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John chooses a 5-digit positive integer and deletes one of its digits to make a 4-digit number. The sum of this 4-digit number and the original 5-digit number is 52713. What is the sum of the digits of the original 5-digit number?

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8 answers

  1. Solved it!!

    Five options possible are:
    1. abcde + abcd
    2. abcde + abce
    3. abcde + abde
    4. abcde + acde
    5. abcde + bcde
    But as last digit of sum is 3 then only (1) is possible. (e + e – cannot be an odd number)
    a must be 4 or 5
    Then work out using 4 and 5. ‘a’ has to be 4.

    47921
    +4792 = 52713
    So answer is 23 (4+7+9+2+1)

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  2. I agree with 47921 plus 4792

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  3. Ha ha.You are cheating in math kangaroo.You searched up the question didn't you?If you did not then you are a good person if you did it is okay.I did that too because I could not solve it.It is really simple.All you have to do is just work it out and your answer is C(23)LoL everyone else explained it so don't worry.

  4. the answer is 23 confirmed.

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  5. I have a awesome response

  6. At least I think so.

  7. Have not given complete solution

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  8. I think it should be 5+7+9+2+1=24

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