Find the largest 4-digit number when divided by 40,48,60 leaves remainder 3 in each case.​

Question

Find the largest 4-digit number when divided by 40,48,60 leaves remainder 3 in each case.​

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Melody 1 week 2021-10-04T21:39:51+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-10-04T21:41:05+00:00

    SOLUTION :

    \bigstar Firstly, we get L.C.M of 40,48 & 60.

    \begin{array}{r|l} 2 & 40,48,60 \\ \cline{2-2} 2 & 20,24,30 \\ \cline{2-2} 2 & 10,12,15 \\ \cline{2-2} 2 & 5 , 6 , 15 \\ \cline{2-2} 3& 5,3,15\\ \cline{2-2} 5 & 5,1,5 \\ \cline{2-2}  & 1,1,1 \end{array}

    ∴ L.C.M = 2 × 2 × 2 × 2 × 3 × 5 = 240

    Now,

    We know greatest four digit = 9999

    240) 9999( 41

            960

    ______________

                 399

                 240

    _______________

                  159

    _______________

    ∴ 9999 – 159 = 9840 which is divisible.

    Thus;

    The required number = 9840 – 3 = 9837

    0
    2021-10-04T21:41:49+00:00

    Answer:

    YOUR ANSWER IS HERE :

    A. 9996 is divisible by all 3 number.

    Step-by-step explanation:

    A. The greatest four digit number is 9999 .

    Now find the HCF of 40 , 48 , 60 –

    40 = 2 × 2 × 5

    48 = 2 × 2 × 2 × 2 × 3

    60 = 2 × 2 × 3 × 5

    HCF = 4

    Divide 9999 by 4

    9999 ÷ 4 = 2499 and remainder is 3

    2499 × 4 + 3

    9999 – 3 = 9996

    The answer is 9996 and the numbers remainder is 3 .

    PLEASE MARK ME AS BRAINLIST.

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