Tong sau co chia het cho3 ko
A= 2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^10
tong sau co chia het cho 3 ko
A= 2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^10
Từ các số đó, ta ghép đc những cặp này :
A có ( 2 + 29 ) + ( 22 + 28 ) + ( 23 + 27 ) + ( 24 + 26 ) + 25 + 210
A = 210 + 210 + 210 + 210 + 210 + 25
Ta có : 1024 + 1024 + 1024 + 1024 + 1024 +32
1024 . 5 + 32
5120 + 32
Vậy : 5152 : 3
= 1717,33
Nên : Tổng sau không chia hết cho 3
khong gia tri xet xem cac tong sau day co chia het cho 9 hay khong 2*3*4*5*6*7+8*9*10*11*12
a) A=2n+22+23+24+25+26+27+28+29+210
Khong tinh gia tri. Hay chung to A chia het cho 3.
b) Tong sau day co chia het cho 5 khong?
B=40+41+4243+44+45+46+47+48+49+410
Tong Say Co chia het cho 3 khong ?
A= 2 + 2^2 +2^3 + 2^4+ 2^5 +2^6 +2^7 +2^8
Đung minh cho 1 tick
\(A=2+2^2+2^3+2^4+2^5+2^6+2^7+2^8\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+\left(2^5+2^6\right)+\left(2^7+2^8\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+2^5\left(1+2\right)+2^7\left(1+2\right)\)
\(=\left(1+2\right)\left(2+2^3+2^5+2^7\right)\)
\(=3\left(2+2^3+2^5+2^7\right)\)\(⋮\)\(3\)
bai 1; cho tong M =126 +213+x. Tim x de M chia het cho 3
bai 2; chung to rang tong ; A= 2 + 2 mu 3 + 2 mu 4 + 2 mu 5 + 2 mu 6 +2 mu 7 +2 mu 9 + 2 mu 10 + 2 mu 12 chia het cho 5
Có : 126 chia hết cho 3, 213 chia hết cho 3
Để được M chia hết cho 3 thì x phải chia hết cho 3
Hay gọi là 3k ( k thuộc N)
2.
Hình như đầu bài bài 2 sai
dung do khong sai dau
a, tong 10^15+8 co chia het cho 2 va 9 khong?
b, tong 10^2010 +8 co chia het cho 9 khong ?
c, 10^2010-4 co chia het cho 3 khong ?
1) Thuc hien phep tinh:
a) 4^6 × 9^2 / 6^3 × 8^2
b) 5^6 × 15^4 /5^4 × 3^3
c) 3^9 - 2^3 × 3^7 + 2^10 × 3^2 - 2^13 / 3^10 - 2^2 × 3^7 + 2^10 × 3^3 - 2^12
d) (1/9)^10 : (1/3)^20
e) (1/64)^5 × (1/4)^7
2) Chung minh rang:
a)2014^100+2014^99 chia het cho 2015
b) 4^13+32^5-8^8 chia het cho 5
c) 81^27-27^9-9^13 chia het cho 405
Help me!! Pleass!!
tong day co chia het cho 5 khong \(A=2+2^3+2^4+2^5+2^6+2^7\)
chứng minh (1+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^10+2^11) chia het cho 9