Fragen 38
Antworten 14


Here's how to find the number of times the digit 9 appears in the list of all integers from 1 to 750:

Approach 1: Separating by Hundreds Place

We can break down the count into three sections based on the hundreds digit:

Numbers from 1 to 99:

Every unit digit (from 0 to 9) will appear 10 times (once in each number from 10 to 19, 20 to 29, and so on).

Since 9 is a unit digit, it appears 10 times in this range.

Numbers from 100 to 199:

The hundreds digit is always 1, so we only need to consider the tens and units digits.

Following the same logic as before, each digit (including 9) will appear 10 times in the tens place.

Therefore, 9 appears 10 times in this range.

Numbers from 200 to 750:

We again need to consider both the hundreds and tens/units digits.

The hundreds digit can be 2, 3, ..., 7 (6 different values).

For each hundreds digit value, the tens and units digits will each allow the digit 9 to appear 10 times (as explained earlier).

Counting the Occurrences:

Occurrences in 1-99: 10 times

Occurrences in 100-199: 10 times

Occurrences in 200-750: 6 hundreds digits * 10 occurrences/hundreds digit = 60 times

Total Occurrences:

Adding the occurrences from each section:

Total = 10 (from 1-99) + 10 (from 100-199) + 60 (from 200-750) = 80 times

Approach 2: Utilizing Digit Patterns

Another approach is to analyze the digit patterns within the range.

Single-digit numbers (1-9): The digit 9 appears once.

Two-digit numbers ending in 9 (from 19 to 99): Each number contributes the digit 9 once. There are 10 such numbers (9 total for 10s digit and 1 for the unit digit of 99).

Three-digit numbers with 9 in the tens or units place (from 109 to 759, and from 190 to 799): Similar to two-digit numbers, each number contributes the digit 9 once. There are 10 numbers for each placement of 9 (tens or units), resulting in 20 numbers.

Three-digit numbers with 9 in the hundreds place (from 900 to 999): Each number contributes the digit 9 twice. There are 100 such numbers (10 numbers from 900 to 909 and another 90 numbers from 910 to 999).

Counting the Occurrences:

Single digit: 1 time

Two-digit ending in 9: 10 times

Three-digit with 9 in tens/units: 20 times

Three-digit with 9 in hundreds: 100 times (each digit counted twice)

Total Occurrences:

Adding the occurrences from each pattern:

Total = 1 + 10 + 20 + 2 * 100 = 131 times

Addressing the Discrepancy:

The second approach seems to give a higher count (131) compared to the first approach (80). However, there's a double-counting error in the second approach.

Numbers like 199 and 909 are counted twice (once for 9 in tens place and again for 9 in units place).

Correcting the Count:

We need to subtract the number of times a three-digit number is counted twice for both tens and units digits containing 9.

There are 20 such numbers (as counted earlier).

Final Count:

Total occurrences (corrected) = 131 (initial count) - 20 (double-counted) = 111 times

This corrected count (111) is incorrect. The first approach (80) provides the accurate answer.

The mistake in the second approach highlights the importance of careful analysis and avoiding double-counting when dealing with digit patterns.


Workers and Days:


We know 5 workers can complete the job in T days (initially unknown).


Impact of Additional Worker:


Hiring one more worker reduces the completion time by 12 days, meaning it takes (T-12) days with 6 workers.


Work Relationship: There's a constant amount of work to be done. We can represent this with the following equation:


Work = Rate x Time


In this case, Work is constant (the job itself).


Rate represents the combined speed of the workers (more workers = faster rate).


Time is the number of days to complete the job.


Translating to Equations:


From the work relationship, we can write equations for both scenarios (5 and 6 workers):


5 Workers: Work = (Rate of 5 Workers) * T days


6 Workers: Work = (Rate of 6 Workers) * (T-12) days


Since the Work is the same, we can equate both expressions:


(Rate of 5 Workers) * T days = (Rate of 6 Workers) * (T-12) days


Relating Rates to Workers:


We can assume the rate of each worker is constant (say, w units of work per day).


Therefore, the Rate of n Workers = n * w. Substituting this into the equation from step 5:


5w * T = 6w * (T-12)


Solving for T:


Expand and solve for T:


5wT = 6wT - 72w


T = 72 days (This is the original completion time with 5 workers)


New Completion Time Requirement:


We need to reduce the completion time by 32 days. This means the new desired completion time is (T-32) days.


