = 27\).
Answer: \(27\)
Suppose that, for some integer \(n \geq
2\), we have \(a_n = x\) and
\(a_{n-1} = y\).
The equation \(a_n + a_{n-1} =
\frac{5}{2}\sqrt{a_n \cdot a_{n-1}}\) can be re-written as \(x + y = \frac{5}{2}\sqrt{xy}\).
Since \(x>0\) and \(y > 0\), squaring both sides of the
equation gives an equivalent equation which is \((x+y)^2 = \frac{25}{4}xy\).
Manipulating algebraically, we obtain the following equivalent
equations: \[\begin{aligned}
(x+y)^2 & = \tfrac{25}{4}xy \\
4(x^2 + 2xy + y^2) & = 25xy \\
4x^2 - 17xy + 4y^2 & = 0 \\
(4x - y)(x - 4y) & = 0\end{aligned}\] Therefore, the given
relationship is equivalent to \(4x =
y\) or \(x = 4y\).
Returning to the sequence notation, we now know that it is the case that
\(4a_n = a_{n-1}\) (that is, \(a_n = \frac{1}{4}a_{n-1}\)) or \(a_n = 4a_{n-1}\).
Putting this another way, each term in the sequence can be obtained from
the previous term either by multiplying by \(4\) or by dividing by \(4\).
We are told that \(a_1 = 4\) and \(a_{11} = 1024\). We note \(\dfrac{a_{11}}{a_1} = \dfrac{1024}{4} = 256 =
4^4\).
We can think of moving along the sequence from \(a_1\) to \(a_{11}\) by making \(10\) "steps", each of which involves either
multiplying by \(4\) or dividing by
\(4\).
If there are \(m\) steps in which we
multiply by \(4\) and \(10-m\) steps in which we divide by \(4\), then \(\dfrac{4^m}{4^{10-m}} = 4^4\) which gives
\(4^{2m-10} = 4^4\) or \(2m - 10 = 4\) and so \(m = 7\).
In other words, the sequence involves \(7\) steps of multiplying by \(4\) and \(3\) steps of dividing by \(4\).
These steps completely define the sequence.
The number of possible sequences, \(S\), equals the number of ways of arranging
these \(10\) steps, which equals \(\displaystyle\binom{10}{3}\).
(If combinatorial notation is unfamiliar, we could systematically count
the number of arrangements instead.)
Therefore, \(S = \dfrac{10 \cdot 9 \cdot 8}{3
\cdot 2 \cdot 1} = 5 \cdot 3 \cdot 8 = 120\). The rightmost two
digits of \(S\) are \(20\).
Answer: \(20\)