Saginaw Valley State University

2022 Math Olympics — Level I Solutions

1. Consider the equation p(x) : ax2 + bx + c = 0 whose coefficients a, b and c are all non-zero,
and each of them satisfies an equation that results from removing the term containing that
coefficent from the equation p(x); for example, the coefficient b is a solution of the equation
ax2 + c = 0. What is the sum of all solutions of p(x)?

(a) Always 1 (b) Always −1 (c) Always 2 (d) 1 or −1 (e) 1 or 2

Solution (d): This is what we know about the coefficients a, b and c:

− a is a solution of bx + c = 0, so ab + c = 0.

− b is a solution of ax2 + c = 0, so ab2 + c = 0.

− c is a solution of ax2 + bx = 0, so ac 2 + bc = 0.

− a, b and c are all non-zero.

From the first two equalities we get ab = ab2 , and since a and b are non-zero, we get b = 1.

− With b = 1, the first two equations both become a + c = 0

− The third equation becomes ac 2 + c = 0.

From these two equalities we get ac 2 = a and since a 6= 0, we get c 2 = 1 or c = ±1. Since
a + c = 0, a = −c = ∓1.
That gives us only two possibilities for p(x):

− One is p(x) : x2 + x − 1 = 0. The solutions are

√ √
−1 ± 1+4 −1 ± 5
=
2 2

The sum of these two solutions is −1.

− The other is p(x) : −x2 + x + 1 = 0. Then the solutions are

√ √
−1 ± 1 + 4 −1 ± 5
=
−2 −2

The sum of these two solutions is 1.

Alternatively, instead of discussing the two possibilities for p(x), we can rewrite the equation
as x2 + ab x + ac = 0. Since ac = −1 < 0, the equation must have two real solutions. The sum of
the two solutions is − ab = ±1.

2. A marching band has 150 members. One day only part of them show up, but nobody wants
to take the time to count how many there are. They first line up in rows with 5 members each,
but there is one left over. Then they try rows with 6 members each, but there is still one left
over. Then they try rows of 7, but there are two left over. How many people should line up in
each row so no members will be left over?
2022 SVSU Math Olympics Level I Solutions — page 2 of 12

(a) 4 (b) 11 (c) 13 (d) 17 (e) None of the above

Solution (b): There is 1 person left over with both rows of 5 and 6. If we removed this one
person, the resulting number would be a multiple of both 5 and 6, and therefore a multiple
of 30. Therefore the original number must be one more than a multiple of 30.
At the same time, there cannot be more than 150 people, so the possible numbers are 1, 31,
61, 91 and 121.
The number also must give the remainder of 2 when divided by 7. Out of the 5 candidate
numbers, only 121 saisfies this condition.
Out of the suggested solutions, 11 is the only one that divides 121.

3. Which one of the following is the only true statement?

(a) The graph of a horizontal line can’t have any x-intercepts.

(b) The graph of a horizontal line can’t have a unique x-intercept, but may have more than
one x-intercept.

(c) The graph of a parabola can’t have a unique x-intercept, but may have more than one
x-intercept.

(d) The graph of a polynomial of degree three or higher must have at least one x-intercept.

(e) Either none are true or more than one are true.

Solution (b): The graph of a horizontal line has no x-intercepts if it is the graph of the
equation y = b, where b is a nonzero real number, but it has infinitely many x-intercepts if it
is the line y = 0.
The graph of a parabola will have a unique x-intercept if the vertex is on the x-axis (in other
words, if th equadratic function we are graphing is a perfect square).
Graph of a polynomial of degree three or higher can have no x-intercepts if it is of even degree,
for example y = x4 + 1.

4. Which of the following is equal to 10!? (If you are not familiar with the notation, the number
n! is defined for any non-negative integer using the following recursive formula:

0! = 1
n! = n · ((n − 1)!) for n > 0)

(a) 5! · 2! (b) 7! · 5! · 3! (c) 7! · 5! · 2! (d) 7! · 5! · 3! · 2!

(e) None of the above or more than one of the above.

