aaabbb = aaa000 + bbb
= a.111.1000 + b.111
= a.3.37.1000 + b.3.37
= 37.(a.3.1000 + b.3) 37
Ta có : \(\overline{aaa}+\overline{bbb}=\left(100a+10a+a\right)+\left(100b+10b+b\right)\)
\(=111a+111b=111\left(a+b\right)=37.3.\left(a+b\right)\)
Vậy ta có đpcm
Ta có : aaa + bbb
= 100a + 10a + a + 100b + 10b + b
= ( 100a + 100b ) + ( 10a + 10b ) + ( a +b )
= 100.(a+b) + 10.(a+b) + ( a+b)
= 111 . ( a + b )
= 37 . 3 . ( a + b ) \(⋮\) 37 ( đpcm)
Ta có:
aaa+bbb
=100a+10a+a+100b+10b+b
=(100a+100b)+(10a+10b)+(a+b)
=100(a+b)+10(a+b)+(a+b)
=111(a+b)
=37.3.(a+b):37( ĐPCM)
Tick cho mik nghen
Chứng tỏ :
( aaa + bbb ) chia hết cho 37
Ta có: aaa + bbb
= a.100 + a . 10 + a +b.100+ b.10 + b
= a. (100+10+1) + b .(100+10+1)
= a. 111 + b .111
= 111. (a+b)
= 37.3 (a+b) (đpcm)