Drill 1 :Become Comfortable with calculations:
You should have a good command on tables and Squares (up to 30).
Now we will check how to do approximate calculations
For example Dividing 283 by 1983 and finding that value as a percentage.
283/1983: whenever denominator is greater than numerator try to approximate denominator. Hence you could take 1983 as 2000.
So now our question looks like 283/2000 which is noting but 0.1415 or 14.15%. Here we change our denominator by 17 which is approximately 1% 0f 1983(19.83).
So in our answer also approximately 1% error is there. 1% of 14.15% is nothing but 0.14%. So our final answer has to be 14.15% + 0.14% = 14.29%.
Now we will check what to do if numerator is greater than denominator, for example Dividing 1132 by 1069.
1132/1069: Here we can write 1132 as 1069 + 63 which will leads to 1 + 63/1069.
So now if we do approximation, it is on very small part. 63/1069≈0.06 and our final answer is 1.06 or 106%.
I hope now the concept of approximation is clear. Just try to solve few questions:
412/6950 a) 5.9% b) 4.7% c) 5.2% d) 6.2% e) 5.72%
511/1680 a) 31.5% b) 29.3% c) 30.1% d) 32.2% e) 30.4%
6.4/6.12 a) 1.045 b) 1.053 c) 1.035 d) 1.023 e) 1.031.
Sunday, October 11, 2009
Basic DI tool
Here are some really basic tools to help you make calculations faster. I am not including the usual "vedic math" or "speed calc" techniques. I found that simpler methods help more!
I am sure you are all aware of that ubiquitous formula (a+b+ab/100). But I am equally certain that you (or even me, for that matter) have discovered even half of the uses of this formula. Well, for now, lets focus on its use in division.
As a really simple starter, lets take a division question:
If 1/3 = 33.33%, what is 1/2? Obviously, its 50%, silly! So, what does it have to do with that formula? Here's what:
Apart from its more popular uses like successive discounts etc, this formula holds true for ANY product of two or more quantities. In fact, this is the most generalized use of the formula. If P=AxB, then an change of a% in A and b% in B will result in a change of (a+b+ab/100)% in P. Conversely, if P has to remain constant, an a% increase/ decrease in A will necessitate a corresponding change of a/(100+a)% or a/(100-a)% in B respectively.
Now,
1/3=33.33% can also be written as 1=3 x 33.33% and
1/2=50% can also be written as 1=2 x 50%.
So, the product has remained constant at 1. 3 has decreased to 2. That's a decrease of 33.33%. According to the formula, the other factor will increase by 33.33/(100-33.33)= 50%. So, 33.33% will increase by 50% to 50%.
This may sound too obvious and easy, but take a look at what you can do with this. How would you calculate, say, 113/256? Now, we know that 113/226 = 1/2=0.50. Using this fact, we can calculate the deviation from 0.5. Since 256 is 30/226 or roughly 1/7.5 or 2/15 or 13.33% more than 226, it means that the final result will be 13.33/(100+13.33)% less than 0.5. That is, approx. 12% less than 0.5, or 0.44. That is fairly close to the correct answer that is 0.4414!
Even if you had taken 13% (instead of dividing it by 113.33), you would have been quite close at 0.45! The key is to round off the denominator to a "convenient" number. And if your denominator deviation is not more than 15%, you can simply take this same deviation as correction in your answer. To summarize, follow these steps:
1. Round off the denominator to either a multiple/ factor of the numerator, or to the nearest hundred. This way, the calculation will be easier for the new fraction.
2. Calculate the percentage deviation from the actual denominator. You may use approximations depending on the need of the hour.
3. Provided that the percentage deviation is less than 15%, adjust your answer by the same percentage in the "other direction".
Obviously, this involves a fairly high amount of calculation, but the good news is that it will get easier with a better knowledge about percentages.
Why don't you go ahead and try out the following:
135/283=? (Hint: Take 283 as 270 or 300)
178/1327=? (Hint: Take 1327 as 1300)
I am sure you are all aware of that ubiquitous formula (a+b+ab/100). But I am equally certain that you (or even me, for that matter) have discovered even half of the uses of this formula. Well, for now, lets focus on its use in division.
