The GRE General Exam - The Definitive Shortcuts Guide NUMBER SYSTEM is conceived as the first chapter of the three we will release every Monday in the coming weeks. It is not only just a collection of simple tricks or shortcuts for the quantitative reasoning portion of the GRE exam, but also a compendium if the students want to have, for example, the divisibility rule at a glance without consulting our GRE Math Essentials - A most comprehensive handout!! [COMPLETED] or the GRE - Math Book which I strongly recommend studying for a complete and robust quant theory knowledge for the GRE.

The subsequent chapters that will be released are:

• ALGEBRA Shortcuts
• Geometry Shortcuts

1.
Method to multiply 2-digit number

\(AB × CD = AC [AD + BC] BD\)

\(63*46 = 24 [36 + 12] 18 \)

Add the middle term = 24 [48] 18

Keep first term intact, form the middle term by adding 2 numbers also keep the last term same, which means \(2 (4+4) (8+1) 8 = 2898\)


2.
Some properties of square and square root

(a) Complete square of a no. is possible if its last digit is 0, 1, 4, 5, 6 & 9. If last digit of a no. is 2, 3, 7, 8 then complete square root of this no. is not possible.
(b) If last digit of a no. is 1, then last digit of its complete square root is either 1 or 9.
(c) If last digit of a no. is 4, then last digit of its complete square root is either 2 or 8.
(d) If last digit of a no. is 5 or 0, then last digit of its complete square root is either 5 or 0.
(e) If last digit of a no. is 6, then last digit of its complete square root is either 4 or 6.
(f) If last digit of a no. is 9, then last digit of its complete square root is either 3 or 7.


3.
Prime Numbers

(a) Find the approx square root of given no. Divide the given no. by the prime no. less than approx square root of no. If a given no. is not divisible by any of these prime no. then the no. is prime otherwise not.

(i) To check 359 is a prime number or not.
Approx sq. root = 19
Prime no. < 19 are 2, 3, 5, 7, 11, 13, 17
359 is not divisible by any of these prime nos. So 359 is a prime no.

(ii) Is \(2^{5001} + 1\) is prime or not?

\(\frac{2^{5001}+1}{2+1} \Rightarrow\) Reminder = 0,

\(2^{5001}+1\) is not prime.


(b) There are 15 prime no. from 1 to 50.
(c) There are 25 prime no. from 1 to 100.
(d) There are 168 prime no. from 1 to 1000.


4.
If a no. is in the form of \(x^n + a^n\), then it is divisible by \((x + a)\); if n is odd.

5.
If \(\frac{x^n}{(x – 1)}\), then the remainder is always 1.

6.
If \(\frac{x^n}{(x + 1)}\)

(a) If n is even, then remainder is 1.
(b) If n is odd, then remainder is x.


7.
(a) Value of \(\sqrt{P+ \sqrt {P+\sqrt{P+..........\infty}=\frac{\sqrt{4P+1}+1}{2} \)

(b) Value of \(\sqrt{P- \sqrt {P-\sqrt{P-..........\infty}=\frac{\sqrt{4P+1}-1}{2} \)

(c) Value of \(\sqrt{P. \sqrt {P.\sqrt{P...........\infty}=P \)

(d) Value of \(\sqrt{P \sqrt {P\sqrt{P..........\infty}=P ^{\frac{(2^n-1)}{2^n}}\)

[Where \(n \rightarrow\) no. of times P repeated]

Note: If factors of P are n & (n + 1) type then value of \(\sqrt{P+ \sqrt {P+\sqrt{P+..........\infty}=(n+1) \) and \(\sqrt{P- \sqrt {P-\sqrt{P-..........\infty}=n \)


8.
Numbers of divisors :

(a) If N is any no. and \(N = a^n × b^m × c^p ×\) .... where a, b, c are prime no.

No. of divisors of \(N = (n + 1) (m + 1) (p + 1) ....\)

e.g. Find the no. of divisors of \(90000\).

