Question

show that of any positive integer in the form of 5 m , 5m + 1 or 5m - 1 for some integer m​

Answer 1

• Answer:-Let 'm' be any positive integer.Apply Euclid's division lemma with 'm'and b=5.wkt, a=bq+r, q>0, 0<r<ba=5q+r. [i. e, r=0,1,2,3,4].If r=o, a=5q→5m

Answer 2

Step-by-step explanation:

Consider that m=2.square of 2=4, (2×2=4).

5m=5 (2)=10

4 not equals to 10

5m+1=5 (2)+1=11

4 not equals to 11

5m+4=5 (2)+4=14

14 is also not equals to 4.

If we take 5 as "m" then the square of 5=25,which satisfies the condition 5m

hence proved.

only 5 is the number which can satisfies the condition 5m. There is no other number which can satisfies the conditions 5m,5m+1,5m+4.