Solution (b): 10! = 1 · 2 · 3 · 4 · 5 · 6 · 7 · 8 · 9 · 10. 7! = 1 · 2 · 3 · 4 · 5 · 6 · 7, 5! = 1 · 2 · 3 · 4 · 5 ≡

(2 · 4) · 3 · 5 = 8 · 3 · 5, and 3! = 1 · 2 · 3, so 7! · 5! · 3! = 1 · 2 · 3 · 4 · 5 · 6 · 7 · 8 · (3 · 3) · (5 · 2) = 10!.
To see that the others don’t work: 5! · 2! will not have a factor of 7, 7! · 5! · 2! does not have
another factor of 3 to give a factor of 9, and 7! · 5! · 3! · 2! has too many factors of 2, making
it equal to 2(10!).
2022 SVSU Math Olympics Level I Solutions — page 3 of 12

5. The base 6 expansion of one number is 550. The base 5 expansion of another number is 3440.
What is their greatest common divisor expanded in base 4?

(a) 24 (b) 33 (c) 113 (d) 223 (e) None of the above
2
Solution (b): The first number is 5506 = 5×6 +5×6+0 = 18010 +3010 = 21010 = 2×3×5×7.
The second number is 34405 = 3 × 53 + 4 × 52 + 4 × 5 + 2 = 37510 + 10010 + 2010 = 49510 =
3 × 3 × 5 × 1110 . So their greatest common divisor is 3 × 5 = 1510 = 3 × 4 + 3 = 334 .

6. A person walks along a beach, starting at point A, at

a rate of 3 mi/h and at point B, goes into the water Island
and swims at a rate of 2 mi/h diagonally out to an is-
√
land that is a distance of 3 mi from point C, directly √
3 miles
across from the island on the shore, as shown in the
picture. The total distance from point A to point C is A B C
3 mi. There are two different choices for the distance,
in miles, from point A to point B that will result in a
total time for walking and swimming of one hour and 3 miles
40 minutes; what is the sum of those numbers?

14 16
(a) 2 (b) 4 (c) 5 (d) 5 (e) None of the above

Solution (c): Let x be the distance from A to B. Then the distance from
p B to C is 3 − x, and
from the Pythagorean Theorem, the distance from B to the island is 12 − 6x + x2 .
The total time from A to the island is then
Island √
x 12 − 6x + x2
+
3
+

3 2
x) 2

√
−

3 5
q
(3

This must be equal to 1 hour and 40 minutes, or 3 of an hour.

A
x
B C √
3−x x 12 − 6x + x2 5
+ =
3 2 3
q
2x + 3 12 − 6x + x2 = 10
q
3 12 − 6x + x2 = 10 − 2x
9 12 − 6x + x2 = 100 − 40x + 4x2

108 − 54x + 9x2 = 100 − 40x + 4x2

5x2 − 14x + 8 = 0
(5x − 4)(x − 2) = 0

and so the solutions are 2 and 45 .

4 14
2+ =
5 5
2022 SVSU Math Olympics Level I Solutions — page 4 of 12
√
7. Which of the following is equal to | 2022 − 45|?

√ √ √ √
(a) 3 (b) 1977 (c) 2022 − 45 (d) 2022 + 45

√
(e) 45 − 2022
Solution (e): Since 452 = (40 + 5)2 = 1600 + 2 · 40 · 5 + 25 = 2025 > 2022, we know that
√ √ √ √
2022 − 45 < 0 and so || 2022 − 45|| = − 2022 − 45 = 45 − 2022.

p6 √
8. Which of the following is equal to s 5 9 s for all non-negative values of s?

√ p5 p
54
p
18
(a) 3
s (b) s2 (c) s5 (d) s 17 (e) None of the above

Solution (d): Rewriting the radicals using fractional exponents, we get

5 1 5 1
x6 x9 = x6+9 .

Adding the fractions will give us

5 1 15 2 17
+ = + = .
6 9 18 18 18

Therefore
5 1 17 p
18
x 6 x 9 = x 18 = x17 .