As a really simple starter, lets take a division question:
If 1/3 = 33.33%, what is 1/2? Obviously, its 50%, silly! So, what does it have to do with that formula? Here's what:
Apart from its more popular uses like successive discounts etc, this formula holds true for ANY product of two or more quantities. In fact, this is the most generalized use of the formula. If P=AxB, then an change of a% in A and b% in B will result in a change of (a+b+ab/100)% in P. Conversely, if P has to remain constant, an a% increase/ decrease in A will necessitate a corresponding change of a/(100+a)% or a/(100-a)% in B respectively.
Now,
1/3=33.33% can also be written as 1=3 x 33.33% and
1/2=50% can also be written as 1=2 x 50%.
So, the product has remained constant at 1. 3 has decreased to 2. That's a decrease of 33.33%. According to the formula, the other factor will increase by 33.33/(100-33.33)= 50%. So, 33.33% will increase by 50% to 50%.
This may sound too obvious and easy, but take a look at what you can do with this. How would you calculate, say, 113/256? Now, we know that 113/226 = 1/2=0.50. Using this fact, we can calculate the deviation from 0.5. Since 256 is 30/226 or roughly 1/7.5 or 2/15 or 13.33% more than 226, it means that the final result will be 13.33/(100+13.33)% less than 0.5. That is, approx. 12% less than 0.5, or 0.44. That is fairly close to the correct answer that is 0.4414!
Even if you had taken 13% (instead of dividing it by 113.33), you would have been quite close at 0.45! The key is to round off the denominator to a "convenient" number. And if your denominator deviation is not more than 15%, you can simply take this same deviation as correction in your answer. To summarize, follow these steps:
1. Round off the denominator to either a multiple/ factor of the numerator, or to the nearest hundred. This way, the calculation will be easier for the new fraction.
2. Calculate the percentage deviation from the actual denominator. You may use approximations depending on the need of the hour.
3. Provided that the percentage deviation is less than 15%, adjust your answer by the same percentage in the "other direction".
Obviously, this involves a fairly high amount of calculation, but the good news is that it will get easier with a better knowledge about percentages.
Why don't you go ahead and try out the following:
135/283=? (Hint: Take 283 as 270 or 300)
178/1327=? (Hint: Take 1327 as 1300)
Perfect Number
In mathematics, a perfect number is defined as a positive integer which is the sum of its proper positive divisors, that is, the sum of the positive divisors not including the number itself. Equivalently, a perfect number is a number that is half the sum of all of its positive divisors, or σ(n) = 2 n.
Example:
The factors of 6 are 1, 2, 3 and 6.1 + 2 + 3 = 6 = 2^ (2-1)*(2^2 – 1)
The factors of 28 are 1, 2, 4, 7, 14 and 28.1 + 2 + 4 + 7 + 14 = 28.
28 = 2^ (3-1)*(2^3 – 1)
The factors of 496 are 1, 2, 4, 8, 16, 31, 62, 124, 248 and 496.1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 = 496.
496= 2^ (5-1)*(2^5 – 1)
The factors of 8128 are 1, 2, 4, 8, 16, 32, 64, 127, 254, 508, 1016, 2032, 4064 and 8128. I'll let you add them up.
8128= 2^ (7-1)*(2^7 – 1)
Formula to calculate Perfect Number:
Perfect Number = 2^ (n-1) * (2^n - 1)
Where n and 2^n – 1 has to be prime numbers.
Note: It is not necessary that when n is prime 2^n – 1 is always prime.
Put n = 11, 2^11 – 1 = 2047= 23*89 .
Example:
The factors of 6 are 1, 2, 3 and 6.1 + 2 + 3 = 6 = 2^ (2-1)*(2^2 – 1)
The factors of 28 are 1, 2, 4, 7, 14 and 28.1 + 2 + 4 + 7 + 14 = 28.
28 = 2^ (3-1)*(2^3 – 1)
The factors of 496 are 1, 2, 4, 8, 16, 31, 62, 124, 248 and 496.1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 = 496.
496= 2^ (5-1)*(2^5 – 1)
The factors of 8128 are 1, 2, 4, 8, 16, 32, 64, 127, 254, 508, 1016, 2032, 4064 and 8128. I'll let you add them up.
