Prove that for any natural numbers ending in 3,its square ends in 9
Answers
Answer:
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. Clearly, we can observe that the square ends in 9.
Answer:
step by step
Step-by-step explanation:
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.
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.
= 100m2 + 10× 8m + (10 + 6)
= 100 m2 + 10× (8m+1) + 6
Clearly, we can observe that the square ends in 6.