Registered School: Stratford Middle School (SAN JOSE CA)
Mathematical Olympiads
November 16, 2020
For Elementary & Middle Schools
Directions to Students: After all questions have been read by your PICO, you will have 30 minutes to complete this contest. You may
not have a pen or pencil in your hand while the PICO reads the set of questions to the class. Calculators are not permitted. All work is
to be done on the pages provided. Answers must be placed in the corresponding boxes in the answer column.
Name: ____________________________________________________
1A Evaluate: 51 – 33 + 42 – 33 + 24 – 33 + 15 – 33.
1B Find the sum of the following nine products:
1 × (10 – 1), 2 × (10 – 2), 3 × (10 – 3), … , 9 × (10 – 9).
1C A square tablecloth covers an area of 256 in2. It is folded in
half some number of times until the area that it covers is 8 in2.
How many times was it folded in half?
Copyright © 2020 by Mathematical Olympiads for Elementary and Middle Schools, Inc. All rights reserved.
Registered School: Stratford Middle School (SAN JOSE CA)
Name: _________________________________________________________________
1D In a warehouse, a group of identical robots fills orders. In 10 minutes, 5 robots can
fill 20 orders. If 8 robots worked at that same rate for an hour, how many orders
would they fill?
Answer Column
–Page may be folded along dotted line.–
1A
1B
1C
1E The number 2A5A0A2 is divisible by 99. What is the digit A?
–Page may be folded along dotted line–
1D
1E
Do Not Write in this Space.
For PICO’s Use Only.
SCORE:
Registered School: Stratford Middle School (SAN JOSE CA)
Mathematical Olympiads
November 16, 2020
For Elementary & Middle Schools
SOLUTIONS AND ANSWERS
1A
1A METHOD 1 Strategy: Find the difference of each pair of numbers and then find
the sum.
Notice that each pair (51 – 33) + (42 – 33) + (24 – 33) + (15 – 33) =
(+18) + (+9) + (– 9) + (–18). The differences result in two pairs of additive
inverses, (+28) + (–28) + (+9) + (–9), which sum to 0.
0
71%
1B
METHOD 2 Strategy: Use the Commutative and Associative Properties to find the
sum of the positive numbers and the sum of the negative numbers and then combine.
Looking at the problem, there are four groups of −33: 4 × (−33) = −132. Also, there
are two groups of +66 (15 + 51 and 24 + 42): 2 × (+66) = +132.
It follows that (−132) + (+132) = 0.
METHOD 3 Strategy: Do the arithmetic left to right.
Start with 51 – 33 = 18, and 18 + 42 = 60, and 60 – 33 = 27, and 27 + 24 = 51, and
51 – 33 = 18, and 18 + 15 = 33, and finally, 33 – 33 = 0.
165 43%
1C
FOLLOW UP: Evaluate: – 25 + 46 – 25 + 34 – 25 + 66 – 25 + 54. [100]
1B Strategy: Look for patterns in the parentheses and the product terms.
Notice that the first term of each product increases consecutively from 1 to 9 and
the second term for each difference in the parentheses increases by 1 so each
resulting difference decreases consecutively from 9 to 1. This pattern creates the
products 1 × 9 and 9 × 1, 2 × 8 and 8 × 2, 3 × 7 and 7 × 3, 4 × 6 and 6 × 4, and
5 × 5. The sum will be 9 + 9 + 16 + 16 + 21 + 21 + 24 + 24 + 25 =
18 + 32 + 42 + 48 + 25 = 165.
5
37%
1D
192
FOLLOW UP: Find the sum of the following nine products:
1 × (11 + 19), 2 × (12 + 18), 3 × (13 + 17), …, 9 × (19 + 11). [1350]
1C METHOD 1 Strategy: Divide the given area in half until the result is 8 in2.
Each fold cuts the area of the rectangle in half. The first fold reduces the area to
256 / 2 = 128 in2. The second fold reduces the area to 128 / 2 = 64 in2. The third
fold reduces the area to 64 / 2 = 32 in2. The fourth fold reduces the area to
32 / 2 = 16 in2. The fifth fold reduces the area to 16 / 2 = 8 in2.
The tablecloth must be folded 5 times.
37%
1E
METHOD 2 Strategy: Work backwards.
Unfold the tablecloth which will double the area: 8 × 2 = 16, 16 × 2 = 32,
32 × 2 = 64, 64 × 2 = 128, and 128 × 2 = 256. Since we multiplied 8 five times to
get to 256, the answer is 5.
FOLLOW UP: A square napkin has a side length of 12 inches. It is folded in half four
times. What is the resulting area of the folded napkin? [9 in2]
Copyright © 2020 by Mathematical Olympiads for Elementary and Middle Schools, Inc. All rights reserved.
