If n is a natural number whuch when divided by4 leaves a remainder 3, then what is the remainder obtained when #n^2 + 6n - 10# is divided by 8?

Question

1246 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-18T18:55:20

Remainder 1

#n equiv_4 3 qquad implies bb(n = 4 lambda + 3) qquad lambda in ZZ#

#n^2 + 6n - 10 = 16 lambda^2 + 24 lambda + 9+ 24 lambda + 18 - 10#

# = 16 lambda^2 + 48 lambda + 17#

#=8( 2 lambda^2 + 6 lambda + 2 + 1/8) #

#=8( 2 lambda^2 + 6 lambda + 2 )+ 1 = bb(8 mu + 1) qquad mu in ZZ#

#implies n^2 + 6n - 10 equiv_8 1 #

Answer 2 · 2018-06-18T18:57:13

# 1#.

Given that #n in NN#, when divided by #4# leaves remainder #3#.

#:. n=4m+3," for some "m in NN#.

Let, #p(n)=n^2+6n-10#.

Subst.ing #n=4m+3# in #p(n)#, we get,

#p(n)=p(4m+3)=(4m+3)^2+6(4m+3)-10#,

#=(16m^2+24m+9)+(24m+18)-10#,

#=16m^2+48m++17#.

#:. p(n)=8(m^2+24m+2)+1#.

Clearly, it follows that #p(n)#, when divided by #8,# will leave

the remainder #1#.

Enjoy Maths.!