\(N = 90000 = 2^2× 3^2× 5^2× 10^2= 2^2× 3^2× 5^2× (2 × 5)^2= 2^4 × 3^2× 5^4\)

So, the no. of divisors \(= (4 + 1) (2 + 1) (4 + 1) = 75\)

(b) \(N = a^n× b^m × c^p\), where a, b, c are prime

Then set of co-prime factors of \(N = [(n + 1) (m + 1) (p + 1) – 1 + nm + mp + pn + 3mnp]\)

(c) If \(N = a^n× b^m × c^p...\), where a, b & c are prime no. Then sum of the divisors \(= \frac{(a^{n+1}-1)(b^{m+1}-1)(c^{p+1}-1)}{(a-1)(b-1)(c-1)}\)


9.
Sum Rules:

• Sum of first n natural numbers\(= \frac{n(n+1)}{2}\)

• Sum of square of first n natural numbers\(=\frac{n(n+1)(2n+1)}{6}\)

• Sum of cubes of first n natural numbers\(= (\frac{n(n+1)}{2})^{2}\)

• Sum of first n odd numbers\(= n^2\)

• Sum of first n even numbers\(= n(n+1)\)


10.

Divisibility Rules
Numbers	IF a Number	Examples
Divisible by 2	End with 0,2,4,6,8 are divisible by 2
Divisible by 3	Sum of its digits is divisible by 3
Divisible by 4	Last two digit divisible by 4
Divisible by 5	Ends with 0 or 5
Divisible by 6	Divides by Both 2 & 3
Divisible by 7	Multiply the last digit with 2 and subtract it to remaining number in given number, result must be divisible by 7	99995 : 9999 – 2 × 5 = 9989 9989 : 998 – 2 × 9 = 980 980 : 98 – 2 × 0 = 98 Now 98 is divisible by 7, so 99995 is also divisible by 7.
Divisible by 8	Last 3 digit divide by 8
Divisible by 9	The sum of the digit of the number is divisible by nine or I.E the result is a multiple of nine	14526: 1+4+5+2+6=18 18 is divisible by nine or is multiple of 9
Divisible by 10	End with 0
Divisible by 11	In a number, if difference of sum of digit at even places and sum of digit at odd places is either 0 or multiple of 11, then no. is divisible by 11.	\(\frac{12342}{ 11}\) Sum of even place digit = 2 + 4 = 6 Sum of odd place digit = 1 + 3 + 2 = 6 Difference = 6 – 6 = 0 12342 is divisible by 11.
Divisible by 12	The number must be divisible by 3 and 4
Divisible by 13	Multiply last digit with 4 and add it to remaining number in given number, result must be divisible by 132	\(\frac{876538 }{ 13}\) 876538: 8 × 4 + 3 = 35 5 × 4 + 3 + 5 = 28 8 × 4 + 2 + 6 = 40 0 × 4 + 4 + 7 = 11 1 × 4 + 1 + 8 = 13 13 is divisible by 13. 876538 is also divisible by 13.
Divisible by 14	The number must be divisible by 2 and 7. Because 2 and 7 are prime factors of 14.
Divisible by 15	The number should be divisible by 3 and 5. Because 3 and 5 are prime factors of 15.
Divisible by 16	The number formed by last four digits in given number must be divisible by 16.
Divisible by 17	Multiply last digit with 5 and subtract it from remaining number in given number, result must be divisible by 17	294678: 29467 – 5 × 8 = 29427 27427: 2942 – 5 × 7 = 2907 2907: 290 – 5 × 7 = 255 255: 25 – 5 × 5 = 0 294678 is completely divisible by 17.
Divisible by 18	The number should be divisible by 2 and 9
Divisible by 19	Multiply last digit with 2 and add it to remaining number in given number, result must be divisible by 19	149264: 4 × 2 + 6 = 14 4 × 2 + 1 + 2 = 11 1 × 2 + 1 + 9 = 12 2 × 2 + 1 + 4 = 9 9 × 2 + 1 = 19 19 is divisible by 19 149264 is divisible by 19.
Divisible by 20	The number formed by last two digits in given number must be divisible by 20.