math

danielle bought a bag of gumdrops.she has fewer than 75 in the bag.if she counts them by 2's, 3's, or 4's she will have 1 left over.if she counts them by 5's she has none left over.how many gumdrops does danielle have in the bag.(explain)

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asked by tanja
  1. N/2 = a + 1 or N = 2a + 2
    N/3 = b + 1 or N = 3b + 3
    N/4 = c + 1 or N = 4c + 4
    N/5 = d or N = 5d

    Can you take it from here?

    Another example:
    What is the least number that leaves a remainder of 3 when divided by 5, a remainder of 2 when divided by 4, a remainder of 1 when divided by 3, and a remainder of 0 when divided by 2?

    From the problem statement, and letting the number we seek equal N, we can write
    N/5 = A + 3/5 or N = 5A + 3
    N/4 = B + 2/4 or N = 4B + 2
    N/3 = C + 1/3 or N = 3C + 1
    N/2 = D + 0 or N = 2D + 0
    1--Combining the first two we get 4B - 5A = 1
    2--Dividing through by 4 yields B - A - A/4 = 1/4 or (A + 1)/4 = B - A
    3--(A + 1)/4 must equal an integer k making A = 4k - 1
    4--Substituting back into (1) yields 4B - 20k + 5 = 1 or B = 5k - 1
    5..k can be any integer from 1 on up
    6--k.....1.....2.....3.....4.....5.....6
    ....A....3.....7....11....15...19...23
    ....B....4.....9....14....19...24...29
    7--Combining the 2nd and 3rd expressions we get 3C - 4B = 1
    8--Dividing through by 3 yields C - B - B/3 = 1/3
    9--Again, ((B + 1)/3 must be an integer k making B = 3k - 1
    10--Substitiuing back into (7) yields 3C - 12k + 4 = 1 or C = 4k - 1
    11--k.....1.....2.....3.....4.....5.....6.....7
    .....B.....2.....5.....8....11...14...17...20
    .....C.....3.....7....11...15...19...23...27
    12--Combining the 3rd and 4th expressions, we get 2D - 3C = 1
    13--Dividing through by 2 yields D - C - C/2 = 1/2
    14--Again, (C + 1)/2 must be an integer k making C = 2k - 1
    15--Substituting back into (12) yields 2D - 6k + 3 = 1 or D = 3k - 1
    16--k....1....2....3....4....5....6....7.....8....9....10....11
    .....C....1....3....5....7....9...11..13....15..17...19....21
    .....D....2....5....8...11..14...17..20...23..26...29....32
    17--Combining the 4th and 1st expressions, we get 2D - 5A = 3
    18--Dividing through by 2 yields D - 2A - A/2 = 1 + 1/2
    19--Again, (A + 1)/2 must be an integer k making A = 2k - 1
    20--Substituting back into (17) yields 2D - 10k + 5 = 3 or D = 5k - 1
    21--k....1....2....3....4....5....6....7....8
    .....D....4....9...14..19..24...29..34..39
    .....A....1....3....5....7....9...11..13..15
    22--The smallest consistant set of values throughout the four tables are the highlightd values of A = 11, B = 14, C = 19 and D = 29.
    23--Therefore, N = 5(11) + 3 = 58 = 4(14) + 2 = 58 = 3(19) + 1 = 58 = 2(29) = 58.

    There are other methods.

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  2. what does that mean? im in fourth grade make in simpler

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    posted by tanja
  3. Since the quantity is less than 75 and evenly divisible by 5, only 70, 65, 60, 55, 50, 45, 40, etc. are candidates.

    Starting with 70, divide by 4. If there is a remainder of 1, divide the 70 by 3. If there is a remainder of 1, divide the 70 by 2. If there is a remainder of 1, you have found your answer.

    70/4 = 17 + 2 NG
    65/4 = 16 + 2 NG
    60/4 = 15 NG
    55/4 = 13 + 3 NG
    and so on to

    25/4 = 6 + 1
    25/3 = 8 + 1
    25/2 = 12 + 1

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  4. THANK YOU SOOOOOO MUCH !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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    posted by tanja

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