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Find a number not greater than 500 which leaves a remainder of 3 when it is divided by 7, which leaves a remainder of 4 when it is divided by 8, and which leaves a remainder of 5 when it is divided by 9.

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  • 3 months ago
    Favourite answer

    Notice if we add 4 to the number, it will have *no* remainder when divided by 7, 8 or 9. Given those numbers have no factors in common (other than 1), the lowest common multiple of 7, 8 and 9 is 7 * 8 * 9 = 504

    Subtracting 4 we get back to the original number 500 (which still matches the criterion of NOT being larger than 500).

    Double-checking:

    500/7 = 71 r 3

    500/8 = 62 r 4

    500/9 = 55 r 5

    Answer:

    500

  • Philip
    Lv 6
    3 months ago

    500/7 = 71 + 3 remainder.;

    500/8 = 62 + 4 remainder.;

    500/9 = 55 + 5 remainder.

  • 7a + 3 = 8b + 4 = 9c + 5

    7a = 9c + 2

    7a = 8b + 1

    8b = 9c + 1

    7a - 2 = 9c

    5 , 12 , 19 , 26 , 33 , 40 , 47 , 54 , 61 , 68 , 75 , ...

    +5 , +3 , +1 , +8 , +6 , +4 , +2 , +0 , +7 , +5 , +3 , ....

    9 * 6 = 7 * 8 - 2

    Every 9 multiples of 6, we should get a number

    7 * 17 - 2 = 117

    7 * 26 - 2 = 180

    7 * 35 - 2 = 243

    7 * 44 - 2 = 306

    7 * 53 - 2 = 369

    7 * 62 - 2 = 432

    7 * 71 - 2 = 495

    Throw out the odd ones, since 8b + 4 will always be even

    180 , 306 , 432

    Throw out the one that isn't divisible by 4, since 8b + 4 is always divisible by 4

    180 , 432

    Throw out the one that is evenly divisible by 8

    180

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