Question

Write the cubes of 5 natural numbers of the form 3n + 2 (i.e. 5 , 8 , 11, ..... ) and verify the ffollowing
'The cube of a natural number of the form 3n + 2 is a natural number of the same form i.e. when it is divideend by 3 the remainder is 2.

Answer 1

Natural numbers of the form 3n + 2,when n is a natural number i.e. 1, 2, 3, 4, 5,.....
If $n=1,then3n+2=3\times 1+2=3+2=5$
If n = 2, then $3n+2=3\times 2+=6+2=8$
If n = 3 then $3n+2=3times3+2=9+2=11$
If n = 4, then $3n+2=3\times 4+2=12+2=14$
and fi n = 5, then $3n+2=3\times 5+2=15+2=17$
Now $(5{)}^3=5\times 5\times 5=125125÷3=41,Remainder=2(8{)}^2=8\times 8\times 8=512512÷3=170,Remainder=2(11{)}^3=11\times 11\times 11=13311331÷3=443,Remainder=2(14{)}^3=14\times 14\times 27442744÷3=914Remainder=2(17{)}^3=17\times 17\times 17=49134913÷3=1637Remainder=2$
We see the cube of the natural number of the form 3n + 2 is also a natural number of the form 3n +2