8128= 2^ (7-1)*(2^7 – 1)
Formula to calculate Perfect Number:
Perfect Number = 2^ (n-1) * (2^n - 1)
Where n and 2^n – 1 has to be prime numbers.
Note: It is not necessary that when n is prime 2^n – 1 is always prime.
Put n = 11, 2^11 – 1 = 2047= 23*89 .
How to calculate Last two digit
Dear student, Agenda is how to calculate last two digits of the number ending on 1.
Last digit or unit digit will be always 1.
Just observe tenth place digit in following example
11^2 = 121 (1 * 2 = 2)
11^3= 1331 (1 * 3 = 3)
21^2 = 441 (2 * 2 = 4)
21^3= 9261 (2 * 3 = 6)
Just apply this technique to the following examples
Example: 31^8
If number is ending on 1 last digit is always 1.Important is how to calculate Tenth place digit. It is also very easy. Tenth place digit will be (3 * 8 =24) so tenth place digit is 4
So last two digits of 31^8 is 41.
Find the last two digits of 41^2789
In no time at all you can calculate the answer to be 61 (4 × 9 = 36. Therefore, 6 will be the tens digit and one will be the units digit)
Find the last two digits of 71^56747
Last two digits will be 91 (7 × 7 gives 9 and 1 as units digit)
Now try to get the answer to this question within 10 s:
Find the last two digits of 51^456 × 61^567
The last two digits of 51^456 will be 01 and the last two digits of 61^567 will be 21. Therefore, the last two digits of 514^56 × 61^567 will be the last two digits of 01 × 21 = 21
Last two digits of numbers ending in 3, 7 or 9
Find the last two digits of 19266.
19^266 = (19^2)133. Now, 19^2 ends in 61 (19^2 = 361) therefore, we need to find the last two digits of (61)133.
Once the number is ending in 1 we can straight away get the last two digits with the help of the previous method. The last two digits are 81 (6 × 3 = 18, so the tens digit will be 8 and last digit will be 1)
Find the last two digits of 33^288.
33^288 = (33^4)72. Now 33^4 ends in 21 (33^4 = 33^2 × 33^2 = 1089 × 1089 = xxxxx21) therefore, we need to find the last two digits of 21^72. By the previous method, the last two digits of 21^72 = 41 (tens digit = 2 × 2 = 4, unit digit = 1)
So here’s the rule for finding the last two digits of numbers ending in 3, 7 and 9:
Convert the number till the number gives 1 as the last digit and then find the last two digits according to the previous method.
Now try the method with a number ending in 7:
Find the last two digits of 87^474.
87^474 = 87^472 × 87^2 = (87^4)^118 × 87^2 = (69 × 69)^118 × 69 (The last two digits of 87^2 are 69) = 61^118 × 69 = 81 × 69 = 89
If you understood the method then try your hands on these questions:
Find the last two digits of:
1. 27^496
2. 79^93
3. 583^512
Last digit or unit digit will be always 1.
Just observe tenth place digit in following example
11^2 = 121 (1 * 2 = 2)
11^3= 1331 (1 * 3 = 3)
21^2 = 441 (2 * 2 = 4)
21^3= 9261 (2 * 3 = 6)
Just apply this technique to the following examples
Example: 31^8
If number is ending on 1 last digit is always 1.Important is how to calculate Tenth place digit. It is also very easy. Tenth place digit will be (3 * 8 =24) so tenth place digit is 4
So last two digits of 31^8 is 41.
Find the last two digits of 41^2789
In no time at all you can calculate the answer to be 61 (4 × 9 = 36. Therefore, 6 will be the tens digit and one will be the units digit)
Find the last two digits of 71^56747
Last two digits will be 91 (7 × 7 gives 9 and 1 as units digit)
Now try to get the answer to this question within 10 s:
Find the last two digits of 51^456 × 61^567
The last two digits of 51^456 will be 01 and the last two digits of 61^567 will be 21. Therefore, the last two digits of 514^56 × 61^567 will be the last two digits of 01 × 21 = 21
Last two digits of numbers ending in 3, 7 or 9
Find the last two digits of 19266.
19^266 = (19^2)133. Now, 19^2 ends in 61 (19^2 = 361) therefore, we need to find the last two digits of (61)133.
