Question

# Write down the units digits of the squares of the following numbers  $(i)24$  $(ii)78$  $(iii)35$

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Hint: When the numbers are small we can easily calculate and write the answer, but when the number is more than 30 it becomes difficult to remember the squares and cubes. That’s when we start using the algebraic expressions and formulae to calculate the value of squares and cubes. For this particular sum we would be using the expression ${(a + b)^2} = {a^2} + {b^2} + 2ab$ .

$(i) - 24$
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.
So, the correct answer is “6”.

$(ii) - 78$
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.
So, the correct answer is “4”.

$(iii) 35$
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
So, the correct answer is “5”.

Note: In such a type of numerical it is always advisable to verify the given answer with the shortcut as illustrated. Also Students have to memorize the algebraic expressions by heart so that it becomes easier to solve much more complex numerical. It is highly beneficial not only in solving the given numericals but also in chapters like Ratio & Proportion, Complex Numbers.