cho abc chia hết cho 3 và 7
chúng ta tổng (a + 19b + 4c) chia hết cho 3 và 7
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NT
0
NT
1
27 tháng 12 2017
Vì a,b và c chia hết cho 7 và 3 => a+b+c chia hết cho 7 và 3.
=>a+b.m+c.n chia hết cho 7 và 3
=> a+19b+4c chia hết cho 7 và 3 😁
TH
2
10 tháng 10 2017
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AK
1
12 tháng 4 2019
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)
31 tháng 1 2021
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
31 tháng 1 2021
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
VT
3
11 tháng 11 2016
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 )
20 tháng 2 2019
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
20 tháng 2 2019
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