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Sagot :
Answer:
[tex]400\; {\rm cm}[/tex].
Step-by-step explanation:
A measuring rod exactly measures a given length if and only if that length is a multiple of the length of the measuring rod.
For example:
- A [tex]40\; {\rm cm}[/tex] measuring rod can exactly measure lengths of [tex]40\; {\rm cm}[/tex], [tex]2 \times 40\; {\rm cm} = 80\; {\rm cm}[/tex], etc.
- A [tex]50\; {\rm cm}[/tex] measuring rod can exactly measure lengths of [tex]50\; {\rm cm}[/tex], [tex]2 \times 50\; {\rm cm} = 100\; {\rm cm}[/tex], etc.
- A [tex]80\; {\rm cm}[/tex] measuring rod can exactly measure lengths of [tex]80\; {\rm cm}[/tex], [tex]2 \times 80\; {\rm cm} = 160\; {\rm cm}[/tex], etc.
If a certain length needs to be exactly measurable using any one of the three measuring rods, that length needs to be a common multiple of all three measuring rod lengths: [tex]40\; {\rm cm}[/tex], [tex]50\; {\rm cm}[/tex], and [tex]80\; {\rm cm}[/tex]. Since the question is asking for the least possible length satisfying these requirements, the goal is equivalent to finding the least common multiple of these three numbers: [tex]40[/tex], [tex]50[/tex], and [tex]80[/tex].
To find the least common multiple of [tex]40[/tex], [tex]50[/tex], and [tex]80[/tex], start by factoring each of the three numbers:
- [tex]40 = 2 \times 2 \times 2 \times 5 = 2^{3} \times 5[/tex].
- [tex]50 = 2 \times 5 \times 5 = 2 \times 5^{2}[/tex].
- [tex]80 = 2 \times 2 \times 2 \times 2 \times 5 = 2^{4} \times 5[/tex].
The prime factors of these three numbers are [tex]2[/tex] and [tex]5[/tex]. Among these three numbers, the maximum power of [tex]2[/tex] is [tex]4[/tex] while the maximum power of [tex]5[/tex] is [tex]2[/tex]. Hence, the least common multiple of these three numbers would be:
[tex]2^{4} \times 5^{2} = 400[/tex].
Therefore, [tex]400\; {\rm cm}[/tex] would be the shortest possible length that can be measured exactly using any one of the three measuring rods.
- [tex](40 \; {\rm cm}) \times 10[/tex].
- [tex](50 \; {\rm cm}) \times 8[/tex].
- [tex](80 \; {\rm cm}) \times 5[/tex].
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