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$ (i)24 $

$ (ii)78 $

$ (iii)35 $

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$ (i) - 24 $

Using the $ {(a + b)^2} = {a^2} + {b^2} + 2ab $ , we will have to define $ a\& b $ .

We can further split $ 24 $ as $ 20 + 4 $ . This is because it is easy to find squares of numbers which have 0 at units place.

After the split we can define $ a = 20 $ and $ b = 4 $

Now using the expression we get

$ \Rightarrow {(24)^2} = {(20)^2} + {(4)^2} + 2 \times 4 \times 20 $

$ \Rightarrow {(24)^2} = 400 + 16 + 160 $

$ \Rightarrow {(24)^2} = 576............(1) $

Thus the Units digit for $ {24^2} $ is $ 6 $ from $ Equation1 $ .

VERIFICATION:

In order to verify we can just take the units digit of the given number and square it. The units’ digit value obtained after squaring the unit digit of the original number is our final answer. In our case we will have to take a unit digit of $ 24 $ i.e. $ 4 $ .

Square of $ 4 $ is $ 16 $ . Thus the units digit of $ 16 $ is $ 6 $ .

$ \therefore $ Units digit of $ {24^2} $ is $ 6 $

Verified.

$ (ii) - 78 $

Using the $ {(a + b)^2} = {a^2} + {b^2} + 2ab $ , we will have to define $ a\& b $ .

We can further split $ 78 $ as $ 70 + 8 $ . This is because it is easy to find squares of numbers which have 0 at units place.

After the split we can define $ a = 70 $ and $ b = 8 $

Now using the expression we get

$ \Rightarrow {(78)^2} = {(70)^2} + {(8)^2} + 2 \times 8 \times 70 $

$ \Rightarrow {(78)^2} = 4900 + 64 + 1120 $

$ \Rightarrow {(78)^2} = 6084............(2) $

Thus the Units digit for $ {78^2} $ is $ 4 $ from $ Equation2 $ .

VERIFICATION:

In order to verify we can just take the units digit of the given number and square it. The units’ digit value obtained after squaring the unit digit of the original number is our final answer. In our case we will have to take a unit digit of $ 78 $ i.e. $ 8 $ .

Square of $ 8 $ is $ 64 $ . Thus the units digit of $ 64 $ is $ 4 $ .

$ \therefore $ Units digit of $ {78^2} $ is $ 4 $

Verified.

$ (iii) 35 $

Using the $ {(a + b)^2} = {a^2} + {b^2} + 2ab $ , we will have to define $ a\& b $ .

We can further split $ 35 $ as $ 30 + 5 $ . This is because it is easy to find squares of numbers which have 0 at units place.

After the split we can define $ a = 30 $ and $ b = 5 $

Now using the expression we get

$ \Rightarrow {(35)^2} = {(30)^2} + {(5)^2} + 2 \times 5 \times 30 $

$ \Rightarrow {(35)^2} = 900 + 25 + 300 $

$ \Rightarrow {(35)^2} = 1225............(3) $

Thus the Units digit for $ {35^2} $ is $ 5 $ from $ Equation3 $ .

VERIFICATION:

In order to verify we can just take the units digit of the given number and square it. The units’ digit value obtained after squaring the unit digit of the original number is our final answer. In our case we will have to take a unit digit of $ 35 $ i.e. $ 5 $ .

Square of $ 5 $ is $ 25 $ . Thus the units digit of $ 25 $ is $ 5 $ .

$ \therefore $ Units digit of $ {35^2} $ is $ 5 $

Verified