euclid division lema for 7m 7m+1 7m+6
Answers
Answer:
Step-by-step explanation:
# here is your answer //
Let take b = 7. (/here b<a. and a , b , q and r are the positive integers)
according to Euclid division lemma
a = bq+r. (0≥r>b)
a = 7q+r (0≥r>7)
so r values are
r -------->0,1,2,3,4,5,6
1) when r value equals to zero (r=0)
a = 7q+r
a = 7q+0
a = 7q
cube both side
a³ = 343q³
a³ = 7(49q³)
a³ = 7m. (/here m is positive integer and m = 49q³)
2) if value of r is equals to 1 (r=1)
a = 7q+r
a = 7q+1
cube both side
(a)³ = (7q+1)³
a³ = 343q³+1³+3(7q)(1)(7q+1)
a³. = 343q³+21q(7q+1) + 1
a³. = 343q³+ 147q² + 21q + 1
a³ = 7(49q³+21q²+3q)+1
a³ = 7m + 1. (/here m is positive integer and m = 49q³+21q²+3q)
3) if r value equals to 6 ( r = 6)
a = 7q+r
a = 7q+6
cube both side
(a)³ = (7q+6)³
a³ = 343q³+216+3(7q)(6)(7q+6)
a³ = 343q³+126q(7q+6)+216
a³ = 343q³+882q²+756q+216
a³ = 7(49q³+127q²+108q)+216
a³ = 7m+216. (/here m is positive integer and m = 49q³+127q²+108q)
I think 7m + 6 is not a cube of any positive integer
// hope it is helpful #R //☺☺