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lect the correct answerA number when divided by D leaves a remainder of 8 and when divided by 3D leaves a remainder of 21 . What is the remainder left, when twice the number is divided by 3D?radio_button_unchecked13radio_button_unchecked42radio_button_unchecked3radio_button_uncheckedcannot be determined

Question

lect the correct answerA number when divided by D leaves a remainder of 8 and when divided by 3D leaves a remainder of 21 . What is the remainder left, when twice the number is divided by 3D?radio_button_unchecked13radio_button_unchecked42radio_button_unchecked3radio_button_uncheckedcannot be determined

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Solution

Let's denote the number as 'a'.

According to the problem, we have two conditions:

  1. a = Dq1 + 8 (where q1 is the quotient when a is divided by D)
  2. a = 3Dq2 + 21 (where q2 is the quotient when a is divided by 3D)

From the first equation, we can express D as D = (a - 8) / q1.

Substituting D in the second equation, we get a = 3((a - 8) / q1)q2 + 21.

Solving this equation, we find that q1 = q2 = 1, and a = 29.

Now, we need to find the remainder when twice the number (2a) is divided by 3D.

2a = 58, and 3D = 63.

So, when 58 is divided by 63, the remainder is 58.

Therefore, the correct answer is 'cannot be determined' because the remainder is larger than the divisor, which contradicts the definition of a remainder.

This problem has been solved

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