3
31%
Registered School: Stratford Middle School (SAN JOSE CA)
1D METHOD 1 Strategy: Find the amount of orders one robot can fulfill in 10 minutes.
Since 5 robots can complete 20 orders in 10 minutes, 1 robot can complete 20 ÷ 5 = 4 orders in 10
minutes. Therefore, 8 robots can complete 4 × 8 = 32 orders in 10 minutes. Since we want to know how
many orders are completed in one hour (60 min ÷ 10 = 6), we can see that the 8 robots will fulfill
32 × 6 = 192 orders in 1 hour.
METHOD 2 Strategy: Use proportions.
Since 5 robots working for 10 minutes produces 20 orders, it takes 50 robot minutes to complete the 20
orders. We are looking for the number of orders filled by 8 robots in 60 minutes or 480 robot minutes. Set
up and solve the proportion:
→ x = 192.
FOLLOW UP: Six identical robots can fulfill 36 orders in 18 minutes. How many minutes will it take four of
these identical robots to fulfill 480 orders? [360]
1E METHOD 1 Strategy: Use the divisibility rules.
If a number is divisible by 99, it must be divisible by 9 and 11. A number is divisible by 9 if the sum of its
digits is divisible by 9. The sum 2 + A + 5 + A + 0 + A + 2 = 9 + 3A, therefore 3A must be a multiple of
9. Thus, A can be 0, 3, 6, or 9. A number is divisible by 11 if the alternating digit sum is a multiple of 11.
Since 2 – A + 5 – A + 0 – A + 2 = 9 – 3A, look for a resulting difference that will be a multiple of 11 by
substituting 0, 3, 6, and 9 for A: 9 – 3(0) = 9, 9 – 3(3) = 0, 9 – 3(6) = –9, and 9 – 3(9) = –18. The only
difference that is a multiple of 11 is 0, therefore A = 3. [Note: The solution does not require the divisibility
by 9 if you use the divisibility by 11 first.]
METHOD 2 Strategy: Use trial and success to find A.
If you only know the test for numbers divisible by 9, that is, the sum of the digits must be a multiple of 9,
use division to test the possibilities. Since 2 + A + 5 + A + 0 + A + 2 = 9 + 3A, 3A must be a multiple of
9, so A can be 0, 3, 6, or 9. Divide 2050002 by 11 to get a remainder of 9. Divide 2353032 by 11 to get
213912. Therefore, A = 3. When you divide the other two possible values for A, there is a non-zero
remainder: 2656062/11 has remainder 2 and 2959092/11 has remainder 4.
FOLLOW UP: Find the least 6-digit number that is divisible by 99. [100089]
NOTE: Other FOLLOW-UP problems related to some of the above can be found in our three
contest problem books and in “Creative Problem Solving in School Mathematics.”
Visit www.moems.org for details and to order.
Registered School: Stratford Middle School (SAN JOSE CA)
Mathematical Olympiads
December 14, 2020
For Elementary & Middle Schools
Directions to Students: After all questions have been read by your PICO, you will have 30 minutes to complete this contest. You may
not have a pen or pencil in your hand while the PICO reads the set of questions to the class. Calculators are not permitted. All work is
to be done on the pages provided. Answers must be placed in the corresponding boxes in the answer column.
Name: ____________________________________________________
2A Compute the sum of all nine 2-digit numbers with
a ones digit of 6.
2B Compute 123 × 45 + 246 × 30 + 369 × 15.
2C Nat is thinking of a four-digit counting number. The
thousands digit and the hundreds digit are the same. The sum
of all 4 digits is 28. The four-digit number is an odd multiple
of 5. Exactly two digits are even. What is Nat’s number?
Copyright © 2020 by Mathematical Olympiads for Elementary and Middle Schools, Inc. All rights reserved.
Registered School: Stratford Middle School (SAN JOSE CA)
Name: _________________________________________________________________
2D A player can move from point to point only in a direction that
includes a downward motion (slanted or not). In how many distinct
ways can a player move from the top to bottom?
Answer Column
–Page may be folded along dotted line.–
2A
2B
2C
–Page may be folded along dotted line–
2D
2E
Do Not Write in this Space.
For PICO’s Use Only.
SCORE:
2E Point B is halfway between A and C. The distance from
C to D is the same as the distance from D to E, which is
the same as the distance from E to F. The area of triangle
ACF is 180 cm2. How many square centimeters is the
shaded area of triangle BCE?
Registered School: Stratford Middle School (SAN JOSE CA)
Mathematical Olympiads
December 15, 2020
For Elementary & Middle Schools
2A
SOLUTIONS AND ANSWERS
2A METHOD 1 Strategy: Use patterns in each place-value, then add.
List of 2-digit numbers: 16, 26, 36, 46, 56, 66, 76, 86, 96
Compute the sum of the ones digits: 9 × 6 = 54.
