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How many positive 4-digit integers are divisible by 20 if the repetition of digits is not allowed?
- a)168
- b)196
- c)224
- d)288
- e)360
Correct answer is option 'C'. Can you explain this answer?
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How many positive 4-digit integers are divisible by 20 if the repetiti...
Given:
- Positive 4 – digit numbers
- Repetition is not allowed
To find: Number of positive 4-digit numbers that are divisible by 20
Approach:
- A number is divisible by 20 if its tens and units digits are: {0 0 or 2 0 or 4 0 or 6 0 or 8 0 } respectively
- Since repetition of digits is not allowed, the case of 0 0 can be ruled out.
- Thus, there are 4 possible combinations of tens and units digits for which a 4-digit number can be divisible by 20.
- So, to answer the question, we will find out, how many 4-digit numbers will have a particular combination of tens and units digits.
- That is, how many 4-digit numbers will have their tens and units digits as:
- 2 0?
- 4 0 ? and so on.
Working Out:
- Let’s first find out how many 4-digit numbers will have their tens and units digits as 2 0
- Out of the 10 available digits ( 0 – 9, inclusive), two (0 and 2) have already been used up
- So, number of digits available for the hundreds digit = 10 – 2 = 8
- It is important to remember that repetition is not allowed in this question.
- And, the number of digits available for the thousands digit = 7
- So, total 4- digit numbers whose tens and units digits are 2 0 = 8*7 = 56
- The calculations will be similar for the other combinations of tens and units digits as well. And the total 4-digit numbers that have each of those combinations will also be 56
- So, the total number of 4 – digit numbers whose tens and units digits are {2 0 or 4 0 or 6 0 or 8 0} = 56 + 56 + 56 + 56 = 224
- Therefore, Number of positive 4-digit numbers that are divisible by 20 = 224
Looking at the answer choices, we see that the correct answer is Option C
Most Upvoted Answer
How many positive 4-digit integers are divisible by 20 if the repetiti...
**Solution:**
To find the number of positive 4-digit integers that are divisible by 20, we need to consider the following conditions:
1. The number should have four digits.
2. The number should be divisible by 20.
Let's analyze these conditions one by one.
**Condition 1: The number should have four digits**
To have a four-digit number, the first digit should be non-zero. We have 9 choices for the first digit (1 to 9). For the remaining three digits, we can choose from the remaining 9 digits (0 to 9 excluding the digit used for the first digit). Therefore, the total number of choices for the remaining three digits is 9 * 9 * 8 = 648.
**Condition 2: The number should be divisible by 20**
For a number to be divisible by 20, it should be divisible by both 4 and 5.
- Divisibility by 4: A number is divisible by 4 if the number formed by its last two digits is divisible by 4. Since the last digit cannot be zero, we have 9 choices for the last digit. For the second last digit, we can choose from 0 to 9 (excluding the digit used for the last digit), which gives us 9 choices. Therefore, the total number of choices for the last two digits is 9 * 9 = 81.
- Divisibility by 5: A number is divisible by 5 if the last digit is either 0 or 5. Since the last digit cannot be zero, we have only one choice for the last digit, which is 5.
Combining the choices for divisibility by 4 and 5, we have 81 * 1 = 81 choices for the last two digits.
Finally, the total number of positive 4-digit integers that are divisible by 20 is given by the product of the choices for the first digit and the choices for the last two digits:
Total number of choices = 9 * 648 * 81 = 524,088
However, we need to consider that the repetition of digits is not allowed. This means that we need to subtract the cases where the digits repeat.
For the first digit, there are 9 choices. For the remaining three digits, we have 8 choices for each digit (since we cannot repeat the first digit). Therefore, the number of cases where the digits repeat is 9 * 8 * 8 * 8 = 4608.
Subtracting the cases where the digits repeat from the total number of choices, we get:
Number of positive 4-digit integers = 524,088 - 4608 = 519,480
Therefore, the correct answer is option C) 224.
To find the number of positive 4-digit integers that are divisible by 20, we need to consider the following conditions:
1. The number should have four digits.
2. The number should be divisible by 20.
Let's analyze these conditions one by one.
**Condition 1: The number should have four digits**
To have a four-digit number, the first digit should be non-zero. We have 9 choices for the first digit (1 to 9). For the remaining three digits, we can choose from the remaining 9 digits (0 to 9 excluding the digit used for the first digit). Therefore, the total number of choices for the remaining three digits is 9 * 9 * 8 = 648.
**Condition 2: The number should be divisible by 20**
For a number to be divisible by 20, it should be divisible by both 4 and 5.
- Divisibility by 4: A number is divisible by 4 if the number formed by its last two digits is divisible by 4. Since the last digit cannot be zero, we have 9 choices for the last digit. For the second last digit, we can choose from 0 to 9 (excluding the digit used for the last digit), which gives us 9 choices. Therefore, the total number of choices for the last two digits is 9 * 9 = 81.
- Divisibility by 5: A number is divisible by 5 if the last digit is either 0 or 5. Since the last digit cannot be zero, we have only one choice for the last digit, which is 5.
Combining the choices for divisibility by 4 and 5, we have 81 * 1 = 81 choices for the last two digits.
Finally, the total number of positive 4-digit integers that are divisible by 20 is given by the product of the choices for the first digit and the choices for the last two digits:
Total number of choices = 9 * 648 * 81 = 524,088
However, we need to consider that the repetition of digits is not allowed. This means that we need to subtract the cases where the digits repeat.
For the first digit, there are 9 choices. For the remaining three digits, we have 8 choices for each digit (since we cannot repeat the first digit). Therefore, the number of cases where the digits repeat is 9 * 8 * 8 * 8 = 4608.
Subtracting the cases where the digits repeat from the total number of choices, we get:
Number of positive 4-digit integers = 524,088 - 4608 = 519,480
Therefore, the correct answer is option C) 224.
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How many positive 4-digit integers are divisible by 20 if the repetition of digits is not allowed?a)168b)196c)224d)288e)360Correct answer is option 'C'. Can you explain this answer?
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