Prove that for any natural number ending in 3, its square ends in 9. What about numbers ending in 5? And numbers ending in 4?

We know that we can express any 2 digit number as 10m+n , where m is the digit at tens place and n is the digit at ones place.

∴ the number ending with 3 can be expressed as 10m+3.

(10m + 3)² = (10m)² + 2×10m×3 + 3² …………using, (x+y)² = x² + 2xy + y²

= 100m² + 10× 6m + 9

Clearly, we can observe that the square ends in 9.

The number ending with 5 can be expressed as 10m+5.

(10m + 5)² = (10m)² + 2×10m×5 + 5² …………using, (x+y)² = x² + 2xy + y²

= 100m² + 100m + (25)

= 100 (m² + m) + (10×2 + 5 )

Clearly, we can observe that the square ends in 5.

The number ending with 4 can be expressed as 10m+4.

(10m + 4)² = (10m)² + 2×10m×4 + 4² …………using, (x+y)² = x² + 2xy + y²

= 100m² + 10× 8m + (16)

= 100m² + 10× 8m + (10 + 6)

= 100 m² + 10× (8m+1) + 6

Clearly, we can observe that the square ends in 6.