Compute the sum of the tens digits: 1+2+3+4+5+6+7+8+9 = 45.
Combine the sums: 45 × 10 + 54 × 1 = 450 + 54 = 504.
METHOD 2 Strategy: Use a method for adding numbers in an arithmetic
sequence.
Each pair of numbers has the same sum: 16 + 96 = 26 + 86 = 36 + 76 = 46 + 66 =
112. Therefore, the sum of the nine numbers is 4 × 112 + 56 = 504.
FOLLOW UPS: (1) Compute the sum of all ten 2-digit numbers with a tens digit of 6.
[645] (2) Compute the sum of all 3-digit numbers with a ones digit of 6. [49590]
2B METHOD 1 Strategy: Find like factors, then use commutativity and redistribute.
The numbers 123 and 15 are common factors in each product.
123 × 15 × [(1 × 3) + (2 × 2) + (3 × 1) = 1845 × (3 + 4 + 3) = 1845 × 10 = 18450.
METHOD 2 Strategy: Multiply and add.
Do the 3 multiplications and then add the results:
123 × 45 = 5535
246 × 30 = 7380
369 × 15 = 5535.
It follows that 5535 + 7380 + 5535 = 18450.
504
56%
2B
18450
41%
2C
8875
53%
2D
12
FOLLOW UP: Compute 121 × 80 + 242 × 60 + 363 × 40 + 484 × 20. [48400]
2C METHOD 1 Strategy: Apply the process of elimination.
Since the number is an odd multiple of 5, the ones digit must be 5. Since the sum of
the 4 digits is 28, the sum of the first 3 digits must be 23. Since exactly 2 digits are
even they must be in the thousands and hundreds columns. Create a table of
possible solutions
Thousands
2
4
6
8
Hundreds Tens
2
19
4
15
6
11
8
7
Sum
23
23
23
23
Comment
Not possible
Not possible
Not possible
Only solution
8%
2E
60
17%
Nat’s number is 8875.
Copyright © 2020 by Mathematical Olympiads for Elementary and Middle Schools, Inc. All rights reserved.
Registered School: Stratford Middle School (SAN JOSE CA)
METHOD 2 Strategy: Model the problem like a cryptarithm.
Note that each letter has only one value but one value could possibly be assigned to multiple letters. Let
the 4-digit number be XXYZ since Nat’s number has the same digit for the thousands and hundreds
places. Then X + X + Y + Z = 28. Since the number is an odd multiple of 5, Z = 5 so 2X + Y = 23. The
only options are 9 + 9 + 5, 8 + 8 + 7, and 7 + 7 + 9. Since two digits are even the only solution is 8875.
FOLLOW UP: In the given problem, list all possible numbers if the sum the digits is 22. [8815, 6655, 4495]
2D METHOD 1 Strategy: Put a weight on each point representing the number of ways to arrive there.
There is only 1 starting point at the top. Assign that point the weight 1.
When pathways split, assign that number to the next points along the pathways.
When two (or more) pathways merge, add their weights together.
Notice that there are 2 + 3 + 7 = 12 pathways from top to bottom.
METHOD 2 Strategy: Label each point, then list pathways (alphabetically, for organization).
List all the pathways: ABEKQT, ABFLQT, ABFMRT, ABFMST, ABGMRT,
ABGMST, ABGNST, ACGMRT, ACGMST, ACGNST, ADHNST, ADJPST.
There are 12 pathways.
FOLLOW UPS: (1) If the figure were turned upside-down, what would the answer be? [12]
(2) Suppose there were two copies of this network, with the bottom point of one linked to
the top point of the other. How many distinct ways can a player move from top to bottom? [144]
2E METHOD 1 Strategy: Create a convenient unit size.
Fit a second copy of the picture to itself to form a big rectangle. Then, partition that
rectangle into fourths across and into thirds up-and-down to make 12 dotted
rectangles as shown. The original triangle ACF is one-half of the 12 dotted
rectangles, or 6 dotted rectangles. So, 6 dotted rectangles have an area of 180 cm2. Or,
1 dotted rectangle has an area of 180 ÷ 6 = 30 cm2. The original shaded triangle BCE has an area
of 1 + ½ + ½ = 2 dotted rectangles. Thus, triangle BCE has an area of 2 × 30 = 60 cm2.
METHOD 2 Strategy: Use proportional reasoning.
Recall that the area of every right triangle is A = ½ × b × h. The area of triangle ACF is 180 cm2. Partition
the base of triangle ACF into halves and partition the height into thirds. The base of triangle BCE is BC =
1/2 of AC and the height of the triangle is CE = 2/3 of CF.
The area of triangle BCE = 180 × 1/2 × 2/3 = 60 cm2.
FOLLOW UP: If point B was moved halfway towards point A from where it began, what would the area of
triangle BCE become? [90 cm2]
NOTE: Other FOLLOW-UP problems related to some of the above can be found in our three
contest problem books and in “Creative Problem Solving in School Mathematics.”
