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Number Theory – Quantitative Ability

By   /   September 5, 2013  /   No Comments

Numbers in QA

Numbers

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

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