Question

What is the units digit in(9^15 - 3^13-6^20)?​

Answer 1

Answer:

Unit digit for 9^15 =9

unit digit for 3^13 =3

unit digit for 6^20 =6

now

9-3-6

9-9

= 0 ans.

Answer 2

Answer:

0 is the required unit digit in $9^{15} -3^{13} -6^{20}$

Step-by-step explanation:

Explanation :

Given , $9^{15} -3^{13} -6^{20}$

Step 1:

Lat digit of 9 is 9

$9^{1} = 9$

$9^{2} = 81$ ⇒1 is the unit digit of 81.

$9^{3} = 729$  ⇒9 is the unit digit of 729.

Here we can see that the pattern[1,9] repeat every time.

So , the cyclicity of 9 is 2 .

Now , on dividing 15 by 2 we get 1 as remainder .

$l^{r} = 9^{1} = 9$ (where l is last digit and r is remainder)

Therefore , 9 is the unit vector of  $9^{15}$ .

Step 2:

For $3^{13}$ .

Last digit is 3

$3^{1} = 3$ which is unit digit .

$3^{2} = 9$  unit digit

$3^{3} = 27$ .

here, 7 is the unit digit of 27 .

$3^{4} = 81$ ⇒ 1 is the unit digit of 81 .

Again ,$3^{5} =243$ ⇒ 3 is the unit digit of 3.

So , the cyclicity of 3 is 4  because (3,9,7,1 pattern repeat every time .)

On dividing 13 by 4 we get 1 as remainder .

$l^{r} = 3^{1} = 3$  is the unit digit of $3^{13}$.

Step3:

For $6^{20}$

Similarly,

The cyclicity of 6 is 6 .

On dividing 20 by 6 we get 2 as a remainder .

Therefore,  $l^{r} = 6^{2} = 36$ .

So , the unit digit is 6 .

Step 4:

$9^{15} -3^{13} -6^{20}$

Put the unit digit of the given number

⇒ 9-3-6 = 0 .

Final answer :

Hence , 0 is the unit digit in ($9^{15} -3^{13} -6^{20}$) .