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5.  A number is prime if it is not divisible by any prime number less than it’s square root.  Ex: Is 179 a prime number ?  Prime Numbers less than 13.3 are 2,3,5,7,11,13  179 is not divisible by any of them, 179 is prime.

7.  Test for divisibility by 7: Double the last digit and subtract it from the remaining leading truncated number. If the result is divisible by 7, then so was the original number. Apply this rule over and over again as necessary.  Example: 826. Twice 6 is 12. So take 12 from the truncated 82. Now 82-12=70. This is divisible by 7, so 826 is divisible by 7.  Test for divisibility by 11: Subtract the last digit from the remaining leading truncated number. If the result is divisible by 11, then so was the first number. Apply this rule over and over again as necessary.  Example: 19151 --> 1915-1 =1914 –>191-4=187 –>18-7=11, so yes, 19151 is divisible by 11.

8.  Test for divisibility by 13: Add four times the last digit to the remaining leading truncated number. If the result is divisible by 13, then so was the first number. Apply this rule over and over again as necessary.  Example: 50661–>5066+4=5070–>507+0=507–>50+28=78 and 78 is 6*13, so 50661 is divisible by 13.  Test for divisibility by 17: Subtract five times the last digit from the remaining leading truncated number. If the result is divisible by 17, then so was the first number. Apply this rule over and over again as necessary.  Example: 3978–>397-5*8=357–>35-5*7=0. So 3978 is divisible by 17.  Test for divisibility by 19: Add two times the last digit to the remaining leading truncated number. If the result is divisible by 19, then so was the first number. Apply this rule over and over again as necessary.  Example:101156–>10115+2*6=10127–>1012+2*7=1026–>102+2*6=114 and 114=6*19, so 101156 is divisible by 19.

9. Divisibility byTest 7Subtract 2 x Last digit 11Subtract 1 x Last digit 13Add 4 x Last Digit 17Subtract 5 x Last Digit 19Add 2 x Last Digit Here is a table using which you can easily remember the previous divisibility rules. Read the table as follows : For divisibility by 7, subtract 2 times the last digit with the truncated number.

10.  Find the unit’s digit of 7 1999 (7 to the power 1999)  Step 1: Divide the exponent by 4 and note down remainder 1999/4 => Rem = 3  Step 2: Raise the unit’s digit of the base (7) to the remainder obtained (3) 7 3 = 343  Step 3: The unit’s digit of the obtained number is the required answer. 343 => Ans 3  If the remainder is 0, then the unit’s digit of the base is raised to 4 and the unit’s digit of the obtained value is the required answer. If Rem = 0, Then 7 4 = XX1 -> Ans 1  Note: For bases with unit’s digits as 1,0,5,6 the unit’s digit for any power will be the 1,0,5,6 itself. Ex: Unit’s digit of 3589494856 453 = 6

11.  We will discuss the last two digits of numbers ending with the following digits in sets :  a) 1  b) 3,7 & 9  c) 2, 4, 6 & 8

12.  a) Number ending with 1 :  Ex : Find the last 2 digits of 31 786  Now, multiply the 10s digit of the number with the last digit of exponent 31 786 = 3 * 6 = 18 -> 8 is the 10s digit.  Units digit is obviously 1  So, last 2 digits are => 81

13.  b) Number Ending with 3, 7 & 9  Ex: Find last 2 digits of 19 266  We need to get this in such as way that the base has last digit as 1 19 266 = (19 2 ) 133 = 361 133  Now, follow the previous method => 6 * 3 = 18  So, last two digits are => 81

14.  b) Number Ending with 3, 7 & 9  Remember :  3 4 = 81  7 4 = 2401  9 2 = 81

15.  Ex 2: Find last two digits of 33 288 Now, 33 288 = (33 4 ) 72 = (xx21) 72 Ten’s digit is -> 2*2 = 04 -> 4 So, last two digits are => 41  Ex 3: find last 2 digits of 87^474 (87 2 )*(87 4 ) 118 => (xx69) * (xx61) 118 (6 x 8 = 48) => (xx69)*(81) So, last two digits are 89

16.  c)Ending with 2, 4, 6 or 8 Here, we use the fact that 76 power any number gives 76. We also need to remember that,  24 2 = xx76  2 10 = xx24  24 even = xx76  24 odd = xx24

17. Ex: Find the last two digits of 2 543 2 543 = ((2 10 ) 54 ) * (2 3 ) = ((xx24) 54 )* 8 = ((xx76) 27 )*8 76 power any number is 76  Which gives last digits as => 76 * 8 = 608  So last two digits are : 08

18.  Highest power of a number that divides the factorial of another number.  What is the highest power of 5 that divides 60!(factorial)  Note: N! = N*(N-1)*(N-2)*(N-3)….(2)*(1)  Now, Continuously divide 60 with 5 as shown 60/5 = 12, 12/5 = 2 (omit remainders) 2/5 = 0 <- stop at 0  Now add up all the quotients => 12+2+0 = 14  So highest power of 5 that divides 60! is 14.

19.  Ex: Find Highest power of 15 that divides 100!  Here, as 15 is not a prime number we first split 15 into prime factors. 15 = 5 * 3  Now, find out highest power of 5 that divides 100! and also highest power of 3 that divides 100!.  For 5 : 100/5 =20 20/5 = 4 4/5 = 0 So, 20 + 4 + 0 = 24  For 3 : 100/3 = 33 33/3 = 11 11/3 = 3 3/3 = 1 1/3 = 0 So, 33 + 11+ 3 + 1 + 0 = 48  Now, the smallest number of these is taken which will be 24.

20.  Ex: Find the number of zeroes in 75!  This means highest power of 10 which can divide 75! 10 = 5*2  If we consider highest power of 5 which can divide 75!, it’s sufficient. 75/5 =15 15/5 =3 3/5 =0 So, 15+3+0 = 18  So, there are 18 zeroes in 75!

21.  If the number N can be expressed as a product of prime factors such that N = (p a )*(q b )*(r c ) where, p,q,r = prime factors a,b,c = powers to which each is raised  Then, No. of factors of N (including 1, N) = (a+1)*(b+1)*(c+1)*….

22.  Even number => Divisible by 2  Odd Number => Not Divisible by 2  Important Results : e x e = e e x o = e o x o = o

23. Download the related exercise here Exercise 1 - Number Systems

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