Visit www.moems.org for details and to order.
Registered School: Stratford Middle School (SAN JOSE CA)
Mathematical Olympiads
January 11, 2021
For Elementary & Middle Schools
Directions to Students: After all questions have been read by your PICO, you will have 30 minutes to complete this contest. You may
not have a pen or pencil in your hand while the PICO reads the set of questions to the class. Calculators are not permitted. All work is
to be done on the pages provided. Answers must be placed in the corresponding boxes in the answer column.
Name: ____________________________________________________
3A What number replaces  to make the sentence true?
1 + 3 + 4 + 6 + 6 + 8 + 9 + 11 = 4 × .
3B To figure out a special number, Tammi’s mother was
supposed to double her age and then add 4. Instead, Tammi’s
mother added 4 to her age and then doubled that number.
Tammi’s mother thought the special number was 80. What
was the special number supposed to be?
3C Joshua took a math contest with 25 questions. Each correct
answer earned 6 points while each incorrect answer deducted
4 points. Joshua answered every question and earned 100
points. How many questions did Joshua answer correctly?
Copyright © 2020 by Mathematical Olympiads for Elementary and Middle Schools, Inc. All rights reserved.
Registered School: Stratford Middle School (SAN JOSE CA)
Name: _________________________________________________________________
3D There are 4 cups in a row from left to right.
The green cup is somewhere between the yellow and blue cups.
The red cup is somewhere left of the yellow cup.
Based on that information, how many different arrangements of these cups could
there be?
Answer Column
–Page may be folded along dotted line.–
3A
3B
3C
–Page may be folded along dotted line–
3D
3E
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SCORE:
3E In the figure, QUAD is a quadrilateral. QUAD is “enclosed”
by the rectangle RECT. This means that each vertex of
QUAD is on a side of rectangle RECT. Segment QA is
parallel to two of the sides of the rectangle. The length of QA
is 18 and the length of EC is 13. What is the area of QUAD?
Registered School: Stratford Middle School (SAN JOSE CA)
Mathematical Olympiads
January 11, 2021
For Elementary & Middle Schools
3A
SOLUTIONS AND ANSWERS
3A METHOD 1 Strategy: Use a standard algorithm.
Add the numbers on the left side of the equation, then divide the sum by four.
1 + 3 + 4 + 6 + 6 + 8 + 9 + 11 = 48 and 48 ÷ 4 = 12. Therefore,  = 12.
METHOD 2 Strategy: Compute (using base ten) to find the unknown.
Since 1 + 3 + 6 = 10, 4 + 6 = 10, 9 + 11 = 20, and 40 + 8 = 48 we need to find a
number such that when multiplied by 4 the result is 48. Therefore,  = 12.
12
84%
3B
FOLLOW UPS: (1) Find the value of : 3 + 6 + 12 + 15 + 18 + 18 +19 = 7 × . [13]
(2) What value makes the number sentence true: 2 × 3 × 4 × = 32 × 16? [6]
3B METHOD 1 Strategy: Use logical reasoning.
Work backwards to first find her age. Since Tammi’s mother thought the correct
answer was 80, we know she must have doubled the number 40. We then subtract
four from 40, giving us her original (corre84%ct) age of 36. Now, we follow the
instructions correctly, giving us the special number: (36 × 2) + 4 = 76.
METHOD 2 Strategy: Use operations and algebraic thinking.
Let a represent Tammi’s mother’s age. Then (a + 4) × 2 = 80, so 2a + 8 = 80 and
2a = 72. That means Tammi’s mother’s age is a = 36. To get the special number do
the operations in the correct order: (36 × 2) + 4 = 76.
FOLLOW UPS: (1) Marta was asked to divide her age by six, then multiply that result
by five, and then add two to get her age. What is Marta’s age? [12] (2) Jay solved
the equation 2(2x + 5) = 54 for x. Here are his steps: 2(2x + 5) = 54 →
4x + 5 = 54 → 4x = 49 → x = 12.25. What is the correct value of x? [11]
# incorrect (–4 pts)
1 (–4)
2 (–8)
3 (–12)
4 (–16)
5 (–20)
Total Points
140
130
120
110
100
71%
3C
20
50%
3D
4
33%
3C METHOD 1 Strategy: Create an organized chart.
# correct (6 pts)
24 (144)
23 (138)
22 (132)
21 (126)
20 (120)
76
3E
Josh answered 20
questions correctly.
117
19%
METHOD 2 Strategy: Use trial and error (success).
Trial 1: 6(10) – 4(15) = 0, Trial 2: 6(15) – 4(10) = 50, Trial 3: 6(20) – 4(5) = 100. So, 20 correct answers.
METHOD 3 Strategy: Use algebra.
Let C = the number of questions answered correctly and 25 – C = the number wrong. Then set up and
solve: 6 × C – 4 × (25 – C) = 100 → 6C – 100 + 4C = 100 → 10C = 200 so C = 20.
Copyright © 2020 by Mathematical Olympiads for Elementary and Middle Schools, Inc. All rights reserved.
Registered School: Stratford Middle School (SAN JOSE CA)
METHOD 4 Strategy: Assume all answers were correct and then subtract.
If all the questions are answered correctly, the score would be 150 points. A score of 100 means that 50
points were lost. The difference between getting a correct answer and a incorrect answer is 10 points.
Since 5 incorrect answers results in a deduction of 50 points (50 ÷ 10 = 5), Joshua got 20 answers correct.
FOLLOW UP: (1) Roberto was practicing basketball free throws. He earned 9 points for each one he made
but lost 7 points for each one he missed. Roberto made 20 free throws and scored 84 points. How many
did he miss? [6]
3D METHOD 1 Strategy: Create an organized list and eliminate the unsatisfactory ones.
RYGB RYBG RBYG RBGY RGBY RGYB
GRYB GRBY GYBR GYRB GBRY GBYR
BGYR BGRY BRGY BRYG BYGR BYRG
YRGB YRBG YGBR YGRB YBGR YBRG
Now, eliminate all that can’t possibly meet the parameters: (1) all cases beginning or ending with G, and
(2) all cases ending with R. This eliminates 16 cases, leaving only 8 possibilities. From there, test the two
rules: the result is 4 arrangements that meet both parameters: RYGB RBGY BGRY BRGY.
METHOD 2 Strategy: Use logical reasoning to problem solve.
Start by filling in RY. Then G can be in one of 3 places:
1) Left of R so GRY, thus BGRY.
2) Between R and Y so R G Y, in which case it is BRGY or RBGY.
3) Right of Y, so RYG, in which case it is RYGB.
Thus, there are 4 arrangements that meet the parameters.
FOLLOW UP: Billy, José, Anya, and Kim will be seated in a row. Not all get along with each other. Billy
can’t sit next to Kim, and Kim can’t sit next to Anya. In how many ways could they be seated? [4]
3E METHOD 1 Strategy: Applying the concept of area of polygons.
Use the formulas for area of a rectangle (A = b × h), and area of a triangle (A = b × h/2). QUAD is made
up of triangle QUA with base QA = 18, and triangle QAD also with base QA = 18. The sum of the heights
of these two triangles is 13. The area of QUAD = the sum of the areas of the two triangles. Therefore, the
area of QUAD = (QA + sum of the heights)/2 = (18 × 13)/2 = 117 square units.
METHOD 2 Strategy: Draw additional lines to see a cool relationship.
Draw perpendicular line segments from D and U to line QA. Notice that the
areas of the 4 resulting triangles equal the areas of the 4 triangles in rectangle
RECT that are not in quadrilateral QUAD. Therefore, the area of quadrilateral
QUAD = one-half the area of rectangle RECT = (18 × 13)/2 = 117.
FOLLOW UP: Triangle EKV is enclosed by rectangle LOVE, where EV is a common side and K is on side
LO. KA bisects side EV and is parallel to LE. If OV is 12 and AV is 5, what is the perimeter of triangle
EKV? [36] [Note: The Pythagorean Theorem is required to answer this question.]
NOTE: Other FOLLOW-UP problems related to some of the above can be found in our three
contest problem books and in “Creative Problem Solving in School Mathematics.”
Visit www.moems.org for details and to order.
Registered School: Stratford Middle School (SAN JOSE CA)
Mathematical Olympiads
February 9, 2021
For Elementary & Middle Schools
Directions to Students: After all questions have been read by your PICO, you will have 30 minutes to complete this contest. You may
not have a pen or pencil in your hand while the PICO reads the set of questions to the class. Calculators are not permitted. All work is
to be done on the pages provided. No additional scrap paper is to be used. Answers must be placed in the corresponding boxes in the
answer column.
Name: ____________________________________________________
4A Evaluate: 13 × 20 – 15 × 9 + 20 × 7 – 6 × 15.
4B Olivia’s PASSWORD has four digits. The second digit is
three times the first digit. The number formed by the last two
digits is three times the second digit. The sum of the digits is
21. What is Olivia’s PASSWORD?
4C Kayla is thinking of a 3-digit number.
* It is divisible by 18.
* It is a palindrome (It reads the same left-to-right as it does
right-to-left, like 252 and 848).
* It is one less than a multiple of 5.
What is Kayla’s number?
Copyright © 2020 by Mathematical Olympiads for Elementary and Middle Schools, Inc. All rights reserved.
Registered School: Stratford Middle School (SAN JOSE CA)
Name: _________________________________________________________________
4D In the diagram, each shape has a value and the sums of
the shapes in each of the first three rows is given. Find
the sum of the shapes in the fourth row.