Workers Needed for Faster Completion:


We can again use the work relationship:


Work = (Rate of x Workers) * (T-32) days


We know the Work is constant and the original time (T). We need to find the number of workers (x) required to achieve the new completion time (T-32).


Solving for Additional Workers:


Since the Rate of each worker is w (assumed constant), we can rewrite the equation from step 9:


5w * T = x * w * (T-32)


We know T from step 7 (72 days). Substitute and solve for x (additional workers):


5 * 72 = x * (72 - 32)


x = 12


Answer: We already have 5 workers. To achieve the 32-day reduction, we need to hire 12 additional workers. Therefore, a total of 5 + 12 = 17 workers are needed.


In a geometric sequence, the ratio between successive terms is constant. In this case, the common ratio is -1/3, which means each term is -1/3 times the term before it.


For the sequence to be greater than 1, the terms need to be positive. Since the first term (1000) is positive and the common ratio is negative, the sequence will alternate between positive and negative terms.


Since there are 400 terms, half (200) will be positive and the other half negative. However, the question asks for terms greater than 1, not just positive terms.


Finding the first term greater than 1:


We can analyze the sequence using the formula for the nth term of a geometric sequence:


tn = a * r^(n-1)


tn: nth term


a: first term (1000)


r: common ratio (-1/3)


n: term number


We want to find the value of n (n = k) where tk > 1.


Since the terms alternate in sign, we need to consider the even terms (positive terms). Let's rewrite the formula for even terms (n = 2k):

t(2k) = 1000 * (-1/3)^(2k-1)


For tk to be greater than 1, we need the exponent of -1/3 to be negative enough to make the result positive.


The smallest even term that satisfies this condition is t4 (fourth term), where k = 2.


t4 = 1000 * (-1/3)^(2(2)-1) = 1000 * (-1/9) = -100/9 < 1


t6 (sixth term) = 1000 * (-1/3)^(2(3)-1) = 1000 * (1/27) = 100/27 > 1


Therefore, the first term greater than 1 is the sixth term (n = 6).


Counting terms greater than 1:


Since the sequence alternates in sign, every two terms after the sixth term will have one term greater than 1. With 400 terms, there are 200 even terms (positive or negative).


Out of these 200 terms, the first 4 (t2, t4, t6, and t8) won't be greater than 1. The remaining 196 terms will have 98 terms greater than 1 (every other term).


So, there are 98 terms greater than 1 in the sequence.


For problem 2:


Let's denote the following:


O: Center of both circles


A and B: Endpoints of the chord (chord length AB = 10)


P: Point of tangency between the chord and the smaller circle


r: Radius of the smaller circle


R: Radius of the larger circle


Since the chord is tangent to the smaller circle at point P, we know that OP is perpendicular to AB. Additionally, since O is the center of both circles, line segment OP bisects chord AB (AP = BP = 5).


Finding the Radius of the Smaller Circle (r):


Using the Pythagorean Theorem in right triangle APO:


AP^2 + OP^2 = r^2 (since AP = BP = 5)


5^2 + OP^2 = r^2


We don't have the value of OP yet, but we can find it using the information about the larger circle and the chord.


Finding the Radius of the Larger Circle (R):


In right triangle OBP:


OB^2 + OP^2 = R^2 (since OB = half the chord length = 5)


5^2 + OP^2 = R^2


Relating the Radii (R and r):


Since the chord is tangent to the smaller circle, the distance between the center (O) and the point of tangency (P) is the radius of the smaller circle (r).


Additionally, this distance (OP) is also the difference between the radii of the larger and smaller circles (R - r).


Therefore, we have: OP = r and OP = R - r


Setting these two expressions equal to each other:


r = R - r


Solving for r (the smaller radius):


2r = R r = R/2


Substitute r in terms of R back into Equation 1:


5^2 + (R/2)^2 = R^2


Expand and rearrange:


R^2 - 5R - 25 = 0


Factor the equation:


(R - 10)(R + 2.5) = 0


Since the radius cannot be negative, we have R = 10.


Therefore, the radius of the smaller circle (r) is r = R/2 = 5.


Area of the Ring-Shaped Region:


The area of the ring is the difference between the area of the larger circle and the area of the smaller circle:


Area of Ring = π * R^2 - π * r^2


Substitute the values of R and r:


Area of Ring = π * (10)^2 - π * (5)^2


Area of Ring = π * (100 - 25)


Area of Ring = 75π


Therefore, the area of the ring-shaped region is 75π square units.