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Element pl-order-blocks-optional: Dragging blocks to form one of the solutions to a problem

Drag and drop SOME of the following blocks to prove the statement:

Construct a python function that computes the sum of two numbers.

Drag from here: 
  • 1
    def my_sum(first, second):

  • 1
    return sum

  • 1
    sum += first + second

  • 1
    sum += first

  • 1
    sum += second

  • 1
    sum = 0
Construct your solution here:
Drag and drop SOME of the following blocks to prove the statement:

If $n$ is an even integer, then $n + 10$ is also an even integer.

Drag from here:
  • Since $m$ is integer, $m-5$ is integer
  • Since $m$ is an integer, $m+5$ is an integer.
  • We can rewrite this as $n = 2m - 9 = 2(m - 5) + 1.$
  • Therefore, $n + 10$ is an even integer.
  • So $n + 10$ can be rewritten as $2$ times some integer
  • Then we know that $n + 10 = 2m + 10 = 2(m + 5)$
  • Suppose by way of contradiction that $n + 10$ is odd.
  • Let $n$ be an arbitrary even integer.
  • Since $n$ can be written as 2 times and integer plus 1, then $n$ must be odd, which is a contradition.
  • By the definition of even, there exists some integer $m$ such that $n = 2m$.
  • Then $n + 10 = 2m + 1$ for some integer $m$
Construct your solution here:

Correct answer

Correct answer (there may be other answers, correct indentation required):
  • 1
    def my_sum(first, second):

  • 1
    sum = 0

  • 1
    sum += second

  • 1
    sum += first

  • 1
    return sum
Correct answer (there may be other answers):
  • Let $n$ be an arbitrary even integer.
  • By the definition of even, there exists some integer $m$ such that $n = 2m$.
  • Since $m$ is an integer, $m+5$ is an integer.
  • Then we know that $n + 10 = 2m + 10 = 2(m + 5)$
  • So $n + 10$ can be rewritten as $2$ times some integer
  • Therefore, $n + 10$ is an even integer.

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QID:
Title:
Element pl-order-blocks-optional: Dragging blocks to form one of the solutions to a problem

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Started at:
2026-09-19 09:08:08 (CDT)
Duration:
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