Question

the unit digit in the sum (124)^372+(124)^373

Answer 1

The answer is 0.

124^372(1+124)

== 124^372(125)

Take all the unit digits.

== 4^2(5)

== 16*5

== 80

The unit digit is 0.

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Answer 2

Answer:

0 is the required  unit digit   of $(124)^{372} +(124)^{373}$.

Step-by-step explanation:

Explanation:

Given, $(124)^{372} +(124)^{373}$

Step1:

Last digit of 124 is 4

Now , $4^{1} = 4$

    $4^{2} = 16$ ⇒6 is the unit digit of 16.

    $4^{3} = 64$ ⇒ 4 is the unit digit of 64

   $4^{4} = 256$ ⇒ 6 is the unit digit of 256

$4^{5} = 1,024$ ⇒ 4 is the unit digit of 1,024.

So, here we can see that Cyclicity of 4 is 2 because it is repeat every time .

Therefore , on dividing 372 by 2 we get remainder 0.

If the remainder become 0 ,then we check the last digit and if the last digit is (l) =2,4,6,8 then the unit digit be 6.

∴The unit digit of $(124)^{372}$ is 6.

Step2:

last digit of 124 is 4 .

Similarly , from step 1 we have Cyclicity of 4 is2

Therefore , on dividing 373 by 2 we get remainder 1 .

Unit digit of any given no. = $l^{r}$

(where l = last digit number and r = remainder )

put the value of l = 4 and r =1

∴ $l^{r} = 4^{1}$ = 4

The unit digit of $(124)^{373}$ is 4

Step 3:

So ,  $(124)^{372} +(124)^{373}$ = 6+4 = 10⇒ 0 is the unit digit of this value .

(where , 6 is the unit digit of  $(124)^{372}$ and 4 is the unit digit of $(124)^{373}$)

Final answer :

Hence , the unit  digit of $(124)^{372} +(124)^{373}$ is 0 .