NUMBER SYSTEM Shortcuts
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:
1.
Method to multiply 2digit 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 \(\sqrt{P \sqrt {P\sqrt{P..........\infty}=P ^{\frac{(2^n1)}{2^n}}\)
[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 coprime 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 \(= \frac{(a^{n+1}1)(b^{m+1}1)(c^{p+1}1)}{(a1)(b1)(c1)}\)
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.  
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