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Thread: remainder left type questions

  1. #1
    huhhhhhh????? what? just joined TestMagic.
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    remainder left type questions

    hey guys,
    came across a few questions and am not sure how to solve them.

    1. Twice the sum of 3 integers x ,y and z when divided by 7 gives remainder 1
    what is the remainder when x+y+z is divided by 7?

    2. x when divided by 7 leaves remainder 4.What is the remainder when 2x + 5 is divided by 7?

    Please tell me the method to solving these, or similar concept questions.

    Thanks.

  2. #2
    TestMagic Guru-in-Training DWarrior 's dreams are becoming reality. DWarrior's Avatar
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    1. I couldn't set it up algebraically, so I had to plug in numbers. let S be x+y+z, so we're looking for:
    2S/7=n/7+1

    This works when 2S=8, because then n=7 and there's a remainder of 1.

    S=4, so 4/7=0+4/7

    So the remainder is 4.

    2. x/7=n/7+4/7
    x=n+4, where n is divisible by 7. Substitute that into the new equation:
    [2(n+4)+5]/7=(2n+8+5)/7=2n/7+13/7=2n/7+7/7+6/7. 13 can't be the remainder, since remainders must be less than the divisor.

    So the remainder is 6

  3. #3
    huhhhhhh????? what? just joined TestMagic.
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    Dwarrior: thanks for the explanation. your answers are right.
    I am still a little confused, but maybe thats just because it is very difficult to put a question like this into an equation. There is no clear cut way to handle this.
    But I do get the idea from your answer. Hope i can use it to figure out a similar problem.

  4. #4
    TestMagic Guru-in-Training DWarrior 's dreams are becoming reality. DWarrior's Avatar
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    I was asking on IRC how to solve problem #1 methodically, and somebody posted an interesting analysis that I think might be more helpful than my solution:

    First, you should realize that we'll always be working with (x+y+z) and never with the individual variables, so you should simplify the problem by substitution (S for x+y+z). That way, there's a greater chance that you'll "see" the solution.

    Second, set up the original problem:
    2S=7p+1, where p is some multiple of 7 divided by 7. We know 2S must be even, and therefore p must be odd (do you see why?).

    So now just plug in any odd number for p and you'll get a valid S. For example, use p=3:
    2S=21+1=22
    S=11
    11/7 leaves remainder of 4.

    This setup actually seems like a good method to solve all problems of this type, where you're given a remainder and then have to find remainder of another related number.

    Set up the original remainder equality. p will always be a positive integer. Look for any restrictions, like it having to be odd/even, or anything else. Then plug a value for p that meets all restrictions to get a sample x, plug the x back into the new number and find the remainder.

    For example, do #2:
    x=7p+4
    In this case there are no other restrictions on p (aside from it being a positive integer).
    x=7*3+4
    x=21+4
    x=25

    Now we need the remainder of 2x+5:
    50+5=55
    55/7 leaves remainder of 6.

  5. #5
    huhhhhhh????? what? just joined TestMagic.
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    Sir that is a Great solution. easy to understand and pretty methodological.

    Thank you very much for sharing.

  6. #6
    Within my grasp! drosera01 just joined TestMagic.
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    These questions are just like plug and play electronic devices now. Plug the number and you will get the answer. right?

  7. #7
    Eager! learner85 just joined TestMagic.
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    drosera hitted the bulls eye. I agreed with him . This are simple plug-in number problems. Practice plugging-in and back-solving, they are really helpful when you are stuck in a problem. Actually almost all of us are supposed to be stucked once or more on GRE exam.

    Regards,

    learner85

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