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A number consists of three digits of which the middle one is zero and their sum is 4. If the number formed by interchanging the first and last digits is greater than the number itself by 198, the difference between the first and last digits is
- a)1
- b)2
- c)3
- d)4
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
A number consists of three digits of which the middle one is zero and ...
Let the hundreds digit be x, the tens digit be 0, and the ones digit be y.
The original number is 100x + 10y, and the number formed by interchanging the first and last digits is 100y + 10x.
The given information can be written as:
100y + 10x - (100x + 10y) = 198
90(y - x) = 198
y - x = 2
The difference between the first and last digits is 2.
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A number consists of three digits of which the middle one is zero and ...
Given: A number consists of three digits of which the middle one is zero and their sum is 4. If the number formed by interchanging the first and last digits is greater than the number itself by 198.
To find: The difference between the first and last digits.
Let's assume the three digits be a, 0, and b, where a and b are the first and last digits respectively.
Sum of the digits = a + 0 + b = 4
=> a + b = 4 --- (Equation 1)
Number formed by interchanging the first and last digits = 100b + 0 + a = 10b + a
Given, 10b + a = ab + 198
=> ab - 10b - a = -198
=> a(b-1) - 10(b-1) = -198
=> (a-10)(b-1) = -198
As a and b are digits, (a-10) cannot be negative.
So, we can write (a-10) as -1 and (b-1) as 198 or vice versa.
Case 1: (a-10) = -1 and (b-1) = -198
=> a = 9 and b = -197
But b cannot be negative. So, this case is not possible.
Case 2: (a-10) = 198 and (b-1) = -1
=> a = 208 and b = 0
Thus, the original number is 208 and the difference between the first and last digits is 2 (8 - 0).
Therefore, the correct answer is option B (2).
To find: The difference between the first and last digits.
Let's assume the three digits be a, 0, and b, where a and b are the first and last digits respectively.
Sum of the digits = a + 0 + b = 4
=> a + b = 4 --- (Equation 1)
Number formed by interchanging the first and last digits = 100b + 0 + a = 10b + a
Given, 10b + a = ab + 198
=> ab - 10b - a = -198
=> a(b-1) - 10(b-1) = -198
=> (a-10)(b-1) = -198
As a and b are digits, (a-10) cannot be negative.
So, we can write (a-10) as -1 and (b-1) as 198 or vice versa.
Case 1: (a-10) = -1 and (b-1) = -198
=> a = 9 and b = -197
But b cannot be negative. So, this case is not possible.
Case 2: (a-10) = 198 and (b-1) = -1
=> a = 208 and b = 0
Thus, the original number is 208 and the difference between the first and last digits is 2 (8 - 0).
Therefore, the correct answer is option B (2).
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