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Math Help - Remainder problem

  1. #1
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    Remainder problem

    (1) A number when divided by a divisor leaves a remainder of 24 . When twice the original number isdivided by the same divisor , the remainder is 11. What is the value of the divisor ?



    (2) When 242 is divided by a certain divisor the remainder obtained is 8 . When 698 is divided by the same divisor the remainder obtained is 9 . However , when the sum of the two numbers 242 and 698 is divided by the divisor , the remainder obtained is 4 . What is the value of the divisor ?
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  2. #2
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    Quote Originally Posted by thereddevils View Post
    (1) A number when divided by a divisor leaves a remainder of 24 . When twice the original number isdivided by the same divisor , the remainder is 11. What is the value of the divisor ?
    Ans-divisor is 37
    let the number be X and divisor be D
    then X=D*q+24 --------------------(1)Euclid division lemma

    2X=2[D*q+24] ------------------muliplying by 2

    =D*2q + 48

    =D*2q + 37 +11

    if we take D=37 then, 37 can be taken common from first 2 terms
    2X =37[2q+1] + 11

    =37[Q] + 11 2q+1=Q (say)

    which is in accordance with 2nd condition.
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  3. #3
    Junior Member AlvinCY's Avatar
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    Since the remainder is 24 and 11 respectively, we know that the divisor must be greater that 24 (otherwise a remainder of 24 is not possible) I used modular arithmetic to find the divisor to be 37, but simple guess and trial would have gotten you there!

    By trial and error:

    61\div37=1r24

    122\div37=3r11

    and...

    By similar methods, the divisor must be greater than 4, 8 and 9... so start testing divisors from 10...

    I know 242\div d= Answer remainder 8...

    divisor of 10 gives remainder of 2
    divisor of 11 gives remainder of 0
    divisor of 12 gives remainder of 2
    divisor of 13 gives remainder of 8

    242\div13=18r8

    698\div13=53r9


    242+698=940

    940\div13=72r4

    Voila!
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