Question

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

Answer 1

Given :  5321³

To Find :  Digit at unit Place

Solution:

0³ = 0  ends with 0

1³  = 1    ends with 1

2³ = 8    end with 8

3³ = 27  ends with 7

4³ = 64  ends with 4

5³ = 125  ends with 5

6³ = 216  ends with 6

7³ = 343  ends with 3

8³ = 512  ends with 2

9³ = 729  ends with 9

As 5321 ends with 1

Hence 5321³  ends with 1

unit place of cube of 5321 is  1

Another way

5321³

= (5320 + 1)³

using (a + b)³ = a³ + b³  + 3ab(a + b)  

a = 5320

b = 1

=  5320³ + 1³  +  3(5320)(1)(5320 + 1)

=  10³ x 532³  + 10 x 3 x  532 (5321)  + 1

= 10 x 100 x 532³ + 10 x 3 x  532 (5321) + 1

= 10 (  100 x 532³ + 3 x  532 (5321) ) + 1

= 10k + 1

Hence Digit at unit place is 1

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Answer 2

SOLUTION

TO DETERMINE

The digit at unit place of cube of 5321

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

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

For some positive integer k

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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