Hi people,

Wanted to share some shortcuts that would be helpful in quant as well as DI sections. The shortcuts have been compiled from various threads of PG itself.

Dunno whether this is the right thread for this purpose but well, let me post.

If any parallelogram can be inscribed in a circle , it must be a rectangle.

If a quadrilateral circumscribes a circle , the sum of a pair of opposite sides is equal to the sum of the other pair .

The ratio of the radii of the circumcircle and incircle of an equilateral triangle is 2:1 .

for similar cones , ratio of radii = ratio of their bases.

the quadrilateral formed by joining the angular bisectors of another f a quadrilateral is always a rectangle.

Let W be any point inside a rectangle ABCD .

Then

WD^2 + WB^2 = WC^2 + WA^2

Let a be the side of an equilateral triangle . then if three circles be drawn inside

this triangle touching each other then each’s radius = a/(2*(root(3)+1))

Area of a regular hexagon : root(3)*3/2*(side)*(side)

the area of a regular n-sided polygon of side ‘a’ is given by

area=(n/4)*a^2*cot(180/n)

In any triangle

a=b*CosC + c*CosB

b=c*CosA + a*CosC

c=a*CosB + b*CosA

————————————————————

————————————————————

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 .

————————————————————

e^x = 1 + (x)/1! + (x^2)/2! + (x^3)/3! + ……..to infinity

————————————————————

(m+n)! is divisible by m! * n!

————————————————————

for squares of numbers between 25 and 50.

let the number be 25+k

first calculate (25-k)^2 and to it add k*100

2)for squares of numbers between 50 and 75.

again let the number 50+k

calculate k^2 and to it add 2500+100*k

————————————————————

3)for squares of numbers between 75 and 100

let the number be 100-k

calculate k^2 and to it add (100-2*k)*100

these formulas are basically a derivation from the formulas (a+b)^2 and (a-b)^2.

————————————————————

2^2n -1 is always divisible by 3

2^2n -1 = (3-1)^2n -1

= 3M +1 -1

= 3M, thus divisible by 3

———————————-

if a number ‘n’ is represented as

n=a^x * b^y * c^z ….

where, {a,b,c,.. } are prime numbers then

Quote:

(a) the total number of factors is (x+1)(y+1)(z+1) ….

(b) the total number of relatively prime numbers less than the number is n * (1-1/a) * (1-1/b) * (1-1/c)…. ???????????????????

(c) the sum of relatively prime numbers less than the number is n/2 * n * (1-1/a) * (1-1/b) * (1-1/c)….

(d) the sum of factors of the number is {a^(x+1)} * {b^(y+1)} * …../(x*y*…)

—————————————–

what is the highest power of 10 in 203!ANS : express 10 as product of primes; 10 = 2*5

divide 203 with 2 and 5 individually

203/2 = 101

101/2 = 50

50/2 = 25

25/2 = 12

12/2 = 6

6/2 = 3

3/2 = 1

thus power of 2 in 203! is, 101 + 50 + 25 + 12 + 6 + 3 + 1 = 198

divide 203 with 5

203/5 = 40

40/5 = 8

8/5 = 1

thus power of 5 in 203! is, 49

so the power of 10 in 203! factorial is 49

——————————————————

n how many ways, 729 can be expressed as a difference of 2 squares?

ANS: 729 = a^2 – b^2

= (a-b)(a+b),

since 729 = 3^5,

total ways of getting 729 are, 1*729, 3*243, 9*81, 27*27.

So 4 ways

Funda is that, all four ways of expressing can be used to findout distinct a,b values,

for example take 9*81

now since 9*81 = (a-b)(a+b) by solving the system a-b = 9 and a+b = 81 we can have 45,36 as soln.

————————————–

if x%a=y%a = r then

(x-y)%r =0 (% stands for modulo operator)

———————————————

total number of primes between 1 and 100 – 25

———————————————-

Reminder Funda

(a) (a + b + c) % n = (a%n + b%n + c%n) %n

EXAMPLE: The reminders when 3 numbers 1221, 1331, 1441 are divided by certain number 9 are 6, 8, 1 respectively. What would be the reminder when you divide 3993 with

9? ( never seen such question though )

the reminder would be (6 + 8 + 1) % 9 = 6

————————————————–

(a*b*c) % n = (a%n * b%n * c%n) %n

————————————–

when a number is divided by 11,7,4 the reminders are 5,6,3 respectively. what would be the reminders when the same number is divided by 4,7,11 respectively?

ANS : whenever such problem is given,

we need to write the numbers in top row and rems in the bottom row like this

11 7 4

| \ \

5 6 3

( coudnt express here properly )

now the number is of the form, LCM ( 11,7,4 ) + 11*(3*7 + 6) + 5

that is 302 + LCM(11,7,4) and thus the rems when the same number is divided by 4,7,11 respectively are,

302 mod 4 = 2

75 mod 7 = 5

10 mod 11 = 10

————————————————————

Plz tell me the answer to remainder when 2^156 / 13 .

