1052 toothpicks can be grouped into 4 groups of third power of 6 ([tex]6^{3}[/tex]), 5 groups of second power of 6 ([tex]6^{2}[/tex]), 1 group of first power of 6 ([tex]6^{1}[/tex]) and 2 groups of zeroth power of 6 ([tex]6^{0}[/tex]).
The number 1052, written as a base 6 number is 4512
Given: 1052 toothpicks
To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number
First we note that, [tex]6^{0}=1,6^{1}=6,6^{2}=36,6^{3}=216,6^{4}=1296[/tex]
This implies that [tex]6^{4}[/tex] exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.
Now, [tex]6^{3}=216[/tex] and [tex]216\times 5=1080, 216\times 4=864[/tex]. This implies that [tex]216\times 5[/tex] exceeds 1052 and thus there can be at most 4 groups of [tex]6^{3}[/tex].
Then,
[tex]1052-4\times6^{3}[/tex]
[tex]1052-4\times216[/tex]
[tex]1052-864[/tex]
[tex]188[/tex]
So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.
Now, [tex]6^{2}=36[/tex] and [tex]36\times 5=180, 36\times 6=216[/tex]. This implies that [tex]36\times 6[/tex] exceeds 188 and thus there can be at most 5 groups of [tex]6^{2}[/tex].
Then,
[tex]188-5\times6^{2}[/tex]
[tex]188-5\times36[/tex]
[tex]188-180[/tex]
[tex]8[/tex]
So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.
Now, [tex]6^{1}=6[/tex] and [tex]6\times 1=6, 6\times 2=12[/tex]. This implies that [tex]6\times 2[/tex] exceeds 8 and thus there can be at most 1 group of [tex]6^{1}[/tex].
Then,
[tex]8-1\times6^{1}[/tex]
[tex]8-1\times6[/tex]
[tex]8-6[/tex]
[tex]2[/tex]
So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.
Now, [tex]6^{0}=1[/tex] and [tex]1\times 2=2[/tex]. This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.
This concludes the grouping.
Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 ([tex]6^{3}[/tex]), 5 groups of second power of 6 ([tex]6^{2}[/tex]), 1 group of first power of 6 ([tex]6^{1}[/tex]) and 2 groups of zeroth power of 6 ([tex]6^{0}[/tex]).
Then,
[tex]1052=4\times6^{3}+5\times6^{2}+1\times6^{1}+2\times6^{0}[/tex]
So, the number 1052, written as a base 6 number is 4512.
Learn more about change of base of numbers here:
https://brainly.com/question/14291917
