4. PRIME NUMBERS
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 ?
179 13.3
Prime Numbers less than 13.3 are
2,3,5,7,11,13
179 is not divisible by any of them, 179 is
prime.
6. SOME MORE DIVISIBILITY RULES
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.
7. SOME MORE DIVISIBILITY RULES
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.
8. REMEMBER IT!
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.
Divisibility by Test
7 Subtract 2 x Last digit
11 Subtract 1 x Last digit
13 Add 4 x Last Digit
17 Subtract 5 x Last Digit
19 Add 2 x Last Digit
9. UNIT’S DIGIT OF A NUMBER
Find the unit’s digit of 71999 (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)
73 = 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 74 = 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.
453
10. LAST TWO DIGITS OF A NUMBER
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
11. LAST TWO DIGITS OF A NUMBER
a) Number ending with 1 :
Ex : Find the last 2 digits of 31786
Now, multiply the 10s digit of the number with
the last digit of exponent
31786 = 3 * 6 = 18 -> 8 is the 10s digit.
Units digit is obviously 1
So, last 2 digits are => 81
12. LAST TWO DIGITS OF A NUMBER
b) Number Ending with 3, 7 & 9
Ex: Find last 2 digits of 19266
We need to get this in such as way that the
base has last digit as 1
19266 = (192)133 = 361133
Now, follow the previous method => 6 * 3 =
18
So, last two digits are => 81
13. LAST TWO DIGITS OF A NUMBER
b) Number Ending with 3, 7 & 9
Remember :
34 = 81
74 = 2401
92 = 81
14. LAST TWO DIGITS OF A NUMBER
Ex 2: Find last two digits of 33288
Now, 33288 = (334)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
(872)*(874)118 => (xx69) * (xx61)118 (6 x 8 = 48)
=> (xx69)*(81)
So, last two digits are 89
15. LAST TWO DIGITS OF A NUMBER
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,
242 = xx76
210 = xx24
24even = xx76
24odd = xx24
16. LAST TWO DIGITS OF A NUMBER
Ex: Find the last two digits of 2543
2543 = ((210)54) * (23)
= ((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
17. HIGHEST POWER
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.
18. HIGHEST POWER
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.
19. NUMBER OF ZEROES
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!
20. NUMBER OF FACTORS OF A NUMBER
If the number N can be expressed as a product of prime
factors such that
N = (pa)*(qb)*(rc)
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)*….
21. EVEN AND ODD
Even number => Divisible by 2
Odd Number => Not Divisible by 2
Important Results :
exe=e
exo=e
oxo=o
22. EXERCISE
Download the related exercise here
Exercise 1 - Number Systems
23. LET ME KNOW!!!
If you liked this presentation, do comment
on
http://nov15.wordpress.com
or
write to Nicky at
nickyswetha20@yahoo.com