a)366;80

Bạn có thể nói cách giải được ko

Đk:a\(\in\)N*

398 chia a dư 38 => 398-38 chia hết a => 360 chia hết a

450 chia a dư 18 => 450-18 chia hết a => 432 chia hết a

=>a=UCLN(360;432). Ta có:

360=2³*3²*5

432=2⁴*3³

=>UCLN(360;432)=2³*3²=72

Suy ra a=72 (thỏa mãn)

Do 398 chia a dư 38; 450 chia a dư 18

=> 398 - 38 chia hết cho a; 450 - 18 chia hết cho a

=> 360 chia hết cho a; 432 chia hết cho a

=> a thuộc ƯC(360; 432)

Mà ƯCLN(360; 432) = 48

=> a thuộc Ư(48)

Mà a > 38 => a = 48

nếu 398:a dư 38 => 360 chia hết cho a => a thuộc Ư(360)

nếu 450:a dư 18 => 432 chia hết cho a => a thuộc Ư(432)

=> a thuộc ƯCLN(360;432)

=>ƯCLN(360;432)=72

=> a=72

À chỗ ƯCLN mk nhầm nha ƯCLN fai = 72 => a = 72

Vì : 398 chia cho a dư 38 \(\Rightarrow398-38⋮a\Rightarrow360⋮a\) ( a > 38 ) (1)

Vì : 450 chia cho a dư 18 \(\Rightarrow450-18⋮a\Rightarrow432⋮a\) ( a > 18 ) (2)

Từ (1) và (2)

\(\Rightarrow a\inƯC\left(360,432\right)\left(a>38\right)\)

\(360=2^3.3^2.5\)

\(432=2^4.3^3\)

\(ƯCLN\left(360,432\right)=2^3.3^2=72\)

\(ƯC\left(360,432\right)=\left\{1;2;3;4;8;24;36;72\right\}\)

Mà : a > 38 => a = 72

Vậy a = 72

398 : a dư 38 =>398-38=360 chia hết cho a

450 : a dư 18 =>450-18=432 chia hết cho a

Điều kiện a thuộc N^*

Ta tìm ƯCLN của 360 và 432:

360=2³.3².5

432=2⁴.3³

Suy ra ƯCLN của 360 và 432 = 2³.3²=8.9=72

Vậy a=72

Zu la la

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