## What are we learning ?

** Competency in Focus:** Number System

This problem from Indian Statistical Institute (ISI Entrance 2012) is based on Number System. It includes finding the remainder when a number is divided by another digit.

## First look at the knowledge graph.

## Next understand the problem

The last digit of \(9!+3^{9966}\) is (A) 3 (B) 9 (C) 7 (D) 1

##### Source of the problem

Indian Statistical Institute (ISI) 2012 Problem 8.

##### Key Competency

### Number system

##### Difficulty Level

4/10

##### Suggested Book

Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics

## Start with hints

Do you really need a hint? Try it first!

Unit digit of the whoe expression will be sum of unit digit of the first term and unit digit of the second term. So if the first term gives last digit 5 and 2nd terms gives 2 then unit digit of whole expresion is (5+2) or 7.

when we expand \(9!\) there will be 5 and 2 in between that when multiplied will give 10 as a factor so the term \(9!\) will have \(0!\) as last digit.

\(3^{1}\) has last digit as 3 \(3^{2}\) has last digit as 9 \(3^{3}\) has last digit as 7 \(3^{4}\) has last digit as 1 \(3^{5} \) has last digit as 3 and the pattern repeats with the power(indxe) at an interval of 4.

9966 when divided by 4 gives 2 as remainder. So, in \(3^{9964} 3^{2} \) , 9 will be the last digit.

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