Once the number is ending in 1 we can straight away get the last two digits with the help of the previous method. The last two digits are 81 (6 × 3 = 18, so the tens digit will be 8 and last digit will be 1)
Find the last two digits of 33^288.
33^288 = (33^4)72. Now 33^4 ends in 21 (33^4 = 33^2 × 33^2 = 1089 × 1089 = xxxxx21) therefore, we need to find the last two digits of 21^72. By the previous method, the last two digits of 21^72 = 41 (tens digit = 2 × 2 = 4, unit digit = 1)
So here’s the rule for finding the last two digits of numbers ending in 3, 7 and 9:
Convert the number till the number gives 1 as the last digit and then find the last two digits according to the previous method.
Now try the method with a number ending in 7:
Find the last two digits of 87^474.
87^474 = 87^472 × 87^2 = (87^4)^118 × 87^2 = (69 × 69)^118 × 69 (The last two digits of 87^2 are 69) = 61^118 × 69 = 81 × 69 = 89
If you understood the method then try your hands on these questions:
Find the last two digits of:
1. 27^496
2. 79^93
3. 583^512
Interesting Maths Fact
1. If you join together the feet of the altitudes of a triangle, they make another triangle called the pedal triangle, and the orthocentre is the incentre of this pedal triangle.
2. The orthocentre, circumcentre and centroid are collinear (Euler’s line) and the centroid is two thirds the distance from the orthocenter to the circumcenter. The incenter is not on the Euler line except when the triangle is isosceles. When the triangle is isosceles, the Euler Line also passes through one of the vertices!
3. Take a triangle with vertices at A, B and C, and let H be its orthocenter. The orthocenter for any of the triangles formed from three of these four points is the fourth point. (4 points which satisfy this condition are called an orthocentric set.)
4. The distance from the orthocenter to a vertex is twice the distance from the circumcenter to the opposite side of the vertex.
One great math site that I love to visit frequently is www.mathforum.org and it is rich with all the knowledge that you will ever need!
2. The orthocentre, circumcentre and centroid are collinear (Euler’s line) and the centroid is two thirds the distance from the orthocenter to the circumcenter. The incenter is not on the Euler line except when the triangle is isosceles. When the triangle is isosceles, the Euler Line also passes through one of the vertices!
3. Take a triangle with vertices at A, B and C, and let H be its orthocenter. The orthocenter for any of the triangles formed from three of these four points is the fourth point. (4 points which satisfy this condition are called an orthocentric set.)
4. The distance from the orthocenter to a vertex is twice the distance from the circumcenter to the opposite side of the vertex.
One great math site that I love to visit frequently is www.mathforum.org and it is rich with all the knowledge that you will ever need!
Points to remember
Quantitative Ability – POINTS TO REMEMBER
1.If an equation (i.e. f(x) = 0) contains all positive co-efficients of any powers of x, it has no positive roots.
Eg: x^3+3x^2+2x+6=0 has no positive roots
2.For an equation, if all the even powers of x have same sign coefficients and all the odd powers of x have the opposite sign coefficients, then it has no negative roots.
3.For an equation f(x)=0 , the maximum number of positive roots it can have is the number of sign changes in f(x) ; and the maximum number of negative roots it can have is the number of sign changes in f(-x)
4.Complex roots occur in pairs, hence if one of the roots of an equation is 2+3i, another has to be
2-3i and if there are three possible roots of the equation, we can conclude that the last root is real. This real root could be found out by finding the sum of the roots of the equation and subtracting (2+3i)+(2-3i)=4 from that sum.
5.For a cubic equation ax^3+bx^2+cx+d=o
Sum of the roots = - b/a
Sum of the product of the roots taken two at a time = c/a
Product of the roots = -d/a
6.For a bi-quadratic equation ax^4+bx^3+cx^2+dx+e = 0
Sum of the roots = - b/a
Sum of the product of the roots taken two at a time = c/a
Sum of the product of the roots taken three at a time = -d/a
Product of the roots = e/a
7.If an equation f(x)= 0 has only odd powers of x and all these have the same sign coefficients or if f(x) = 0 has only odd powers of x and all these have the same sign coefficients, then the equation has no real roots in each case (except for x=0 in the second case)
8.Consider the two equations
a1x+b1y=c1
a2x+b2y=c2
Then,
If a1/a2 = b1/b2 = c1/c2, then we have infinite solutions for these equations.
