cho abc chia hết cho 3 và 7
chúng ta tổng (a + 19b + 4c) chia hết cho 3 và 7
1,choabc chia hết cho 3;7. Chứng tỏ rằng a+19b+4c chia hết cho 3 và 7
Chứng minh rằng nếu số a,b và c chia hết cho7 và 3 thì a+19b+4c chia hết cho 7 và3
cho abc chia het cho 3 va 7
chung ta tong ( a + 19b + 4c ) chia het cho 3 va 7
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chứng minh nếu abc chia hết cho 21 thì a+19b+4c chia hết cho 21
abc\(⋮\)21=> 100a+10b+c\(⋮\)21
=> 16a+10b+c\(⋮\)21(vì 84a\(⋮\)21)
=> 64a+40b+4c\(⋮\)21
mà 64a+40b+4c-(a+19b+4c)=63a+21b\(⋮\)21
=> a+19b+4c\(⋮\)21(đpcm)
Cho abc là số tự nhiên có 3 chữ số. Chứng minh rằng abc chia hết cho 21 khi và chỉ khi a - 2b + 4c chia hết cho 21.
Ta có :
4 . abc = 400a + 40b + 4c = 399a + 42b + a - 2b + 4c
= 21 ( 19a + 2b ) + ( a - 2b + 4c ) chia hết cho 21
( Do 21 chia hết cho 21 và a - 2b + 4c chia hết cho 21 )
=> 400a + 40b + 4c chia hết cho 21
=> 4 ( 100a + 10b + c ) chia hết cho 21
=> 100a + 10b + c chia hết cho 21
=> abc chia hết cho 21
Vậy nếu a-2b+4c chia hết cho 21 thì abc chia hết cho 21
cho abc là số có 3 chữ số . CMR: abc chia hết cho 21 khi và chỉ khi (a-2b+4c)chia hết cho21
Ta có :
4 . abc = 400a + 40b + 4c = 399a + 42b + a - 2b + 4c
= 21 ( 19a + 2b ) + ( a - 2b + 4c ) chia hết cho 21
( Do 21 chia hết cho 21 và a - 2b + 4c chia hết cho 21 )
=> 400a + 40b + 4c chia hết cho 21
=> 4 ( 100a + 10b + c ) chia hết cho 21
=> 100a + 10b + c chia hết cho 21
=> abc chia hết cho 21
Vậy nếu a-2b+4c chia hết cho 21 thì abc chia hết cho 21
Bài 1 : Cho 9a + 4b + 5 c chia hết cho 11 . CMR : 9a + b + 4c chia hết cho 11
Bài 2 : Tìm số A = abc biết A chia hết 7 và a + b + c chia hết cho 7
PLe hãy giải cho e
Ta có a+3b chia hết cho 7 ( a,b thuộc N) Chứng minh 4a+19b chi hết cho 7
Ta có : 4a + 19b
<=> 4a + 12b + 7b
<=> 4( a + 3b ) + 7b
Vì a + 3b ⋮ 7 => 4 ( a + 3b ) ⋮ 7 (1)
7b có 7 ⋮ 7 => 7b ⋮ 7 (2)
Từ (1) ; (2) => 4 ( a + 3b ) + 7b ⋮ 7
=> 4a + 19 b ⋮ 7 ( đpcm )
ta có : a+3b chia hết cho 7 suy ra a và 3b chia hết cho 7 và b chia hết cho 7
suy ra 4a cũng chia hết cho 7
mà 19b cũng chia hết cho 7
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Biết a+b chia hết cho 3, CM:
a, a+7b chia hết cho3
b, 2a- 7b chia hết cho 3
c, 13a+ 19b+ 2016 chia hết cho 3
a,ta có:
a+7b=(a+b)+6b
vì \(\hept{\begin{cases}\left(a+b\right)⋮3\\6b⋮3\end{cases}}\)
=>a+7a chia hết cho 3 với a+b chia hết cho 3
b,ta có:
2a-7b=2(a+b)-9b
vì\(\hept{\begin{cases}2\left(a+b\right)⋮3\\-9b⋮3\end{cases}}\)
=>2a-7b chia hết cho 3 với a+b chia hết cho 3
c, ta có:
13a+19b+2016=13(a+b)+6b+2016
vì\(\hept{\begin{cases}13 \left(a+b\right)⋮3\\6b⋮3\\2016⋮3\end{cases}}\)
=>13a+19b+2016 chia hết cho 3 với a+b chia hết cho 3