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 o the digit 3, how many digits are in the number? a. 29 b. 33 c. 39 d. 48 e. 57
#1
n(n+1)/2 >= 2000
n^2+n-4000 > 0
n > 62.7
So, 63 days
#2 clearly the sum must be even.
sum = n/2 (24+(n-1)(-2))
http://www.wolframalpha.com/input/?i=n%2F2+%2824%2B%28n-1%29%28-2%29%29+for+n%3D1+to+50
#3
1-digit #'s: 3
teens: 13
20's: 23
30's: 30-33
1. To solve this problem, we need to find the sum of an arithmetic series. The formula for the sum of an arithmetic series is S = (n/2)(a + l), where S is the sum, n is the number of terms, a is the first term, and l is the last term.
In this case, we know that the first term is 1 penny and the last term is n pennies. So the sum of the series is S = (n/2)(1 + n).
We need to find the value of n when the sum is at least 20 dollars, which is 20 * 100 = 2000 pennies. We can set up the equation (n/2)(1 + n) >= 2000.
Simplifying the inequality, we get n^2 + n - 4000 >= 0. Now we can solve this quadratic inequality to find the range of n that satisfies the condition.
Using factoring or the quadratic formula, we find that the solutions to the equation n^2 + n - 4000 = 0 are approximately -64.28 and 63.28. Since n must be a positive integer, the first day on which he has at least 20 dollars in the jar is day 64. Therefore, the correct answer is (e) 64.
2. To find the sum of the sequence, we can use the formula for the sum of an arithmetic series, which is S = (n/2)(a + l), where S is the sum, n is the number of terms, a is the first term, and l is the last term.
In this case, the first term is 12, and the difference between each term is -2. So the sequence becomes 12, 10, 8, 6, ...
Let's consider the sum of the first n terms. We can write it as S = (n/2)(a + l) = (n/2)(12 + (12 + (-2)(n - 1))) = (n/2)(12 + 12 - 2n + 2) = (n/2)(26 - 2n).
To determine which of the given options could be the sum of the first n terms, substitute each option into the formula and solve for n.
For option (a) -14, we have -14 = (n/2)(26 - 2n). Solving this equation, we find that n = 8.
For option (b) -2, we have -2 = (n/2)(26 - 2n). Solving this equation, we find that n = 4.
For option (c) 28, we have 28 = (n/2)(26 - 2n). Solving this equation, we find that n = 6.
For option (d) 31, we have 31 = (n/2)(26 - 2n). Solving this equation, we do not get an integer value for n.
For option (e) 81, we have 81 = (n/2)(26 - 2n). Solving this equation, we do not get an integer value for n.
Therefore, the possible sums of the first n terms are (a) -14, (b) -2, and (c) 28. So the correct answer is (a, b, c).
3. To find the number of digits in the number "123...91011...3" until the eighth occurrence of the digit 3, we need to count the digits from 1 to 911 until we reach the eighth 3.
Let's break down the problem into smaller parts. From 1 to 9, there are 9 digits. From 10 to 99, there are 90 two-digit numbers, resulting in 180 digits. From 100 to 199, there are 100 three-digit numbers, resulting in 300 digits. From 200 to 299, there are 100 two-digit numbers, resulting in 200 digits. From 300 to 399, there are 100 three-digit numbers, resulting in 300 digits. From 400 to 499, there are 100 three-digit numbers, resulting in 300 digits. From 500 to 599, there are 100 three-digit numbers, resulting in 300 digits. From 600 to 699, there are 100 three-digit numbers, resulting in 300 digits. From 700 to 799, there are 100 three-digit numbers, resulting in 300 digits. From 800 to 899, there are 100 three-digit numbers, resulting in 300 digits. From 900 to 909, there are 10 two-digit numbers, resulting in 20 digits. From 910 to 911, there are 2 digits.
Summing up these digits, we get 9 + 180 + 300 + 200 + 300 + 300 + 300 + 300 + 300 + 20 + 2 = 2331.
However, we need to subtract the digits beyond the eighth occurrence of the digit 3. In this case, the number 911 occurs after the eighth occurrence of the digit 3, so we need to exclude the digits from 912 to 911, which is 2 digits.
Therefore, the total number of digits in the number "123...91011...3" until the eighth occurrence of the digit 3 is 2331 - 2 = 2329.
So the correct answer is (e) 2329.