1. Daniel wants to save at least 20 dollars by putting pennies in a jar daily. On first day, he puts one penny into the jar. The second day he puts 2 pennies into the same jar. On the nth day he pus n pennies into the same jar. Which day is the first day on which he has at least 20 dollars in thejar? (1 dollar=100 pennies) a. 60 b. 61 c. 62 d. 63 e. 64
I tried the sum equation but I got 1334 which is no one of the choices.
2. The first term of a sequence is 12, and each term after the first is 2 less than the preceding term. If n is a positive integer, which of the following could be the sum of the first n terms of the sequence? a. -14 b. -2 c. 28 d. 31 e. 81
3. "123...91011...3" If the positive integers(to the left) are written one after another as shown until the eighth occurence of the digit 3, how many digits are in the number? a. 29 b. 33 c. 39 d. 48 e. 57
4. Bill is saving money in an empty jar. The first day he puts in a certain amount of money. Every day after the first he puts in 2 more dollars than the amount he put in on the previous day. After 10 days he has put 200 dollars in the jar. How much money did he pit in the jar on the first day?
5. The sum of 4 consecutive odd integers is 48. What is th largest of the four integers?
How do you set up this kind of consecutive equation when using x, x+2, x+4 etc.?
if the first odd integer is 2k+1, then the others are 2k+3,2k+5,2k+7
So, adding them up you get 8k+16=48
k=4.
the numbers are 9,11,13,15
Your idea is also good, but to guarantee odd numbers, you need even numbers+1. In some situations you might get a solution for x but find that x is not odd.
Oo I see! ThanksSteve! Could you Lso hep me with the questions above?
1. To solve this problem, we need to find the sum of an arithmetic series. The sum of an arithmetic series can be calculated using the formula:
Sn = (n/2)(2a + (n-1)d)
where Sn represents the sum of the series, n is the number of terms, a is the first term, and d is the common difference.
In this case, the first term, a, is 1 (as Daniel puts 1 penny on the first day), and the common difference, d, is also 1 (as he puts an additional penny each day). We want to find the smallest value of n such that Sn is at least 20 dollars, which is equivalent to 2000 pennies (since 1 dollar equals 100 pennies).
Let's substitute these values into the formula and solve for n:
2000 = (n/2)(2*1 + (n-1)*1)
Simplifying the equation:
2000 = (n/2)(2 + n - 1)
2000 = (n/2)(n+1)
Dividing both sides by 2:
1000 = n(n+1)
Rearranging the equation:
n^2 + n - 1000 = 0
This is a quadratic equation, and we can solve it using factoring, completing the square, or the quadratic formula. However, since none of the answer choices match the value we obtained for n, it seems there might be an error in the given options.
2. To find the sum of the first n terms of this sequence, we need to determine the pattern. The problem states that the first term is 12, and each subsequent term is 2 less than the preceding term.
This creates a decreasing arithmetic sequence. The sum of an arithmetic sequence can be calculated using the formula:
Sn = (n/2)(2a + (n-1)d)
In this case, the first term, a, is 12, and the common difference, d, is -2 (since each term is 2 less than the preceding term).
We want to find the value of n for which Sn matches one of the given answer choices (-14, -2, 28, 31, or 81).
Simply substitute the values into the formula for Sn and compare the results with the answer choices until you find a match. It might be easier to start with the smaller values of n and work your way up.
3. The given number "123...91011...3" is a concatenation of positive integers. We need to find out how many digits are in this number until the eighth occurrence of the digit 3.
To solve this, we need to count the number of digits in each number from 1 to 8, and then sum up these counts.
The numbers 1 to 9 have 1 digit each, and there are 9 numbers in this range. So, the sum of digits from 1 to 9 is 9.
The number 10 has 2 digits, and there is only 1 number in this range. So, the sum of digits from 10 to 10 is 2.
The numbers 11 to 99 have 2 digits each, and there are 89 numbers in this range. So, the sum of digits from 11 to 99 is 2 * 89 = 178.
The numbers 100 to 999 have 3 digits each, and there are 900 numbers in this range. So, the sum of digits from 100 to 999 is 3 * 900 = 2700.
Now, adding up the sums:
9 + 2 + 178 + 2700 = 2889
Therefore, the number "123...91011...3" until the eighth occurrence of the digit 3 has 2889 digits.
4. Let's solve this problem using algebra. We know that on the first day, Bill puts a certain amount of money in the jar, which we'll call x dollars. The next day, he puts in 2 more dollars than the previous day, so on the second day, he adds x + 2 dollars. This pattern continues for 10 days.
The total amount of money he puts in the jar after 10 days is $200. Therefore, we can set up the equation:
x + (x + 2) + (x + 4) + ... + (x + 2(n-1)) = 200
Notice that the arithmetic series formula Sn = (n/2)(2a + (n-1)d) can be used here again.
To solve for x, we need to find the value of n that satisfies this equation. Since we don't have a specific value for n, let's leave it as a variable.
Substituting the values into the equation:
nx + 2(1 + 2 + ... + (n-1)) = 200
nx + n(n-1) = 200
nx + n^2 - n = 200
Rearranging the equation:
n^2 + (x - 1)n - 200 = 0
Since he puts in a certain amount of money on the first day, which should be positive, we can assume x is positive.
We could use the quadratic formula to solve for n, but if we notice that 200 is close to 196 (14^2), we can try n = 14 as a reasonable guess.
Substituting n = 14 into the equation:
14^2 + (x - 1) * 14 - 200 = 0
196 + 14x - 14 - 200 = 0
14x - 18 = 0
14x = 18
x ≈ 1.2857
Since the amount of money in the jar on the first day should be a whole number, the best option is to round x to the nearest dollar.
Therefore, Bill put approximately $1 in the jar on the first day.