Question

The digit in the unit place of the number 7^295*3^158 is

Answer 1

Answer:

Hi friend

Hope this answer helps you

Step-by-step explanation:

Thus, the last digit of 7295 is equal to the last digit of 73 i.e. 3. Let's divide 158 by 4, the remainder is 2. Hence the last digit will be 9. Therefore, unit's digit of (7925 X 3158) is unit's digit of product of digit at unit's place of 7925 and 3158 = 3 * 9 = 27.

Answer 2

Correct Question

The digit in the unit place of the number $7^{295}\times 3^{158}$.

Answer:

The digit in the unit place of the number $7^{295}\times 3^{158}$ is 27.

Step-by-step explanation:

Given:

The digit in the unit place of the number $7^{295}\times 3^{158}$.

To Find:

The digit in the unit place of the number $7^{295}\times 3^{158}$.

Solution:

As given, the digit in the unit place of the number $7^{295}\times 3^{158}$.

The following is the  Cyclicity table for 7.

$7^{1} =7$    

$7^{2} =49$

$7^{3} =343$

$7^{4} =2401$

Let’s divide 295 by 4 and  we get the remainder is 3.

Thus, the last digit of $7^{295}$ is equal to the last digit of $7^{3}$ that is 3.

The following is the Cyclicity table for 3.

$3^{1} =3\\3^{2} =9\\3^{3} =27\\3^{4} =81\\3^{1} =243$

The Cyclicity table for 3 is as follows:

31 =3

32 =9

33 = 27

34 = 81

35 = 243

Let’s divide 158 by 4, the remainder is 2. Hence the last digit will be 9.

Therefore, the digit in the unit place of the number $7^{295}\times 3^{158}$

=  unit’s digit of the product of digit at unit’s place of $7^{295\ and \ 3^{158}$

$=3\times 9=27$

Thus, the digit in the unit place of the number $7^{295}\times 3^{158}$ is 27.

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