This is in continuation from our previous articles on Quantitative Ability. Once an aspirant has learnt the technique of calculating fast by practicing the shortcuts of addition, subtraction, multiplication & division, from the article ‘Crack Quanta with us’, he may proceed with the following concepts used in number theory.

**Calculating number of factors under number theory:**

We can find out the number of factors that a number has, without actually checking how many numbers divide it apart from 1 and itself. The method is as follows.

Step 1 – Factorize the number and express it as a product of prime numbers.

Example, 24 = (2^3)*(3^1)

Step 2 – Add one to the power of each prime factor and multiply them.

Number of factors = (3+1)*(1+1) = 4*2 = 8

Therefore, 24 has 8 factors.

**Calculating sum of factors under number theory:**

The sum of all factors of a number can also be calculated without actually finding out the factors.

Step 1 – Factorize the number and express it as a product of prime numbers.

Example, 24 = (2^3)*(3^1)

Step 2 – Add each prime factor from the prime factor raised to the power 0 to the prime factor raised to the power as found in step 1.

For 2 -> (2^0)+(2^1)+(2^2)+(2^3) = 1+2+4+8 = 15

For 3 -> (3^0)+(3^1) = 1+3 = 4

Step 3 – Multiply the sums found in step 2.

Sum of factors = 15*4 = 60

Therefore, the sum of all factors of 24 is 60.

**Wilson’s Theorem under number theory:**

If n is a prime factor, then (n-1)! + 1 is divisible by n.

Cross checking the theorem – Let n=3. So, (3-1)! + 1 = 2! + 1 = 3, which is divisible by 3.

**Fermat’s Theorem under number theory:**

If p is a prime number and n is prime to p, then (n^(p-1)) – 1 is divisible by p.

Cross checking – Let p=5 and n=3. So, (3^(5-1)) – 1 = (3^4) – 1 = 81 – 1 = 80, which is divisible by 5.

**Remainder Theorem under number theory:**

The product of any 2 or more than 2 natural numbers has the same remainder when divided by any natural numbers, as the product of their remainders.

Example – Find the remainder of (12*13)/7.

Step 1 – Find the remainder of each natural number

Remainder of 12/7 = 5

Remainder of 13/7 = 6

Step 2 – Multiply the remainders and divide by the divisor

Remainder = Remainder of (5*6)/7 = Remainder of 30/7 = 2

Therefore, remainder of (12*13)/7 = 2

**Theorem on prime numbers under number theory:**

All prime squares (for p>3) are of the form k+1

**Rule of Cyclicity under number theory:**

The digit at the units place of a number raised to any power can be easily found using rule of cyclicity.

Example – Find the units place digit for (7^95)*(3^58)

Step 1 – For each number, find the cyclicity of its power for the units digit.

For 7 -> (7^1) = 7, (7^2) = 9, (7^3) = 3, (7^4) = 1, (7^5) = 7….and the cycle repeats. So the cycle for 7 is 7->9->3->1, which is a cycle of 4.

Similarly, for 3, the cycle is 3->9->7->1, which is a cycle of 4.

Step 2 – Find the remainder for each power divided by the cycle number (usually 4).

Remainder of 95/4 = 3

Remainder of 58/4 = 2

Step 3 – Find the unit’s digit for each number raised to the given powers by associating the remainders found in step 2 with the corresponding cycles.

For (7^95), unit digit is 7^3 = 3

For (3^58), unit digit is 3^2 = 9

Step 4 – Multiply the unit digits found in step 3 to get the final digit in units place. We are multiplying because there is a multiplication sign in the original problem.

Unit digit = unit digit for (3*9) = 1

Therefore, unit digit for (7^95)*(3^58) = 1