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10. The H.C.F of 120, 150, and 210 is [tex]k^2 - 6[/tex]. What is the value of [tex]k[/tex]?

A. 5
B. 7
C. [tex]2^3 \times 3^2 \times 5[/tex]
D. [tex]2 \times 3 \times 5[/tex]

11. The H.C.F of 17, 23, and 29 is:

A. 1
B. 7
C. 13
D. 11

Answer :

We start by finding the highest common factor (H.C.F.) for each set of numbers.

________________________________________
1. For the numbers [tex]\(120\)[/tex], [tex]\(150\)[/tex], and [tex]\(210\)[/tex]:

 a) First, find the H.C.F. of [tex]\(120\)[/tex] and [tex]\(150\)[/tex]. It turns out that
  [tex]\[
  \text{H.C.F.}(120,150)=30.
  \][/tex]

 b) Next, find the H.C.F. of [tex]\(30\)[/tex] (the result above) and [tex]\(210\)[/tex]:
  [tex]\[
  \text{H.C.F.}(30,210)=30.
  ]

Thus, the overall H.C.F. is:
\[
\text{H.C.F.}(120,150,210)=30.
\][/tex]

According to the problem, this H.C.F. is expressed as:
[tex]\[
k^2-6=30.
\][/tex]

To find [tex]\(k\)[/tex], solve the equation:
[tex]\[
\begin{aligned}
k^2 - 6 &= 30,\\[1mm]
k^2 &= 30 + 6 = 36,\\[1mm]
k &= \sqrt{36}=6.
\end{aligned}
\][/tex]

So, the value of [tex]\(k\)[/tex] is [tex]\(6\)[/tex].

________________________________________
2. For the numbers [tex]\(17\)[/tex], [tex]\(23\)[/tex], and [tex]\(29\)[/tex]:

Since these are all prime numbers and do not have any common factors other than [tex]\(1\)[/tex]:
[tex]\[
\text{H.C.F.}(17,23,29)=1.
\][/tex]

________________________________________
Final Answers:

10. The value of [tex]\(k\)[/tex] is [tex]\(\boxed{6}\)[/tex].

11. The H.C.F. of [tex]\(17\)[/tex], [tex]\(23\)[/tex], and [tex]\(29\)[/tex] is [tex]\(\boxed{1}\)[/tex].

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