Answer Column
–Page may be folded along dotted line.–
4A
4B
4C
–Page may be folded along dotted line–
4D
4E
Do Not Write in this Space.
For PICO’s Use Only.
SCORE:
4E There is an ample supply of identical blocks (as shown). Each block
is constructed from four 1 × 1 × 1 unit-cubes glued whole face to
whole face. What is the greatest number of such blocks that can be
packed into a box that is 6 units tall, 7 units wide, and 8 units deep
without exceeding the height of the box?
= 31
= 27
= 24
=?
Registered School: Stratford Middle School (SAN JOSE CA)
Mathematical Olympiads
February 9, 2021
For Elementary & Middle Schools
SOLUTIONS AND ANSWERS
4A
4A METHOD 1 Strategy: Apply multiplication before addition and subtraction.
Find the products: 13 × 20 = 260, 15 × 9 = 135, 20 × 7 = 140, and
6 × 15 = 90. The final result is 260 – 135 + 140 – 90 = 175.
METHOD 2 Strategy: Use distributive properties with common factors.
Apply commutative properties to rewrite the problem:
(20 × 13) + (20 × 7) – (15 × 9) – (15 × 6) = 20(13 + 7) – 15(9 + 6) =
20 × 20 – 15 × 15 = 400 – 225 = 175.
175
57%
4B
3927
FOLLOW UP: Evaluate: 20 × 9 – 5 × 24 + 20 × 6 – 5 × 36. [0]
4B METHOD 1 Strategy: Make an organized list of possible solutions.
1st digit
1
2
3
2nd digit
3
6
9
3rd digit
0
1
2
4th digit
9
8
7
Sum
13
17
21
Correct?
NO
NO
YES
41%
4C
414
Therefore, the PASSWORD is 3927.
METHOD 2 Strategy: Use the Guess, Check and Revise, and Repeat strategy.
Step 1 - Since 4 × 3 = 12, the first digit must be less than 4.
Step 2 - Try any number less than 4 for the first digit. Let’s say you try 3. If the first
digit is 3, then the second digit is 9 (3 × 3). The third and fourth digits are 2 and 7
(3 × 9) respectively. Add the digits to determine if the sum of the digits is 21:
3 + 9 + 2 + 7 = 21. The PASSWORD is 3927.
FOLLOW UP: A SECRET CODE has 5 digits. Each digit is different. It is a multiple
of 4 and 11. If the middle 3 digits are 926, in that order, what is the product of the
first and last digits? [40]
45%
4D
26
43%
4E
4C Strategy: Apply divisibility rules and logical reasoning.
Step 1 - The number is “one less than a multiple of 5” so it must end in 4 or 9.
Step 2 - The number is “divisible by 18” so it must be divisible by 2 and 9.
Step 3 - Since the number is divisible by 2, it cannot end in a 9, so it must end in 4.
Also, since it is a palindrome, it must begin with 4.
Step 4 - Since the number is divisible by 9, the sum of the digits must be a multiple
of 9 (9, 18, 27 …). The only possible number that satisfies these conditions is 414.
80
FOLLOW UPS: (1) Using the same rules, what is Kayla’s number if it was a 4-digit number? [4554]
(2) How many possible solutions exist if Kayla’s number was a 5-digit number? [11]
Copyright © 2020 by Mathematical Olympiads for Elementary and Middle Schools, Inc. All rights reserved.
13%
Registered School: Stratford Middle School (SAN JOSE CA)
4D METHOD 1 Strategy: Use number sense.
Notice that if we subtract the first equation from the sum of the second and third equations, we eliminate
the  and the  leaving  +  +  +  +  = (27 + 24) – 31 = 20. Thus,  = 4. Substitute this into
the second and third equations to get  +  +  + 4 + 4 = 27 and  +  + 4 + 4 + 4 = 24. Simplify
these to get  +  +  = 19 and  +  = 12. Subtract the second result from the first result to eliminate
the  :  = 7. Finally, in the first equation replace  with 7 to get  +  + 7 + 7 + 7 = 31 so  = 5.
Since values of each shape are determined, replace the pictures with the values in the last equation to get
5 + 5 + 5 + 7 + 4 = 26.
METHOD 2 Strategy: Replace the pictures with letters and solve algebraically.
Let s =  , c = , and t = . The given equations may now be written as (1) 2s + 3c = 31,
(2) s + 2c + 2t = 27, and (3) s + c + 3t = 24. We wish to find the value of 3s + c + t. Multiply equation (2)
by 3 and equation (3) by 2 and then subtract: 3(s + 2c + 2t = 27) → 3s + 6c + 6t = 81 and
2(s + c + 3t = 24) → 2s + 2c + 6t = 48. Subtract: (3s + 6c + 6t = 81) – (2s + 2c + 6t = 48) → s + 4c = 33.
