prime numbers between 50 and 100

by seeing the prime number table we know that 25 prime numbers are existing in between 1 to 100
the following table explains

	Prime numbers
1-25	2,3,5,7,11,13,17,19,23	9
26-50	29,31,37,41,43,47	6
51-75	53,59,71,73,61,67	6
75-100	79,83,89,97	4

in the above table shows that from 1 to 25 numbers there are 9 prime numbers

similarly from 25 to 50 there are 6 prime numbers are there.like wise from 51-75 there are 6 prime numbers and from 75 to 100 there are 4 prime numbers there.

since the prime numbers between 50 and 100 is 53,59,71,73,61,67,79,83,89,97.

so there are 10 prime numbers are there between 50 and 100

Comments

average of first 100 odd numbers

Odd numbers are 1,3,5,7,9,…. Average = sum of elements/ no of elments Now we will see the sum of elements: Sum of first 2 odd numbers: 1+3= 4= 2 2 Sum of first 3 odd numbers: 1+3+5= 9= 3 2 Sum of first 4 odd numbers: 1+3+5+7=16= 4 2 Sum of first 5 odd numbers: 1+3+5+7+9 = 25 = 5 2 Sum of first n odd numbers: 1+3+5+7+9+. . . . . n numbers= n 2 Now we will see average: Average of first 2 odd numbers= (1+3)/2 = 4/2 =2 Average of first 3 odd numbers= (1+3+5)/3 = 9/3 =3 Average of first 4 odd numbers= (1+3+5+7)/4 = 16/4 =4 Average of first 5 odd numbers= (1+3+5+7+9)/5

LOGIC BEHIND MAGIC SQUARE

Assume a magic square of 3x3 A B C D E F G H I The above figure shows a 3x3 magic square. In the magic square sum of colum, row, diagnol numbers should be same A+B+C=D+E+F=G+H+I=A+D+G=B+E+H=C+F+I=A+E+I=C+E+G=sum A+B+C D+E+F G+H+I A+D+G B+E+H C+F+I A+E+I C+E+G By seeing the above there is 8 type of sum is available. Row wise-3, column wise-3, diagnol wise-2. Condition#1: There are 8 different combinations are there with same sum in a magic square. In the above 8 combinations A=2 times, B= 3 times, C= 2 times, D=3 times E=4 times, F=3 times, G=2 times, H=3 times, I= 2 times Hence the above forms the condition#2. In simply condition#2, middle number E appears 4 times in 8 combinations. Hence row, column, diagnol sum should be equal. In 3x3 magic square we are taking numbers from 1 to 9. From 9 numbers we need to take 3 numbers. From combination f

average of first n natural numbers

The natural numbers are 1,2,3,4,5,. . . . . . now we will see the sum of natural numbers sum of first 3 natural numbers= 1+2+3=6 the above can be written as sum of first 3 natural numbers= 1+2+3=6=3(3+1)/2 sum of first 4 natural numbers= 1+2+3+4=4(4+1)/2 then sum of n natural numbers=1+2+3+......n terms=n(n+1)/2 then the average formula = sum of elements/no of elements the average of n natural numbers= sum of elements/ no of elements =(n(n+1)/2)/n =n(n+1)/2xn =(n+1)/2 THE AVERAGE OF FIRST N NATURAL NUMBERS=(n+1)/2 The average of first 5 natural numbers sum of 5 natural numbers=1+2+3+4+5=15

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