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Element pl-order-blocks: Dragging blocks to form the solution of a problem

This pl-order-blocks has become increasingly complex, with many different possible combinations of options. Here we test a bunch of these different configurations.

Drag from here:
  • def my_sum(first, second):
  • return first + second
Construct your solution here:
Drag from here:
  • def my_sum(first, second):
  • return first + second
Construct your solution here:
Drag from here:
  • Assume $A \land B \land C$.
  • Then $A$ is true.
  • Then $C$ is true.
  • Since $A$ and $C$, we know $A \land C$
Construct your solution here:
Drag from here:
  • Then $A$ is true.
  • Since $A$ and $C$, we know $A \land C$
  • Assume $A \land B \land C$.
  • Then $C$ is true.
Construct your solution here:
Drag from here:
  • def my_sum(first, second):
  • return sum
  • sum + first
  • sum + second
  • sum += first
  • sum += second
  • sum = 0
Construct your solution here:
Drag from here:
  • 2
  • 5
  • 6
  • 1
  • 4
  • 10
  • 7
  • 11
  • 3
  • 9
  • 8
Construct your solution here:
Drag from here:
  • A
  • AA
  • B
  • C
  • D
Construct your solution here:
Drag from here:
  • first
  • second
Construct your solution here:
Drag from here:
  • all or nothing first
  • all or nothing second
  • all or nothing third
Construct your solution here:
Drag from here:
  • first
  • second
  • third
Construct your solution here:
Drag from here:
  • Assume that $a|b$ and $b|c.$
  • Let $a, b, c$ be arbitrary positive integers.
  • By definition $a|c.$
  • By definition, there exists $q_1$ such that $b=a\cdot q_1$ and there exists $q_2$ such that $c = b\cdot q_2.$
  • We can see that $c = b\cdot q_2 = (a\cdot q_1)\cdot q_2 = a \cdot (q_1\cdot q_2).$
Construct your solution here:
Drag from here:
  • Some correct block before the distractor group
  • Distractor 2
  • Distractor 1
  • Correct code
Construct your solution here:
Drag from here:
  • Some correct block before the distractor group
  • Correct code
  • Distractor 1
  • Distractor 2
Construct your solution here:
Drag from here:
  • Some correct block before the distractor group
  • Some correct block after the distractor group
  • Some correct block between the distractor group
  • Distractor 1 for correct code 1
  • Distractor 1 for correct code 2
  • Correct code 2
  • Distractor 2 for correct code 2
  • Distractor 2 for correct code 1
  • Correct code 1
Construct your solution here:
Drag from here:
  • $2^n = 2 \cdot 2^{n-1}$
  • $n=1: 2^1 \gt 1$
  • Let $n$ be an arbitrary integer.
  • Assume $2^k \gt k$ for some integer $1 \leq k \lt n$
  • Since $2(n-1) \gt 2n-n$, then $2^n \gt n$
  • $2^n \gt 2(n-1)$ by IH
Construct your solution here:
  • Base Case
  • Induction Hypothesis
  • Inductive Step

Correct answer

Correct answer (there may be other correct orders, correct indentation required):
  • def my_sum(first, second):
  • return first + second
Correct answer (there may be other correct orders, correct indentation required):
  • def my_sum(first, second):
  • return first + second
Correct answer (there may be other correct orders):
  • Assume $A \land B \land C$.
  • Then $A$ is true.
  • Then $C$ is true.
  • Since $A$ and $C$, we know $A \land C$
Correct answer (there may be other correct orders):
  • Assume $A \land B \land C$.
  • Then $A$ is true.
  • Then $C$ is true.
  • Since $A$ and $C$, we know $A \land C$
Correct answer (there may be other correct orders, correct indentation required):
  • def my_sum(first, second):
  • sum = 0
  • sum += second
  • sum += first
  • return sum


Incorrect Blocks (these blocks should not be included in your solution):
  • sum + first
  • sum + second
Correct answer (there may be other correct orders):
  • 1
  • 4
  • 2
  • 3
  • 5
  • 6
  • 8
  • 7
  • 9
  • 10
  • 11
Correct answer (there may be other correct orders):
  • A
  • AA


Incorrect Blocks (these blocks should not be included in your solution):
  • B
  • C
  • D
Correct answer (there may be other correct orders):
  • first
  • second
Correct answer (in the specified order):
  • all or nothing first
  • all or nothing second
  • all or nothing third
Correct answer (in the specified order):
  • first
  • second
  • third
Correct answer (there may be other correct orders):
  • Let $a, b, c$ be arbitrary positive integers.
  • Assume that $a|b$ and $b|c.$
  • By definition, there exists $q_1$ such that $b=a\cdot q_1$ and there exists $q_2$ such that $c = b\cdot q_2.$
  • We can see that $c = b\cdot q_2 = (a\cdot q_1)\cdot q_2 = a \cdot (q_1\cdot q_2).$
  • By definition $a|c.$
Correct answer (there may be other correct orders):
  • Some correct block before the distractor group
  • Correct code


Incorrect Blocks (these blocks should not be included in your solution):
  • Distractor 2
  • Distractor 1
Correct answer (there may be other correct orders):
  • Some correct block before the distractor group
  • Correct code


Incorrect Blocks (these blocks should not be included in your solution):
  • Distractor 1
  • Distractor 2
Correct answer (there may be other correct orders):
  • Some correct block before the distractor group
  • Correct code 1
  • Some correct block between the distractor group
  • Correct code 2
  • Some correct block after the distractor group


Incorrect Blocks (these blocks should not be included in your solution):
  • Distractor 1 for correct code 1
  • Distractor 1 for correct code 2
  • Distractor 2 for correct code 2
  • Distractor 2 for correct code 1
Correct answer (there may be other correct orders):
  • Base Case
  • $n=1: 2^1 \gt 1$
  • Induction Hypothesis
  • Assume $2^k \gt k$ for some integer $1 \leq k \lt n$
  • Inductive Step
  • Let $n$ be an arbitrary integer.
  • $2^n = 2 \cdot 2^{n-1}$
  • $2^n \gt 2(n-1)$ by IH
  • Since $2(n-1) \gt 2n-n$, then $2^n \gt n$

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Question

QID:
Title:
Element pl-order-blocks: Dragging blocks to form the solution of a problem

Variant

Started at:
2026-08-04 20:44:51 (CDT)
Duration:
0s
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      "inner_html": "Inductive Step",
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    },
    {
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      "inner_html": "$2^n = 2 \\cdot 2^{n-1}$",
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      "inner_html": "$2^n \\gt 2(n-1)$  by IH",
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    },
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      "inner_html": "sum += second",
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}
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