Question

If you had 1052 toothpicks and were asked to group them in powers of 6, how many groups of each power of 6 would you have? Put those together to write that number as a base 6 number.

Answer 1

1052 toothpicks can be grouped into 4 groups of third power of 6 ($6^{3}$), 5 groups of second power of 6 ($6^{2}$), 1 group of first power of 6 ($6^{1}$) and 2 groups of zeroth power of 6 ($6^{0}$).

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, $6^{0}=1,6^{1}=6,6^{2}=36,6^{3}=216,6^{4}=1296$

This implies that $6^{4}$ exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.

Now, $6^{3}=216$ and $216\times 5=1080, 216\times 4=864$. This implies that $216\times 5$ exceeds 1052 and thus there can be at most 4 groups of $6^{3}$.

Then,

$1052-4\times6^{3}$

$1052-4\times216$

$1052-864$

$188$

So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.

Now, $6^{2}=36$ and $36\times 5=180, 36\times 6=216$. This implies that $36\times 6$ exceeds 188 and thus there can be at most 5 groups of $6^{2}$.

Then,

$188-5\times6^{2}$

$188-5\times36$

$188-180$

$8$

So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.

Now, $6^{1}=6$ and $6\times 1=6, 6\times 2=12$. This implies that $6\times 2$ exceeds 8 and thus there can be at most 1 group of $6^{1}$.

Then,

$8-1\times6^{1}$

$8-1\times6$

$8-6$

$2$

So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.

Now, $6^{0}=1$ and $1\times 2=2$. 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 ($6^{3}$), 5 groups of second power of 6 ($6^{2}$), 1 group of first power of 6 ($6^{1}$) and 2 groups of zeroth power of 6 ($6^{0}$).

Then,

$1052=4\times6^{3}+5\times6^{2}+1\times6^{1}+2\times6^{0}$

So, the number 1052, written as a base 6 number is 4512.

Learn more about change of base of numbers here:

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