2^156/13 = 2^(6*26)/13 (156 = 26 x 6)

= 64^26/13 = (65-1)^26/13 (since 65 is multiple of 13)

= (-1)^26=1

u just have to bring close to the multiple of the number which divides the numerator..(like 65 in this case)…..

———————————————————–

In how many ways can 2310 be expressed as a product of 3 factors?

ANS: 2310 = 2*3*5*7*11

When a number can be expressed as a product of n distinct primes,

then it can be expressed as a product of 3 numbers in (3^(n-1) + 1)/2 ways

(I guess this is incorrect)

————————————————————-

Multiplication by 7: Going from right to left, we use

this rule: Double each number and add half the

neighbor (digit to the right, dropping any fraction);

add 5 if the number (not the neighbor) is odd. And of

course, we have to deal with carries:

3852 x 7 = 26964

Starting at the right (2), we double the first number

(it has no neighbor) and write down the right-most

digit of that (4) and we have no carry. Then we double

the next number (2×5=10), add five (+5=15), and add

half the neighbor (+1=16), and write down the right

digit (6) of that and carry the 1. Then we double the

next number (2×8=16), and add half the neighbor

(+2=1, and add the carry (+1=19). Then we double the

next number (2×3=6), add five (+5=11), add half the

neighbor (+4=15), and add the carry (+1=16). Now we

double a zero off to the left of our 3852

(Trachtenberg wrote the zero out there: 03852) and add

half the neighbor (0+1=1), and add the carry (+1=2).

And we have our answer.

Notice that the carries are smaller than they were in

normal multiplication by 7. The above rule is not

simple, but once mastered, it is easy to use. It

should be about as fast as multiplying normally (which

requires memorizing the multiplication table).

Multiplication by other small numbers (3 through 12)

uses similar rules.

————————————-

Square 2-digit

Rule:

1. Square the second digit

2. Multiply the two digits and double

3. Square the first digit

example

49

81 L.D. 1 cy=8

4*9*2=72 +8 = 80 cy=8

4*4=16 + 8=24

ans=2401

——————————————-

Multiply just-under numbers

Example:

98

x 99

100 – 98 = 2 picture 98 2

100 – 99 = 1 picture 99 1

97 98 – 1 (upper left – lower right)

02 2 x 1 (multiply two right numbers)

9702

98 x 99 = 9702

99*94

99 1

94 6

(99-6)1*6=9306

————————————————————

Multiply just-over numbers

Example:

104

x106

104 -100 = 4 picture 104 4

106 -100 = 6 picture 106 6

110 104+6 (upper left + lower right)

24 4 x 6 (multiply two right numbers)

11024

104 x 106 = 11024

————————————————————–

Multiply just over and just under numbers

97*104

97 -3

104 4

101 -12

100 88 (subtract 1 from the result of cross addition. take the complement of the right multiplication)

10088

103*96

103 3

96 -4

99 -12

9888

————————————————————

38*32=1216.ie,3*4=12 and 8*2=16.only criteria is that the units digit number should add to 10 and that the ten’s digit number shud b the same

————————————————————

3)multiply 57 by 63

method :

5 7

| \ / |

| / \ |

6 3

__________

35 9 1

There are 3 steps:

a)Multiply vertically on the right: 7 x 3 = 21

This gives the last figure of the answer 1 and 2 as carry.

b) Multiply crosswise and add: (5 x 3 + 6 x 7) + 2(carry) = 59

This gives the middle figure 9 and 5 as carry.

c) Multiply vertically on the left: (5 x 6) + 5(carry) = 35.

This gives the first figure of the answer.

for 3 digits

127*93

12 7

9 3

118 1 1

——————————————————

Square just-under numbers

Example:

96^2

100 – 96 is 4

92 96 – 4 is 92

16 4 squared is 16

9216

962 = 9216

—————————————

Square just-over numbers

Example:

106^2

106 – 100 is 6

112 106 + 6

36 6^2

Ans: 11236

—————————————-

Square of number between 50 and 60

Example:

56^2

31 25 + last digit (25+6)

36 sqare of last digit (62)

3136

562 = 3136

———————–

Find remainder when 2222^5555 + 5555^2222 is divided by 7

2222 mod 7=3

5555 mod 7 = 4

so prob reduces to 3^5555 +4^2222 remainder 7

now this can b written as 243^1111 + 16^1111 remainder 7

now use direct formula tht x^n + y^n is divisible by x+y if n is odd

given expression is divisible by 243+16=259

so it is also divisible by 7

and hence remainder=0

anoter method:

