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Tìm số tự nhiên a nhỏ nhất biết a chia cho 3,5,7 được số dư là 2,3,4

Đề bài: Tìm số tự nhiên a nhỏ nhất sao cho a chia cho 3, 5, 7 được số dư theo thứ tự 2, 3, 4

LÀM ƠN GIÚP MÌNH VỚI, MAI THI RỒI MÀ MÌNH CHƯA LÀM ĐƯỢC,
CẢM ƠN CÁC BẠN NHIỀU NHA!

Theo dõi Vi phạm
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  • Theo đề ta có : a chia 3 dư 3 => 2a - 1 chia hết cho 3

    a chia 5 dư 5 => 2a -1 chia hết cho 5

    a chia 7 dư 4 => 2a-1 chia hết cho 7

    Vì 2a-1 chia hết cho 5;3;7 và a nhỏ nhất => 2a-1 thuộc BCNN(3;5;7)

    5=5 ; 3=3; 7=7 | BCNN(3;5;7)= 5.3.7= 105

    Ta có: 2a -1= 105

    2a = 105 +1

    2a = 106

    a = 106 : 2

    a = 53

    Vậy a= 53

      bởi thunga nga 25/12/2018
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