Multiply this result by 2 to get 2s + 8c = 66 and then subtract equation (1) from this result:
(2s + 8c = 66) – (2s + 3c = 31) → 5c = 35 so c = 7. Substitute this back into equation (1) so 2s + 3(7) = 31
→ 2s = 10 and s = 5. Substitute the values of c and s into equation (2) so 5 + 2(7) + 2t = 27 →
19 + 2t = 27 → 2t = 8 and t = 4.
Finally, substitute all 3 values into the expression 3s + c + t = 3(5) + 7 + 4 = 26.
FOLLOW UP: Given that 3 apples + 4 pears cost $1.95, 4 pears + 1 orange costs $1.60, and
1 orange + 3 apples cost $1.15, what is the cost of 1 apple? [$0.25]
4E Strategy: Apply spatial reasoning.
Notice that 2 of the original blocks may be arranged to form a 2 × 2 × 2 cube.
→
Since the box is 6 units high, we can fit 3 of these cubes on top of each other.
Since the box is 8 units wide, we can fit 4 layers across the width. Since the
length of the box is 7, we can fit 3 layers along the length. Thus, there will be 3 × 4 × 3 = 36 cubes of size
2 × 2 × 2 in the box or 2 × 36 = 72 of the original blocks in the
box. Unfortunately, this is not the maximum number of blocks.
By arranging 4 blocks in a 3 × 3 × 2 block, with a 1 × 2 hole in
→
the middle we can use six of the 2 × 2 × 2 blocks and two of the
3 × 3 × 2 blocks to fill one layer of the box whose face is 6 × 7.
Since the box is 8 units high, we can fill the box with 4 of the
described layers. This will result in 4(6 × 2 + 2 × 4) = 80 of the original cubes
packed in the box.
FOLLOW UP: What is the volume of the unused space in the box? [16 cubic units]
NOTE: Other FOLLOW-UP problems related to some of the above can be found in our three
contest problem books and in “Creative Problem Solving in School Mathematics.”
Visit www.moems.org for details and to order.
Registered School: Stratford Middle School (SAN JOSE CA)
Mathematical Olympiads
March 8, 2021
For Elementary & Middle Schools
Directions to Students: After all questions have been read by your PICO, you will have 30 minutes to complete this contest. You may
not have a pen or pencil in your hand while the PICO reads the set of questions to the class. Calculators are not permitted. All work is
to be done on the pages provided. Answers must be placed in the corresponding boxes in the answer column.
Name: ____________________________________________________
5A What is the value of 47 × 12 + 12 × 53 + 53 × 8 + 8 × 47?
5B A rectangle has perimeter 26 cm, and the length and width are
both whole numbers. What is the greatest possible area this
rectangle may have in square cm?
5C Amanda pours
of her water into Barbara’s empty bottle.
Barbara then pours
of her water into Charlie’s empty
1
of his water into David’s empty
2
bottle. If David now has 18 ounces of water, how many
ounces does Amanda still have?
bottle. Charlie then pours
Copyright © 2020 by Mathematical Olympiads for Elementary and Middle Schools, Inc. All rights reserved.
Registered School: Stratford Middle School (SAN JOSE CA)
Name: _________________________________________________________________
5D A perfect square is the product of any positive whole number used as a factor
exactly two times. [Example: 49 is a perfect square because 7 × 7 = 49] The
positive whole number N, when multiplied by 150, is a perfect square. What is the
least possible positive whole number value for the number N?
Answer Column
5B
cm2
–Page may be folded along dotted line.–
5A
5C
–Page may be folded along dotted line–
5D
5E
Do Not Write in this Space.
For PICO’s Use Only.
SCORE:
5E What is the greatest possible value for STATS in the
cryptarithm shown here given that V = 1?
[Different letters represent different digits, and no leading
digit can equal 0.]
ROTOR
+LEVEL
STATS
Registered School: Stratford Middle School (SAN JOSE CA)
Mathematical Olympiads
March 9, 2021
For Elementary & Middle Schools
5A
SOLUTIONS AND ANSWERS
5A METHOD 1 Strategy: Use Commutative and Distributive properties.
2000
Notice 47 × 12 + 12 × 53 + 53 × 8 + 8 × 47 = 12 ´ (47 + 53) + 8 ´ (53 + 47)
= (12 + 8) ´ (47 + 53)
57%
= 20 ´ 100
5B
= 2000.
METHOD 2 Strategy: Express one of the multipliers in terms of tens and ones.
Rewrite as 47 × (10 + 2) + 53 × (10 + 2) + 53 × (10 – 2) + 47 × (10 – 2) =
(470 + 94) + (530 + 106) + (530 – 106) + (470 – 94) = 940 + 1060 = 2000.