If a1/a2 = b1/b2 <> c1/c2, then we have no solution.
If a1/a2 <> b1/b2, then we have a unique solution.
9.a + b = a + b if a*b>=0
else, a + b >= a + b
10.The equation ax^2+bx+c=0 will have max. value when a<0>0. The max. or min. value is given by (4ac-b^2)/4a and will occur at x = -b/2a
11.If for two numbers x + y=k (a constant), then their PRODUCT is MAXIMUM if x=y (=k/2). The maximum product is then (k2)/4.
12. If for two numbers x*y=k (a constant), then their SUM is MINIMUM if x=y (=root(k)). The minimum sum is then 2*root (k).
13.Product of any two numbers = Product of their HCF and LCM. Hence product of two numbers = LCM of the numbers if they are prime to each other.
14. For any 2 numbers a, b where a>b
a>AM>GM>HM>b (where AM, GM ,HM stand for arithmetic, geometric , harmonic means respectively)
(GM)^2 = AM * HM
15.For three positive numbers a, b, c
(a + b + c) * (1/a + 1/b + 1/c)>=9
16.For any positive integer n
2<= (1 + 1/n)^n <=3
17. a^2 + b^2 + c^2 >= ab + bc + ca
If a=b=c, then the case of equality holds good.
18.a^4 + b^4 + c^4 + d^4 >= 4abcd (Equality arises when a=b=c=d=1)
19.(n!)^2 > n^n
20.If a + b + c + d=constant, then the product a^p * b^q * c^r * d^s will be maximum if a/p = b/q = c/r = d/s
21.x^n -a^n = (x-a)(x^(n-1) + x^(n-2) + .......+ a^(n-1) ) ......Very useful for finding multiples. For example (17-14=3 will be a multiple of 17^3 - 14^3)
22.e^x = 1 + (x)/1! + (x^2)/2! + (x^3)/3! + ........to infinity
Note: 2 < e < 3
23.log(1+x) = x - (x^2)/2 + (x^3)/3 - (x^4)/4 .........to infinity [Note the alternating sign . .Also note that the logarithm is with respect to base e]
24.(m + n)! is divisible by m! * n!
25.When a three digit number is reversed and the difference of these two numbers is taken, the middle number is always 9 and the sum of the other two numbers is always 9.
26.Any function of the type y=f(x)=(ax-b)/(bx-a) is always of the form x=f(y)
The sum of first n natural numbers = n(n+1)/2
The sum of squares of first n natural numbers is n(n+1)(2n+1)/6
The sum of cubes of first n natural numbers is (n(n+1)/2)2/4
The sum of first n even numbers= n (n+1)
The sum of first n odd numbers= n^2
1.If an equation (i.e. f(x) = 0) contains all positive co-efficients of any powers of x, it has no positive roots.
Eg: x^3+3x^2+2x+6=0 has no positive roots
2.For an equation, if all the even powers of x have same sign coefficients and all the odd powers of x have the opposite sign coefficients, then it has no negative roots.
3.For an equation f(x)=0 , the maximum number of positive roots it can have is the number of sign changes in f(x) ; and the maximum number of negative roots it can have is the number of sign changes in f(-x)
4.Complex roots occur in pairs, hence if one of the roots of an equation is 2+3i, another has to be
2-3i and if there are three possible roots of the equation, we can conclude that the last root is real. This real root could be found out by finding the sum of the roots of the equation and subtracting (2+3i)+(2-3i)=4 from that sum.
5.For a cubic equation ax^3+bx^2+cx+d=o
Sum of the roots = - b/a
Sum of the product of the roots taken two at a time = c/a
Product of the roots = -d/a
6.For a bi-quadratic equation ax^4+bx^3+cx^2+dx+e = 0
Sum of the roots = - b/a
Sum of the product of the roots taken two at a time = c/a
Sum of the product of the roots taken three at a time = -d/a
Product of the roots = e/a
7.If an equation f(x)= 0 has only odd powers of x and all these have the same sign coefficients or if f(x) = 0 has only odd powers of x and all these have the same sign coefficients, then the equation has no real roots in each case (except for x=0 in the second case)
8.Consider the two equations
a1x+b1y=c1
a2x+b2y=c2
Then,
If a1/a2 = b1/b2 = c1/c2, then we have infinite solutions for these equations.
