Let a number of three digits have for its middle digit the sum of the other two digits. Then it is a multiple of

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Q: 25 (CAPF/2008)
Let a number of three digits have for its middle digit the sum of the other two digits. Then it is a multiple of—

question_subject: 

Maths

question_exam: 

CAPF

stats: 

0,23,4,23,3,1,0

keywords: 

{'middle digit': [0, 0, 1, 0], 'digits': [0, 0, 3, 2], 'sum': [0, 2, 5, 4], 'number': [0, 0, 0, 2], 'multiple': [0, 0, 1, 0]}

To solve this problem, we can let the three-digit number be represented as ABC, where A, B, and C are the three digits. According to the question, the middle digit (B) is the sum of the other two digits (A and C).

We know that a number is divisible by 11 if the difference between the sum of its digits in odd positions and the sum of its digits in even positions is divisible by 11. In our case, the odd-positioned digits are A and C, and the even-positioned digit is B. So, we can express the number as (A + C) - B.

Since B is the sum of A and C, we can replace B with A + C in the expression, giving us (A + C) - (A + C). This simplifies to 0, which means that the number is divisible by 11.

Therefore, the correct answer is option 1, 11, as the number is always divisible by 11.

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