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.
  • Assume $A \land B \land C$.
  • Then $C$ is true.
  • Since $A$ and $C$, we know $A \land C$
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:
  • 9
  • 5
  • 11
  • 8
  • 2
  • 10
  • 6
  • 1
  • 7
  • 4
  • 3
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:
  • By definition, there exists $q_1$ such that $b=a\cdot q_1$ and there exists $q_2$ such that $c = b\cdot q_2.$
  • Assume that $a|b$ and $b|c.$
  • Let $a, b, c$ be arbitrary positive integers.
  • By definition $a|c.$
  • 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
  • Correct code
  • Distractor 1
  • Distractor 2
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 between the distractor group
  • Some correct block after the distractor group
  • Some correct block before the distractor group
  • Correct code 1
  • Distractor 1 for correct code 1
  • Distractor 1 for correct code 2
  • Distractor 2 for correct code 1
  • Distractor 2 for correct code 2
  • Correct code 2
Construct your solution here:
Drag from here:
  • $2^n = 2 \cdot 2^{n-1}$
  • 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
  • $n=1: 2^1 \gt 1$
  • Let $n$ be an arbitrary integer.
Construct your solution here:
  • Base Case
  • Induction Hypothesis
  • Inductive Step

Correct answer

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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 20:44:49 (CDT)
Duration:
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    },
    {
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      "inner_html": " 8",
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      "inner_html": " 7",
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    },
    {
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      "depends": [
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      "ranking": -1,
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      "inner_html": " 9",
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    {
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    {
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      "depends": [],
      "ranking": -1,
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      "inner_html": "Some correct block before the distractor group",
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    },
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      "inner_html": "Correct code",
      "distractor_bin": "f30e8ba8-1c10-4ced-a738-a01a83901b50",
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      "indent": null,
      "depends": [],
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      "inner_html": "Some correct block before the distractor group",
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    },
    {
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      "inner_html": "Correct code",
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    {
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      "inner_html": "Correct code 1",
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      "inner_html": "Some correct block between the distractor group",
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    {
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      "inner_html": "Correct code 2",
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      "inner_html": "Some correct block after the distractor group",
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  "test-ordered-grading-method-all-or-nothing": [
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    {
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  "test-ordered-grading-method-partial-credit": [
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    {
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      "inner_html": "third",
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  ]
}
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