If a1/a2 = b1/b2 <> c1/c2, then we have no solution.
If a1/a2 <> b1/b2, then we have a unique solution.
9.a + b = a + b if a*b>=0
else, a + b >= a + b
10.The equation ax^2+bx+c=0 will have max. value when a<0>0. The max. or min. value is given by (4ac-b^2)/4a and will occur at x = -b/2a
11.If for two numbers x + y=k (a constant), then their PRODUCT is MAXIMUM if x=y (=k/2). The maximum product is then (k2)/4.
12. If for two numbers x*y=k (a constant), then their SUM is MINIMUM if x=y (=root(k)). The minimum sum is then 2*root (k).
13.Product of any two numbers = Product of their HCF and LCM. Hence product of two numbers = LCM of the numbers if they are prime to each other.
14. For any 2 numbers a, b where a>b
a>AM>GM>HM>b (where AM, GM ,HM stand for arithmetic, geometric , harmonic means respectively)
(GM)^2 = AM * HM
15.For three positive numbers a, b, c
(a + b + c) * (1/a + 1/b + 1/c)>=9
16.For any positive integer n
2<= (1 + 1/n)^n <=3
17. a^2 + b^2 + c^2 >= ab + bc + ca
If a=b=c, then the case of equality holds good.
18.a^4 + b^4 + c^4 + d^4 >= 4abcd (Equality arises when a=b=c=d=1)
19.(n!)^2 > n^n
20.If a + b + c + d=constant, then the product a^p * b^q * c^r * d^s will be maximum if a/p = b/q = c/r = d/s
21.x^n -a^n = (x-a)(x^(n-1) + x^(n-2) + .......+ a^(n-1) ) ......Very useful for finding multiples. For example (17-14=3 will be a multiple of 17^3 - 14^3)
22.e^x = 1 + (x)/1! + (x^2)/2! + (x^3)/3! + ........to infinity
Note: 2 < e < 3
23.log(1+x) = x - (x^2)/2 + (x^3)/3 - (x^4)/4 .........to infinity [Note the alternating sign . .Also note that the logarithm is with respect to base e]
24.(m + n)! is divisible by m! * n!
25.When a three digit number is reversed and the difference of these two numbers is taken, the middle number is always 9 and the sum of the other two numbers is always 9.
26.Any function of the type y=f(x)=(ax-b)/(bx-a) is always of the form x=f(y)
The sum of first n natural numbers = n(n+1)/2
The sum of squares of first n natural numbers is n(n+1)(2n+1)/6
The sum of cubes of first n natural numbers is (n(n+1)/2)2/4
The sum of first n even numbers= n (n+1)
The sum of first n odd numbers= n^2
Prime Nos.
A prime numbers is a number greater than one whose factors are only one and itself. In other words, 4 is not prime, because its factors are 1, 2 and 4 (1x4, 2x2). But 5 is prime, because the only way you can get a product of 5 is by multiplying 1 and 5 (1x5). Composite numbers are those who has more than two factors. The number 1 is not prime OR composite because it has only one factor.
Here are all the prime numbers up to 500:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, ...
How to find big Prime Numbers:
Following is a somewhat celebrated result of Euler:
The function f(x) = x^2 + x + 41 is always prime for x = 0, 1 ,..., 39.
This quadratic was the record holder for centuries as a consecutive,
Distinct quadratic prime-producer for an initial range of input
Values. It is not, however, the new function is there that can give more range of prime numbers than Euler’s:
The Function:
f(x) = 36x^2 - 810x + 2753
Which is always prime for x = 0, 1, ... , 44.
Here are all the prime numbers up to 500:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, ...
How to find big Prime Numbers:
Following is a somewhat celebrated result of Euler:
The function f(x) = x^2 + x + 41 is always prime for x = 0, 1 ,..., 39.
This quadratic was the record holder for centuries as a consecutive,
Distinct quadratic prime-producer for an initial range of input
Values. It is not, however, the new function is there that can give more range of prime numbers than Euler’s:
The Function:
f(x) = 36x^2 - 810x + 2753
Which is always prime for x = 0, 1, ... , 44.
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