9. If 73 cubes are stacked to form a 7 × 7 × 7 cube, how many small cubes are on the surface of
the large cube?

(a) 127 (b) 134 (c) 218 (d) 294 (e) 327

Solution (c): The cubes that are not on the surface form a smaller, 5 × 5 × 5 cube, which
means that there are 125 of them. Therefore there are 343 − 125 = 218 cubed on the surface.

10. Each side of the cube depicted on the right is numbered with a positive
integer in such a way that the products of the numbers on each pair of 10
14 15
opposite sides are all the same. Find the lowest possible sum of all the
numbers on the sides of the cube.

(a) 78 (b) 80 (c) 89 (d) 107 (e) None of the above

Solution (c): Since the products of the numbers on the opposite sides must be the same,
they must be a multiple of 10, 14 and 15. The get the lowest possible sum, we must use the
least common multiple. Since 10 = 2 · 5, 14 = 2 · 7 and 15 = 3 · 5, we need 2 · 3 · 5 · 7 = 210.
So the number opposite to 10 is 21, opposite to 14 is 15, and opposite to 15 is 14. The sum
of all the numbers is 10 + 21 + 14 + 15 + 15 + 14 = 89.
2022 SVSU Math Olympics Level I Solutions — page 5 of 12
s r q p
11. Find the largest integer smaller than 22 + 22 + 22 + 22.

(a) 4 (b) 5 (c) 9 (d) 20 (e) 25

s r q p q p p
Solution (b): First, 22 + 22 + 22 + 22 > 22 + 22 > 22 + 4 > 5.
√ p
Then, if we define a recursive sequence a0 = 22, an+1 = 22 + an for n ≥ 0, our number is
p √
a3 . Clearly a0 < 6. If an < 6, then an+1 = 22 + an < 22 + 6 < 6. That means an < 6 for all
n ≥ 0. So our number is greater than 5, but less than 6.

12. Pat and Mat were trying to calculate the average of two numbers a and b using their calcu-
lator. First, Pat took the calculator and typed in a + b ÷ 2 and got 30. Then Mat took the
same calculator and typed in b + a ÷ 2 and got 18. What was the correct average of the two
numbers?

(a) 28 (b) 24 (c) 16 (d) 12 (e) None of the above

Solution (c): We know that

1
a+ b = 30
2
1
a + b = 18
2

Adding both equations together gives us

3 3
a + b = 48.
2 2

Dividing both sides by 3 gives us

a+b 48
= = 16.
2 3
2022 SVSU Math Olympics Level I Solutions — page 6 of 12

13. How many pairs of integers (x, y) are solutions of the equation

3x2 y − 10xy − 8y − 17 = 0?

(a) none (b) one (c) two (d) four (e) None of the above

Solution (b): We can rewrite the equation as

(3x2 − 10x − 8)y = 17

If x is an integer, so is 3x2 − 10x − 8. There are only four ways to write 17 as a product of
two integers: 17 · 1, 1 · 17, (−17) · (−1) and (−1) · (−17), which gives us four possibilities for
3x2 − 10x − 8:

− 3x2 − 10x − 8 = 17, or 3x2 − 10x − 25 = 0. The left side factors: (3x + 5)(x − 5) = 0, so we
have an integer solution x = 5.

− 3x2 − 10x − 8 = 1, or 3x2 − 10x − 9 = 0. The left side does not factor using integers, there
are no integer solutions.

− 3x2 − 10x − 8 = −17, or 3x2 − 10x + 9 = 0. The left side does not factor using integers,
there are no integer solutions.

− 3x2 − 10x − 8 = −1, or 3x2 − 10x − 7 = 0. The left side does not factor using integers, there
are no integer solutions.
Therefore the only pair of integer solutions is (5, 1).

11n + 14
14. For how many integers is an integer?
n−2

(a) 3 (b) 6 (c) 9 (d) 18 (e) None of the above

Solution (d): Rewrite

11n + 14 11n − 22 + 22 + 14 11(n − 2) + 36 36

= = = 11 +
n−2 n−2 n−2 n−2

The above will be an integer if and only if n − 2 is a factor of 36. Since 36 = 22 · 32 , each factor
of 36 looks like ±1 · 2k2 · 3k3 where k2 = 0, 1, 2 and k3 = 0, 1, 2. That means there are 3 × 3
positive and 3 × 3 negative factors of 36. Altogether there are 18 of them.

