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Find the missing digit in each of the following numbers to make them divisible by 11: a) 27...3, b) 1...574, c) 939...?

Mathematics

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Answers

4 parts
a

8\text{8}

b

3\text{3}

c

4\text{4}

d

4\text{4}

Step-by-step solution

8 steps

To find the digit in the blank space such that the number is divisible by 11, we use the divisibility rule for 11: a number is divisible by 11 if the alternating sum of its digits (starting from the rightmost digit) is divisible by 11.

Part a· 2 steps

27.......3

Let the missing digit be xx. The number is 27x327x3.

  1. Calculate the alternating sum of the digits.

    3−x+7−2=8−x3 - x + 7 - 2 = 8 - x
  2. For the number to be divisible by 11, the alternating sum must be a multiple of 11 (e.g., 0, 11, -11, etc.).

    Since xx is a single digit (0-9), the only multiple of 11 possible for 8−x8-x is 0.

    8−x=08 - x = 0 x=8x = 8

    The number is 2783.

    8\boxed{\text{8}}

Part b

3\text{3}

1....... 574

Let the missing digit be xx. The number is 1x5741x574.

  1. Calculate the alternating sum of the digits.

    4−7+5−x+1=3−x4 - 7 + 5 - x + 1 = 3 - x
  2. For the number to be divisible by 11, the alternating sum must be a multiple of 11.

    Since xx is a single digit (0-9), the only multiple of 11 possible for 3−x3-x is 0.

    3−x=03 - x = 0 x=3x = 3

    The number is 13574.

    3\boxed{\text{3}}

Part c

4\text{4}

939.......

Let the missing digit be xx. The number is 939x939x.

  1. Calculate the alternating sum of the digits.

    x−9+3−9=x−15x - 9 + 3 - 9 = x - 15
  2. For the number to be divisible by 11, the alternating sum must be a multiple of 11.

    Since xx is a single digit (0-9), the only multiple of 11 possible for x−15x-15 is -11.

    x−15=−11x - 15 = -11 x=15−11x = 15 - 11 x=4x = 4

    The number is 9394.

    4\boxed{\text{4}}

Part d

4\text{4}

10....... 5

Let the missing digit be xx. The number is 10x510x5.

  1. Calculate the alternating sum of the digits.

    5−x+0−1=4−x5 - x + 0 - 1 = 4 - x
  2. For the number to be divisible by 11, the alternating sum must be a multiple of 11.

    Since xx is a single digit (0-9), the only multiple of 11 possible for 4−x4-x is 0.

    4−x=04 - x = 0 x=4x = 4

    The number is 1045.

    4\boxed{\text{4}}
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