ONE-EIGHTH STAGE 2024
Keta SHTS: 57 Points
Kumasi Academy: 46 Points
Acherensua SHS: 22 Points
ROUND 5 - RIDDLE
RIDDLE
I am a kind of triangle
I appear Isosceles
ANSWER: Pascal’s Triangle
EXPLANATION:
Pascal's triangle is a symmetric, isosceles-shaped array of the binomial coefficients $\binom{n}{r}$.
Each entry is the sum of the two entries directly above it.
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PRACTICE QUESTIONS
1. RIDDLE:
I am a mathematician.
I lived in France in the seventeenth century.
A unit of pressure is named after me.
I built one of the first mechanical calculators.
With Pierre de Fermat, I helped start the theory of probability.
A triangle of binomial coefficients carries my name.
Who am I?
ANSWER: BLAISE PASCAL
EXPLANATION:
Blaise Pascal (1623–1662) studied the triangle of binomial coefficients now named after him.
2. RIDDLE:
I come from Pascal's triangle.
I double from one row to the next.
For row 3, I am 8.
I count all the subsets of a set with n elements.
I equal $\binom{n}{0} + \binom{n}{1} + \cdot s + \binom{n}{n}$.
For row $n$, I am $2^n$.
Who am I?
ANSWER: ROW SUM ($2^n$)
EXPLANATION:
Row 3 is 1, 3, 3, 1 and $1 + 3 + 3 + 1 = 8 = 2^3$.
3. RIDDLE:
I am a sequence found in Pascal's triangle.
I lie along a diagonal, next to the edge of 1s.
My terms increase by 1 each time.
Each of my terms is $\binom{n}{1}$.
I am the set of counting numbers.
My terms are 1, 2, 3, 4, 5, …
Who am I?
ANSWER: NATURAL NUMBERS
EXPLANATION:
$\binom{n}{1} = n$, so the second diagonal is 1, 2, 3, 4, …
4. RIDDLE:
I am a rule behind Pascal's triangle.
I explain how each row is built from the one above it.
I involve three neighbouring entries.
I link binomial coefficients in consecutive rows.
I say that $\binom{n}{r} + \binom{n}{r + 1} = \binom{n + 1}{r + 1}$.
I say that each entry is the sum of the two entries above it.
Who am I?
ANSWER: PASCAL'S RULE
EXPLANATION:
$\binom{4}{1} + \binom{4}{2} = 4 + 6 = 10 = \binom{5}{2}$
5. RIDDLE:
I am linked to Pascal's triangle.
The first five rows of the triangle, read as numbers, give me.
My first five values are 1, 11, 121, 1331 and 14641.
After row 4, carrying spoils the simple pattern.
I am the powers of a two-digit number.
I am $11^0, 11^1, 11^2, \ldots$
Who am I?
ANSWER: POWERS OF 11
EXPLANATION:
$11^4 = 14641$, which is row 4: 1, 4, 6, 4, 1.
6. RIDDLE:
I am an algebraic result.
Pascal's triangle gives my coefficients.
I have $n + 1$ terms when the power is $n$.
My terms have powers that add up to n.
For $n = 2$, I am $a^2 + 2ab + b^2$.
I am the expanded form of $(a + b)^n$.
Who am I?
ANSWER: BINOMIAL EXPANSION
EXPLANATION:
$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$, using row 3: 1, 3, 3, 1.