Answer:
Step-by-step explanation:
Let .
and
The graph of the two equations are shown in the attachment.
The points where the two graphs intersected are:
The x-coordinates of the intersection points are the solutions to .
Therefore the solutions are:
1052 toothpicks can be grouped into 4 groups of third power of 6 (), 5 groups of second power of 6 (
), 1 group of first power of 6 (
) and 2 groups of zeroth power of 6 (
).
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,
This implies that exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.
Now, and
. This implies that
exceeds 1052 and thus there can be at most 4 groups of
.
Then,
So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.
Now, and
. This implies that
exceeds 188 and thus there can be at most 5 groups of
.
Then,
So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.
Now, and
. This implies that
exceeds 8 and thus there can be at most 1 group of
.
Then,
So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.
Now, and
. 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 (), 5 groups of second power of 6 (
), 1 group of first power of 6 (
) and 2 groups of zeroth power of 6 (
).
Then,
So, the number 1052, written as a base 6 number is 4512.
Learn more about change of base of numbers here: