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If the number 10*47* is divisible by both 5 and 11, then the missing digits are respectively
- a)1 and 5
- b)6 and 0
- c)5 and 0
- d)2 and 5
- e)None of these
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
If the number 10*47* is divisible by both 5 and 11, then the missing d...
Check the options in the number 10x47y
all numbers will be divisible by 5 because in end it is 5 and 0
for number to be divisible by 11, (y+4+0) – (7+x+1) should be divisible by 11
from option A, y = 5, x = 1 gives (y+4+0) – (7+x+1) as 0 which is divisible by 11
all numbers will be divisible by 5 because in end it is 5 and 0
for number to be divisible by 11, (y+4+0) – (7+x+1) should be divisible by 11
from option A, y = 5, x = 1 gives (y+4+0) – (7+x+1) as 0 which is divisible by 11
Most Upvoted Answer
If the number 10*47* is divisible by both 5 and 11, then the missing d...
Solution:
To find the missing digits, we need to know the divisibility rules of 5 and 11.
Divisibility rule of 5: If a number ends with 0 or 5, then it is divisible by 5.
Divisibility rule of 11: If the difference between the sum of the digits in the odd places and the sum of the digits in the even places is either 0 or a multiple of 11, then the number is divisible by 11.
Let us check the divisibility by 5 first.
10*47* is divisible by 5 because it ends with 0.
Now, we need to check the divisibility by 11.
The product 10*47* can be written as:
10*47* = 100*47 + *
Here, * represents the missing digits.
The sum of the digits in the odd places is 4 + 7 = 11.
The sum of the digits in the even places is 0 + * = *.
For 10*47* to be divisible by 11, the difference between the sum of the digits in the odd places and the sum of the digits in the even places must be either 0 or a multiple of 11.
So, we have two cases:
Case 1: The difference is 0
In this case, the missing digits must be such that the sum of the digits in the even places is 11. This is only possible if the missing digits are 1 and 5.
The product 10*47* = 100*47 + 15 is divisible by both 5 and 11.
Case 2: The difference is a multiple of 11
In this case, the missing digits must be such that the sum of the digits in the even places is either 0, 11, 22, 33, 44, or 55. But, since the sum of the digits in the odd places is already 11, the sum of the digits in the even places cannot be 11. Therefore, this case is not possible.
Hence, the missing digits are 1 and 5. Option (a) is the correct answer.
To find the missing digits, we need to know the divisibility rules of 5 and 11.
Divisibility rule of 5: If a number ends with 0 or 5, then it is divisible by 5.
Divisibility rule of 11: If the difference between the sum of the digits in the odd places and the sum of the digits in the even places is either 0 or a multiple of 11, then the number is divisible by 11.
Let us check the divisibility by 5 first.
10*47* is divisible by 5 because it ends with 0.
Now, we need to check the divisibility by 11.
The product 10*47* can be written as:
10*47* = 100*47 + *
Here, * represents the missing digits.
The sum of the digits in the odd places is 4 + 7 = 11.
The sum of the digits in the even places is 0 + * = *.
For 10*47* to be divisible by 11, the difference between the sum of the digits in the odd places and the sum of the digits in the even places must be either 0 or a multiple of 11.
So, we have two cases:
Case 1: The difference is 0
In this case, the missing digits must be such that the sum of the digits in the even places is 11. This is only possible if the missing digits are 1 and 5.
The product 10*47* = 100*47 + 15 is divisible by both 5 and 11.
Case 2: The difference is a multiple of 11
In this case, the missing digits must be such that the sum of the digits in the even places is either 0, 11, 22, 33, 44, or 55. But, since the sum of the digits in the odd places is already 11, the sum of the digits in the even places cannot be 11. Therefore, this case is not possible.
Hence, the missing digits are 1 and 5. Option (a) is the correct answer.
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Community Answer
If the number 10*47* is divisible by both 5 and 11, then the missing d...
Check the options in the number 10x47y
all numbers will be divisible by 5 because in end it is 5 and 0
for number to be divisible by 11, (y+4+0) – (7+x+1) should be divisible by 11
from option A, y = 5, x = 1 gives (y+4+0) – (7+x+1) as 0 which is divisible by 11
all numbers will be divisible by 5 because in end it is 5 and 0
for number to be divisible by 11, (y+4+0) – (7+x+1) should be divisible by 11
from option A, y = 5, x = 1 gives (y+4+0) – (7+x+1) as 0 which is divisible by 11
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If the number 10*47* is divisible by both 5 and 11, then the missing digits are respectivelya)1 and 5b)6 and 0c)5 and 0d)2 and 5e)None of theseCorrect answer is option 'A'. Can you explain this answer?
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