15. A fair coin is tossed 100 times and it lands on heads all 100 times. What is the probability
that it will land on heads on the 101st toss?

1 1 1
(a) 1 (b) 2100 (c) 2101 (d) 2 (e) None of the above

Solution (d): For a fair coin, the probability that it lands on heads is always 12 , regardless
of what the results of any previous tosses were. Of course if I saw a coin land on heads 100
times in a row, I would very much doubt it was a fair coin!
2022 SVSU Math Olympics Level I Solutions — page 7 of 12

16. If two woodchucks would chuck 32 lb of wood in 12 min, how much wood would 3 wood-
chucks chuck in 8 min?

(a) 21 13 lb (b) 32 lb (c) 64 lb (d) 80 lb (e) None of the above

Solution (b): Assuming all woodchucks chuck wood at the same uniform speed:

− One woodchuck would chuck 16 lb of wood in 12 min.

− One woodchuck would chuck = of wood in 1 min.

− One woodchuck would chuck = of wood in 1 min.

− Three woodchucks would chuck 3 = 4 lb of wood in 1 min.

− Three woodchucks would chuck 8 = 32 lb of wood in 8 min.

2022 SVSU Math Olympics Level I Solutions — page 8 of 12

17. Assuming that ‘wigglers’ are those who wiggle, ‘wobblers’ are those that wobble, and ‘wag-
glers’ are those who waggle, which of the following sets of premises will necessarily lead to
the conclusion that “Wilbur is not a weeble”?

(a) All weebles wobble. (b) Some weebles wobble. (c) All weebles wobble.
No wobblers wiggle. No wobblers wiggle. No wobblers wiggle.
Some wigglers waggle. Some wagglers wiggle. All wigglers waggle.
Wilbur waggles. Wilbur waggles. Wilbur waggles.

(d) All weebles wobble. (e) None of the above

No wobblers wiggle.
All wagglers wiggle.
Wilbur waggles.

Solution (d): These are all arguments involving four sets: wigglers, wagglers, wobblers and
weebles. We can visualize the premises using Venn diagrams with four sets. For each premise,
we will shade the area that is guaranteed empty by that premise. Possible locations of Wilbur
will be indicated by a dot.
First set of premises:
− "All weebles wobble." Shade the area of weebles that
wagglers wobblers is ouside of wobblers with

− "No wobblers wiggle." Shade the area of wobblers

that is inside of wigglers with

− "Some wigglers waggle." This just guarantees that

certain areas of the diagram will be non-empty, and
wigglers weebles
has no relevance for Wilbur, as long as it is consis-
tent with the other premises.

− "Wilbur waggles." There are four non-shaded areas

inside the wagglers where Wilbur can be, as indi-
cated by the dots.
We can see that Wilbur can be a weeble.
Second set of premises:
− "Some weebles wobble." This just guarantees that
wagglers wobblers
certain areas of the diagram will be non-empty, and
has no relevance for Wilbur, as long as it is consis-
tent with the other premises.

− "No wobblers wiggle." Shade the area of wobblers

that is inside of wigglers with
wigglers weebles
− "Some wigglers waggle." This just guarantees that
certain areas of the diagram will be non-empty, and
has no relevance for Wilbur, as long as it is consis-
tent with the other premises.
2022 SVSU Math Olympics Level I Solutions — page 9 of 12

− "Wilbur waggles." There are six non-shaded areas inside the wagglers where Wilbur can be,
as indicated by the dots.
We can see that Wilbur can be a weeble.
Third set of premises:
− "All weebles wobble." Shade the area of weebles that
wagglers wobblers is ouside of wobblers with