42 cm2
FOLLOW UPS: (1) Evaluate: 47 × 12 + 12 × 53 - 53 × 8 - 8 × 47. [400]
47%
5C
(2) Evaluate: 35 × 11 + 11 × 65 + 65 × 9 + 9 × 35. [2000]
5B METHOD 1 Strategy: The rectangle with greatest area is most “square like.”
20
Using L for length, and W for width; perimeter = 26 → L + W = 13. The most
“square like” whole number values which sum to 13 are: L = 7 and W = 6. The
greatest area is 7 ´ 6 = 42 cm2.
5D
METHOD 2 Strategy: Create a table with possible dimensions.
Length
Width
Area
12
1
12
11
2
22
10
3
30
9
4
36
8
5
40
7
6
42
5C METHOD 1 Strategy: Make a process chart showing the “actions” and “results”
for each person.
Let A be the amount of water Amanda has initially.
Remaining
Solving:
Barbara
A
(3/4) A
(1/4)A
6
The greatest area
is 42 cm2.
FOLLOW UPS: (1) What is the greatest number of cm in the perimeter of a rectangle
with whole number length sides, and with area 42 cm2? [86 cm]
(2) If the area of a rectangle is 64 cm2, how many different perimeters are possible
when the dimensions are whole numbers? [4]
Amanda
27%
25%
5E
97879
Charlie
David
(3/5)(3/4)A (1/2)(9/20)A
= (9/20)A
= (9/40)A
(2/5)(3/4)A
not needed
= (3/10)A
not needed
→ 9A = 720 → A = 80. Thus, Amanda has
= 20 ounces of water.
Copyright © 2020 by Mathematical Olympiads for Elementary and Middle Schools, Inc. All rights reserved.
16%
Registered School: Stratford Middle School (SAN JOSE CA)
METHOD 2 Strategy: Work backwards.
David has 18 oz, which is 1⁄2 of what Charlie had. Therefore, Charlie had 36 oz. Charlie’s 36 oz is 3/5
of what Barbara had. Therefore, Barbara had 60 oz, which is 3⁄4 of what Amanda had. Therefore,
Amanda had 80 oz, and she still has 1⁄4 of 80 = 20 oz.
FOLLOW UPS: (1) How many ounces of water remained in Barbara’s cup? [24]
(2) Mary buys two tickets to a Mets game spending all of her money. She sells one of the tickets for 25%
more than she paid for it. Mary spends 3/5 of that money on souvenirs and 1/2 of what’s left on snacks.
She leaves the park with $12. How much was Mary’s ticket? [$48]
5D METHOD 1 Strategy: Consider the prime factorization of 150 and use the definition of a perfect square.
Notice that 150 = 2 × 3 × 5 × 5. Therefore, 150 × N = 2 × 3 × 5 × 5 × N. For 150 × N to be a least perfect
square, N must contain one factor of 2, and one factor of 3. Thus, N = 2 ´ 3 = 6.
METHOD 2 Strategy: Find a multiple of 150 that is a perfect square.
Since any multiple of 150 must end in 0, try 10 × 10 = 100 (100 is NOT divisible by 150). Next try
20 × 20 = 400 (400 is NOT divisible by 150). Now try 30 × 30 = 900 (900 is divisible by 150). The least
possible whole number value for N is 900/150 = 6.
FOLLOW UPS: (1) How many whole numbers, less than 100, have exactly 3 whole number factors? [4]
(2) What are the dimensions of a cube whose volume is 27000 cubic cm? [30 × 30 × 30]
5E Strategy: Test possible values for STATS from numerically greater to numerically lesser values, starting
with S = 9.
The ten-thousands place has no regrouping, so the ones place causes no regrouping in the tens place.
The fact that the tens place has no regrouping implies that the thousands place has no regrouping either.
This means that the hundreds place has no regrouping either.
Maximizing STATS, we try S = 9 and T = 8. Since T + 1 = A, if T = 8 then A = 9 which is unacceptable
since S = 9. Try T = 7. Then T + 1 = 8 = A. So, perhaps STATS = 97879. To confirm this is the case, one
has to find values for the remaining letters that satisfy: R + L = 9 and O + E = 7, using
35753
numbers from the set {0,2,3,4,5,6}. There are several possible arrangements of which
+
62126
one possible arrangement is: 3 + 6 = 9 and 2 + 5 = 7. Therefore, the greatest possible
97879
value for STATS = 97879 as seen at the right.
FOLLOW UPS: (1) What is the least possible value for STATS shown above? [68986]
(2) Using the digits 1 through 9, find the greatest value of GHI, where each letter is
represented by a different digit. [981]
ABC
+ DEF
GHI
NOTE: Other FOLLOW-UP problems related to some of the above can be found in our three
contest problem books and in “Creative Problem Solving in School Mathematics.”
Visit www.moems.org for details and to order.