3A. The decimal representation of 5/11 is 0.454545.
(This is equal to 0.454545….)
What is the sum of the first 2023 digits after the decimal point?
3B. The polygon in the diagram has a perimeter of
100cm.
All pairs of adjacent sides have the same length and
meet at right angles.
What is the area of the polygon in square cm?
3C. Adam, Beth and Carol are each thinking of a prime number.
The sum of Adam’s prime and Beth’s prime is 21.
The sum of Beth’s prime and Carol’s prime is 25.
What is the sum of Adam’s prime, Beth’s prime and Carol’s prime?
3D. Spencer writes down three consecutive counting numbers.
The first is a multiple of 5, the second is a multiple of 3, and the third is
a multiple of 4.
What is a the least possible sum of Spencer’s three numbers?
3E. Given the graph shown, Abigail moves a
piece from the top point to the bottom
point.
She must move downwards, in either a
straight or sideways direction.
How many different pathways are
possible?
The correct answer and explanation is:
3A. The sum of the first 2023 digits after the decimal point of 511\frac{5}{11}:
The decimal representation of 511=0.454545…\frac{5}{11} = 0.454545\ldots is repeating, with “45” being the repeating block of two digits.
Each pair of digits (45) contributes 4+5=94 + 5 = 9 to the total sum.
- There are 20232=1011\frac{2023}{2} = 1011 complete repetitions of “45” in the first 2022 digits, contributing 1011×9=90991011 \times 9 = 9099.
- The 2023rd digit is the first digit of the next repetition, which is 4.
Thus, the sum of the first 2023 digits is:
9099+4=91039099 + 4 = 9103
Answer: 9103\mathbf{9103}
3B. The area of the polygon:
The polygon is made of sides of equal length meeting at right angles, forming a rectilinear shape. Since the perimeter is 100 cm100 \, \text{cm}, the total length of all sides is 100100.
If the polygon has nn sides and is rectilinear, its area is maximized when it forms a square. The side length of such a square would be: Side length=1004=25 cm\text{Side length} = \frac{100}{4} = 25 \, \text{cm}
The area of this square is: Area=25×25=625 cm2\text{Area} = 25 \times 25 = 625 \, \text{cm}^2
Answer: 625 cm2\mathbf{625 \, \text{cm}^2}
3C. The sum of Adam’s, Beth’s, and Carol’s primes:
Let Adam’s prime be AA, Beth’s prime be BB, and Carol’s prime be CC.
From the problem, we know:
- A+B=21A + B = 21
- B+C=25B + C = 25
From (1) and (2):
C−A=4⇒C=A+4C – A = 4 \quad \Rightarrow \quad C = A + 4
Now solve for A,B,CA, B, C:
- The prime numbers that sum to 2121 are A=2A = 2, B=19B = 19 (only valid pair).
- Substituting B=19B = 19 into B+C=25B + C = 25:
19+C=25⇒C=619 + C = 25 \quad \Rightarrow \quad C = 6
However, C=6C = 6 is not prime! Checking again: Only valid solution matches 2,192, 19—so, write the prime adjustment.
Let me provide you with the corrected final answers and explanations for all questions.
3C. The sum of Adam’s, Beth’s, and Carol’s primes:
Given:
- A+B=21A + B = 21
- B+C=25B + C = 25
From these equations: C−A=4soC=A+4C – A = 4 \quad \text{so} \quad C = A + 4
Checking valid combinations for AA, BB, and CC, where all are prime numbers:
- For A=2A = 2, B=19B = 19:
C=A+4=2+4=23C = A + 4 = 2 + 4 = 23, and both 19 and 23 are prime.
Thus, the primes are A=2A = 2, B=19B = 19, C=23C = 23.
Their sum is: 2+19+23=442 + 19 + 23 = 44
Answer: 44\mathbf{44}
3D. The least possible sum of Spencer’s three consecutive numbers:
Spencer writes three consecutive integers n,n+1,n+2n, n+1, n+2:
- nn is a multiple of 5, so n=5kn = 5k.
- n+1n+1 is a multiple of 3.
- n+2n+2 is a multiple of 4.
We find the smallest nn satisfying all conditions.
- n+1n+1 must be divisible by 3: 5k+1≡0(mod3)⇒5k≡−1≡2(mod3)5k + 1 \equiv 0 \pmod{3} \Rightarrow 5k \equiv -1 \equiv 2 \pmod{3}.
Solving 5k≡2(mod3)5k \equiv 2 \pmod{3}: k≡1(mod3)k \equiv 1 \pmod{3}. - n+2n+2 must be divisible by 4: 5k+2≡0(mod4)⇒5k≡−2≡2(mod4)5k + 2 \equiv 0 \pmod{4} \Rightarrow 5k \equiv -2 \equiv 2 \pmod{4}.
Solving 5k≡2(mod4)5k \equiv 2 \pmod{4}: k≡2(mod4)k \equiv 2 \pmod{4}.
Combining k≡1(mod3)k \equiv 1 \pmod{3} and k≡2(mod4)k \equiv 2 \pmod{4}:
Using the Chinese Remainder Theorem, k≡10(mod12)k \equiv 10 \pmod{12}.
Thus, k=10k = 10, and n=5k=50n = 5k = 50.
The three numbers are 50,51,5250, 51, 52. Their sum is: 50+51+52=15350 + 51 + 52 = 153
Answer: 153\mathbf{153}
3E. The number of different pathways Abigail can take:
This is a classic combinatorics problem based on paths in a grid. Abigail must move from the top to the bottom, moving either straight or diagonally sideways.
Let the number of levels in the graph be nn. At each level:
- Abigail has multiple paths from each vertex to vertices in the next row.
- The number of paths at each point corresponds to summing contributions from previous rows.
If the graph has, for example, 4 levels and each level is connected to 3-4 vertices, apply combinatorics (binomial sums).
To solve fully, I’ll analyze the edges shown for pathways if given!