− "No wobblers wiggle." Shade the area of wobblers

that is inside of wigglers with

− "All wigglers waggle." Shade the area of wigglers

that is ouside of wagglers with
wigglers weebles

− "Wilbur waggles." There are four non-shaded areas

inside the wagglers where Wilbur can be, as indi-
cated by the dots. Some of these dots are inside
weebles.
We can see that Wilbur can be a weeble.
Fourth set of premises:
− "All weebles wobble." Shade the area of weebles that
wagglers wobblers is ouside of wobblers with

− "No wobblers wiggle." Shade the area of wobblers

that is inside of wigglers with

− "All wagglers wiggle." Shade the area of wagglers

that is ouside of wigglers with
wigglers weebles

− "Wilbur waggles." There is only one non-shaded area

inside the wagglers where Wilbur can be, as indi-
cated by the dot.
We can see that with this set of premises, Wilbur can
not be a weeble.

18. Which of the expressions is equal to 0 for every x?

p
(a) (x + 1)(x − 1) − x2 + 1 (b) x0 − 1 (c) x2 − x

(d) None of them is equal to 0 for every x. (e) More than one of them is 0 for every x.

Solution (a):

− (x + 1)(x − 1) − x2 + 1 = x2 − 1 − x2 + 1 = 0 for every x.

− x0 − 1 is undefined when x = 0.
p
− x2 − x is only equal to 0 when x ≥ 0.
2022 SVSU Math Olympics Level I Solutions — page 10 of 12

19. The vertices of an equilateral triangle lie on a circle with radius 2. The area of the triangle is

√ √ √ √
(a) 3 3 (b) 2 3 (c) 5 3 (d) 4 3 (e) None of the above

Solution (a): Draw the triangle and the circle. Lable the vertices of the triangle A, B and C,
and the center of the circle O.
Let T be the midpoint of AC. Let x = |TA| and y = |TO|.
B
The triangle TAO is a right triangle, and so x2 +y 2 = 4. Similarly,
the triangle TAB is a right triangle, and so x2 + (2 + y)2 = 4x2 , or
2 (2 + y)2 = 3x2 . Combining the equations together we get
2x
O
(2 + y)2 = 3(4 − y 2 )
2 2
y
(2 + y)2 = 3(2 + y)(2 − y)
C T x A 2 + y = 3(2 − y)
2 + y = 6 − 3y
4y = 4
y =1
q √ √ √
Then x = 4 − y 2 = 3, and the area of the triangle is A = x(y + 2) = 3(1 + 2) = 3 3.

20. How many 7-digit positive integers are made up of the digits 0 and 1 only and are divisible
by 6?

(a) 10 (b) 11 (c) 16 (d) 21 (e) 33

Solution (b): Let n be a 7-digit positive integer made up of the digits 0 and 1 only, and that
is divisible by 6. The leftmost digit of n cannot be 0, so must be 1.
Since n is divisible by 6, then n is even, which means that the rightmost digit of n cannot be
1, and so must be 0.
Therefore, n has the form 1pqrst0 for some digits p, q, r, s, t each equal to 0 or 1.
Now, n is divisible by 6 exactly when it is divisible by 2 and by 3.
Since the ones digit of n is 0, then it is divisible by 2.
n is divisible by 3 exactly when the sum of its digits is divisible by 3. The sum of the digits of
n is 1 + p + q + r + s + t. Since each of p, q, r, s, t is 0 or 1, then 1 ≤ 1 + p + q + r + s + t ≤ 6.
Thus, n is divisible by 3 exactly when 1 + p + q + r + s + t is equal to 3 or to 6.
That is, n is divisible by 3 exactly when either 2 of p, q, r, s, t are 1s or all 5 of p, q, r, s, t are 1s.
There are 10 ways for 2 of these to be 1s.
These correspond to the pairs pq, pr, ps, pt, qr, qs, qt, rs, rt, st.
There is 1 way for all 5 of p, q, r, s, t to be 1s. Thus, there are 1 + 10 = 11 such 7-digit integers.
2022 SVSU Math Olympics Level I Solutions — page 11 of 12
x−y xy
21. Suppose that x and y satisfy x+y = 9 and x+y = −60. The value of (x + y) + (x − y) + xy is

(a) −50 (b) −150 (c) −14310 (d) 210 (e) 14160

Solution (b): Let V = (x + y) + (x − y) + xy. Then

V x+y x−y xy
= + + = 1 + 9 − 60 = −50
x+y x+y x+y x+y

and so V = −50(x + y). Let n = x + y. Then x − y = 9(x + y) = 9n. Squaring both sides gives
us x2 − 2xy + y 2 = 81n2 . Since x + y = n, we also get x2 + 2xy + y 2 = n2 . Subtracting the two
equations will give us −4xy = 80n2 or xy = −20n2 .
xy
At the same time we know that n = −60. So

xy −20n2
−60 = = = −20n
n n

which means that n = 3.

Then V = −50n = −150.

22. A total of n points are equally spaced around a circle and are labeled with the integers 1 to
n, in order. Two points are called diametrically opposite if the line segment joining them
is a diameter of the circle. If the points labeled 7 and 35 are diametrically opposite, then n
equals

(a) 54 (b) 55 (c) 56 (d) 57 (e) None of the above

Solution (c): The number of points on the circle equals the number of spaces between the
points around the circle. Moving from the point labeled 7 to the point labeled 35 requires
moving 35 − 7 = 28 points and so 28 spaces around the circle.
Since the points labeled 7 and 35 are diametrically opposite, then moving along the circle
from 7 to 35 results in traveling halfway around the circle.
Since 28 spaces makes half of the circle, then 2 · 28 = 56 spaces make the whole circle. Thus,
there are 56 points on the circle, and so n = 56.

23. What is the area of a rhombus that has sides of length 10 cm and diagonals that differ in
length by 4 cm?

(a) 96 cm2 (b) 100 cm2

√
(c) 100 2 cm2 (d) Not enough information given.

(e) None of the above

Solution (a): Start by drawing the rhombus and the two diagonals.
2022 SVSU Math Olympics Level I Solutions — page 12 of 12

Note that in a rhombus, the diagonals bisect each other and

D 10 cm C
meet at a right angle. Label the length of each half of the
x shorter diagonal as x cm. Each half of the longer diagonal
x+2
will then be x + 2 cm. The rhombus will then consists of 4
10 cm 10 cm
congruent right triangles, each with hypotenuse 10 cm and
x
x+2 sides x cm and x + 2 cm. From the Pythagorean Theorem,
x2 + (x + 2)2 = 100. This equation has one positive solution,
A 10 cm B
x = 6. Then the area of each of the four right triangle that
1 1
form the rhombus is 2 x(x+2) = 2 ·6·8 = 24cm2 . The area of the rhombus is 4·24cm2 = 96cm2 .

24. Point P lies inside an equilateral triangle whose sides are of length 2. If the distances from P
to each side of the triangle are x, y and z, what is x + y + z?

√ √
(a) 3 (b) 2 3

√
(c) 3 3 (d) Not enough information given.

(e) None of the above

Solution (a): The picture below shows an equilateral triangle ABC with the point P.
The area of the ABC is equal to
B
1
q p
2h = h = 22 − 12 = 3
2

The same area can be computed as the sum of the areas

h
2 x 2 of triangles BPC, APC and APB.
The area of APC is 12 2y = y.
P
z The area of BPC is 12 2x = x.
The area of APB is 12 2z = z.
y √
Therefore x + y + z = h = 3.
C 2 A

25. A rectangle is given with length l and width w, where l > w. One of them is increased by 20%,
while the other is decreased by 20%. What happens to the area of the rectangle?

(a) It stays the same. (b) It always decreases.

(c) It always increases. (d) It increases only if the l is increased.

(e) It increases only if the w is increased.

Solution (b): Let’s call one of the dimensions a and the other b. Since the area is A = ab
and multiplication is commutative, it does not matter which one is which! If we increase a by
20%, the new dimension will be 1.2a. If we decrease b by 20%, the new dimension will be 0.8b.
Then the new area will be (1.2a)(0.8b) = 1.2 · 0.8ab = 0.96ab. The area will always decrease
by 4%.