Anthony earned $131 in tips last night while working at the restaurant. If his tips consisted of $1 bills and $5 bills only and
1 answer:
Answer:
21
Step-by-step explanation:
Let x = $1 bill
Let y = $5bill
<em>Translating the word problem into an algebraic equation;</em>
<em>For the total number of bills;</em>
.........equation 1
<em>For the total number of tips;</em>
..........equation 2
<em>*Solving the linear equation by using the substitution method*</em>
<em>Making x the subject in equation 1:</em>
.......equation 3
<em>Substituting "x" into equation 2;</em>
<em>Simplifying the equation, we have;</em>
y = 21 (For the $5 bills).
<em>Therefore, Anthony received 21 pieces of the $5 bills.</em>
To find x;
Substituting "y" into equation 3;
x = 26 (For the $1 bill).
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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.
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