Question

Write the cubes of 5 natural numbers which are of hte form 3n +1(e.g. 4,7,10,...) and erify the following :
'The cube of a natural number of the form 3n + 1 is a natural number of the same form i.e. when divided by 3 it leave the remainder 1 '.

Answer 1

3n + 1
Let n = 1 , 2 , 3 , 4, 5, hten
If n = 1, then $3n+1=3\times +1=4$
If n = 2, then $3n+1=3\times 2+1=7$
If n = 3, then $3n+1=3\times 3+1=9+1=10$
If n = 4, then $3n+1=3\times 4+1=12+1=13$
If n = 5, then $3n+1=3\times 5+1=15+1=16$
Now $(4{)}^3=4\times 4\times 4=64$
Which is $\frac{64}{3}=21,$ remainder = 1
$(7{)}^3=7\times 7\times 7=343$
Which is $\frac{343}{3}=114,$ Remainder = 1
$(10{)}^3=10\times 10\times 10=1000÷3=333,Remainder=1(13{)}^3=13\times 13\times 13=2197÷3=732,Remainder=1(16{)}^3=16\times 16\times 16=4096÷3=1365,$
Remainder = 1
Hence cube ofd antural number of the form, 3n + 1 , is a mnatural of the form 3n + 1