Question

find the digit at unit place of cube of 5321
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Answer 1

Find the digit at the unit place of the cube of 5321.

Step-by-step explanation:

• The statement to find the unit place of a cube is,

The unit digit of the cube of any number will be the unit digit of the cube of its last digit.  

• The last digit of 5321 is 1.

• The cube of the last digit is = 1³ = 1

• Therefore, the digit at the unit place of the cube of 5321 is 1.  

• Some other examples are:

A) 121  

• Unit digit of 1³ = 1

• The digit at the unit's place of the cube is also 1.

B) 25  

• Unit digit of 5³ = 125  

• The digit at the unit's place of the cube is also 5.

C) 77

• Unit digit of 7³ = 343

• The digit at the unit's place of the cube is also 3.

Answer 2

SOLUTION

TO DETERMINE

The digit at unit place of cube of 5321

FORMULA TO BE IMPLEMENTED

We are aware of the identity that

(a + b)³ = a³ + 3a²b + 3ab² + b³

EVALUATION

Here the given number is 5321

The number 5321 can be rewritten as

5321 = 5320 + 1

Now we find Cube of 5321 as below

$\sf{ {(5321)}^{3} }$

$= \sf{ {(5320 + 1)}^{3} }$

$= \sf{ {(5320)}^{3} + 3 \times {(5320)}^{2} \times 1 + 3 \times 5320 \times {(1) }^{2} + {(1)}^{3} }$

$= \sf{ \bigg[ {(5320)}^{3} + 3 \times {(5320)}^{2} + 3 \times 5320 \bigg]+ 1 }$

$\sf{ since \: all \: of \: {(5320)}^{3} \: , \: 3 \times {(5320)}^{2} \: , \: 3 \times 5320 }$

are divisible by 10

So

$\sf{ \bigg[ {(5320)}^{3} + 3 \times {(5320)}^{2} + 3 \times 5320 \bigg] = 10k }$

Hence from above

$\sf{ {(5321)}^{3} } = 10k + 1$

Hence the digit at unit place of cube of 5321 is 1

FINAL ANSWER

The digit at unit place of cube of 5321 is 1

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