Tables : Tables How many…….. Till 10…..??
till 20……
till 30……
But 30 X10……no its for kids….
so what we must always remember …
30X30
Squares : Squares 1-10 1=1
2=4
3=9
4=16
5=25
6=36
7=49
8=64
9=81
10=100 11-20. 11=121
12=144
13=169
14=196
15=225
16=256
17=289
18=324
19=361
20=400.
Factorisation : Factorisation If a no can be represented in the powers of its prime factors as
ap *bq*cr where a b and c are the prime numbers and p q and r are their powers, then the total no of divisors of the number is given by
(p+1)(q+1)(r+1)
Eg 45=3X3X5=32X5
No of divisors=(2+1)(1+1)
Which are 1,3,5,9,15,45
And sum of divisors is given by
(ap+1-1).(bq+1-1).(cr+1-1)
a-1 b-1 c-1
Applying this formula to the no 45 we get the sum of the factors as 78 which is equal to
Sum of 1,3,5,9,15,45
Slide 4 : Question 1 If both 112 and 33 are factors of the number a * 43 * 62 * 1311, then what is the smallest possible value of 'a'?
Question 2 Find the greatest number of five digits, which is exactly divisible by 7, 10, 15, 21 and 28.
Slide 5 : Question 3 Anita had to do a multiplication. Instead of taking 35 as one of the multipliers, she took 53. As a result, the product went up by 540. What is the new product?
Question 4 Let x, y and z be distinct integers. x and y are odd and positive, and z is even and positive. Which one of the following statements cannot be true?
1) (x-z)2 y is even(2) (x-z)y2 is odd(3) (x-z)y is odd(4) (x-y)2z is even
Question 5 When a number is divided by 36, it leaves a remainder of 19. What will be the remainder when the number is divided by 12?
Slide 6 : Question 6 The sum of the first 100 numbers, 1 to 100 is divisible by
1) 2, 4 and 8(2) 2 and 4(3) 2 only(4) None of these
Question 7 How many different factors are there for the number 48, excluding 1 and 48?
Question 8 1025 - 7 is divisible by
Slide 7 : Question 9 Find the G.C.D of 12x2y3z2, 18x3y2z4, and 24xy4z3
Question 10 What is the value of M and N respectively? If M39048458N is divisible by 8 & 11; Where M & N are single digit integers?
Slide 8 : Ans 4. x and y are odd and positive and z is even and positive(x - z)2 y is even cannot truex - z is odd and y is oddTherefore, (x - z)2 will be odd and (x - z)2 y will be odd
Ans 8. 1025 - 7 = (1025 - 1) - 6The number 1025 - 1 = 99.....9 (25 digits) is divisible by 3 and 4.Therefore, (1025 - 1) - 6 = (24 nines and unit digit is 3) 99.......93.This number is only divisible by 3 (from the given choices).
Question : Question I go to ISBT to catch a bus to Jaipur, there I come to know that there is a bus leaving at 10 am for jaipur but there is a bus leaving at 10 am for chandigarh, ambala and ajmer also, further I come to know that bus to jaipur leaves in every 10 min, to chd every 15 min to ambala every 20 min and to ajmer every 30 min. When will all the buses to all these cities leave together again.
Slide 10 : A , b and c start running on a circular track of 200 meters with a speed of 25m/s ,20 m/s and 10 m/s respectively from the same point . After how much time they will meet again at the starting point.?
HCF &LCM : HCF &LCM The greatest number which divides all the given numbers is called the Highest common Factor or Least common multiple.
The smallest number which is divisible by the given set of numbers is called Least common multiple
Properties : Properties HCF X LCM=Product of the numbers
For any given set of numbers HCF is necessarily a factor of LCM.
LCM of a set of numbers would be greater than or equal to the largest number in that set.
HCF of a given set of numbers would be smaller than or equal to the smallest number in that set
Methods : Methods Factorisation.
To find HCF of 136,144, 168
136=2x2x2x17=23X17
144=2x2x2x2x3x3=24X32
168=2X2X2X3X7=23x3X7
If we look at these numbers the greatest power of the common factor it is 23
Division method
Suppose 2 numbers are given, divide the greater number by lesser one ;divide the lesser by the remainder ; divide the first remainder by the new remainder and so on till there is no remainder .The last divisor required is HCF
Methods( LCM) : Methods( LCM) Factorisation:
Resolve each number into prime factors , then their LCM is the Product of highest power of all factors that occur in the number or else…
Use short cut…