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If the sum of the digits of any integer lying between 100 and 1000 is subtracted from the number, the result always is (SSC CHSL 2013)
- a)divisible by 5
- b)divisible by 6
- c)divisible by 2
- d)divisible by 9
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
If the sum of the digits of any integer lying between 100 and 1000 is ...
(100x + 10y + z) – (x + y + z) = 99x + 9y
= 9 (11x + y)
= 9 (11x + y)
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Community Answer
If the sum of the digits of any integer lying between 100 and 1000 is ...
Explanation:
To find the answer to this question, let's consider any number between 100 and 1000. For example, let's take the number 256.
Step 1: Find the sum of the digits:
The sum of the digits of 256 is 2 + 5 + 6 = 13.
Step 2: Subtract the sum of the digits from the number:
256 - 13 = 243.
Step 3: Check if the result is divisible by 9:
To determine if a number is divisible by 9, we need to check if the sum of its digits is divisible by 9. In this case, the sum of the digits of 243 is 2 + 4 + 3 = 9, which is divisible by 9.
Step 4: Conclusion:
Since the result of subtracting the sum of the digits from the number is divisible by 9, we can conclude that the answer to the question is option D - divisible by 9.
Why is the answer divisible by 9?
The reason the answer is divisible by 9 is due to the properties of divisibility. When we subtract the sum of the digits from the number, we are essentially subtracting a multiple of 9 from the number. This is because any number between 100 and 1000 can be expressed as 100x + 10y + z, where x, y, and z are the digits of the number.
For example, in the number 256, x = 2, y = 5, and z = 6. So, 256 can be expressed as 100(2) + 10(5) + 6.
When we subtract the sum of the digits (13 in this case) from the number (256), we are subtracting 9x from the hundreds place, 9y from the tens place, and 9z from the units place. This results in a number that is divisible by 9.
Therefore, for any number between 100 and 1000, the result of subtracting the sum of the digits from the number is always divisible by 9.
To find the answer to this question, let's consider any number between 100 and 1000. For example, let's take the number 256.
Step 1: Find the sum of the digits:
The sum of the digits of 256 is 2 + 5 + 6 = 13.
Step 2: Subtract the sum of the digits from the number:
256 - 13 = 243.
Step 3: Check if the result is divisible by 9:
To determine if a number is divisible by 9, we need to check if the sum of its digits is divisible by 9. In this case, the sum of the digits of 243 is 2 + 4 + 3 = 9, which is divisible by 9.
Step 4: Conclusion:
Since the result of subtracting the sum of the digits from the number is divisible by 9, we can conclude that the answer to the question is option D - divisible by 9.
Why is the answer divisible by 9?
The reason the answer is divisible by 9 is due to the properties of divisibility. When we subtract the sum of the digits from the number, we are essentially subtracting a multiple of 9 from the number. This is because any number between 100 and 1000 can be expressed as 100x + 10y + z, where x, y, and z are the digits of the number.
For example, in the number 256, x = 2, y = 5, and z = 6. So, 256 can be expressed as 100(2) + 10(5) + 6.
When we subtract the sum of the digits (13 in this case) from the number (256), we are subtracting 9x from the hundreds place, 9y from the tens place, and 9z from the units place. This results in a number that is divisible by 9.
Therefore, for any number between 100 and 1000, the result of subtracting the sum of the digits from the number is always divisible by 9.
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If the sum of the digits of any integer lying between 100 and 1000 is subtracted from the number, the result always is (SSC CHSL 2013)a)divisible by 5b)divisible by 6c)divisible by 2d)divisible by 9Correct answer is option 'D'. Can you explain this answer?
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