Well fermat’s rule states that

a^(p-1) whenever divided by p where p is a prime no. and a is coprime to p, then the remainder would be 1.

as the divisor is 7 and both 2222 and 5555 are coprime to 7 thus any1 of them raised to the power of 6 or a multiple of it would give remainder as 1

2222^5555 + 5555^2222 is equivqlent to

3^5555 + 4^2222 which is

3^(6k+5) + 4^(6m+2)

which is 3^5 + 4^2 (since 3^6 and 4^6 will give 1 as remainder)

——————————————————–

Some pythagorean triplets:

3,4,5 (3^2=4+5)

5,12,13 (5^2=12+13)

7,24,25 (7^2=24+25)

8,15,17 (8^2 / 2 = 15+17 )

9,40,41 (9^2=40+41)

11,60,61 (11^2=60+61)

12,35,37 (12^2 / 2 = 35+37)

16,63,65 (16^2 /2 = 63+65)

20,21,29(EXCEPTION)

—————————————————-

For a set of fractions ( num<den), if the difference between the numerator and the denominator is the same, then the fraction with the largest numerator is the largest and the one with the smallest numerator is the smallest

e.g. 21/27, 23/29, 19/26

largest: 23/29

smallest: 19/26

———————————————————–

For a set of fractions ( num>den), if the difference between the numerator and the denominator is the same, then the fraction with the smallest numerator is the largest and the one with the largest numerator is the smallest

e.g. 27/21, 29/23, 26/19

smallest: 29/23

largest: 26/19

——————————————————————-

i guess we could find sqrt for numbers whose nearby sqrt we already know.

steps are :-

1. take the difference from the nearest smaller perfect square. for eg. sqrt (39) nearest perfect square is 36(whose sqrt we know as 6). and the difference is 39-36=3.

2. divide the difference with product of 2 and sq root of the number which we know.

ie difference 3/(2* sqrt(36) )

3/(2*6)=.25

3.sum the above with the sqrt which we know

ie .25+6=6.25

also works for subtraction

sqrt(33) = 6-.25 = 5.75 (aprox)

4. there is a great chance of getting the answer right with even the first decimal

is correct in most of the cases.

——————————————————————

Finding square root:

Splitting the difference:

Sqrt(3125) : group the digits taking two at a time from the left (like in the usual way of finding sq roots)

First pair: 31

1)We know 5*5 =25

2)rough estimate sqrt is 50

3)divide 3125by 50 =62.5

4)subtracton: 62.5-50 =12.5

5)divide 12.5 by 2 =6.25

round it off downwards :6

rough estimate 50+6=56

try this one:

sqrt(93560)

9,35,60

roughly :300

div 93560 by 300

311.8667-300=11.87

11.87/2=5.935

approx 6

ans 306

sqrt(3847214

38 47 21 48

approx 6000

38472148/6000

6412.014-6000=412.014

412.114/2=206.05

6206.05…(has considerable error)

In most cases error is just 1%

——————————————————————

————————————–

when a number is divided by 11,7,4 the reminders are 5,6,3 respectively. what would be the reminders when the same number is divided by 4,7,11 respectively?

ANS : whenever such problem is given,

we need to write the numbers in top row and rems in the bottom row like this

11 7 4

| \ \

5 6 3

( coudnt express here properly )

now the number is of the form, LCM ( 11,7,4 ) + 11*(3*7 + 6) + 5

that is 302 + LCM(11,7,4) and thus the rems when the same number is divided by 4,7,11 respectively are,

302 mod 4 = 2

75 mod 7 = 5

10 mod 11 = 10

————————————————————

Plz tell me the answer to remainder when 2^156 / 13 .

2^156/13 = 2^(6*26)/13 (156 = 26 x 6)

= 64^26/13 = (65-1)^26/13 (since 65 is multiple of 13)

= (-1)^26=1

u just have to bring close to the multiple of the number which divides the numerator..(like 65 in this case)…..

———————————————————–

In how many ways can 2310 be expressed as a product of 3 factors?

ANS: 2310 = 2*3*5*7*11

When a number can be expressed as a product of n distinct primes,

then it can be expressed as a product of 3 numbers in (3^(n-1) + 1)/2 ways

(I guess this is incorrect)

————————————————————-

Multiplication by 7: Going from right to left, we use

this rule: Double each number and add half the

neighbor (digit to the right, dropping any fraction);

add 5 if the number (not the neighbor) is odd. And of

course, we have to deal with carries:

3852 x 7 = 26964

Starting at the right (2), we double the first number

(it has no neighbor) and write down the right-most

digit of that (4) and we have no carry. Then we double

the next number (2×5=10), add five (+5=15), and add

half the neighbor (+1=16), and write down the right

digit (6) of that and carry the 1. Then we double the

next number (2×8=16), and add half the neighbor

(+2=1, and add the carry (+1=19). Then we double the

next number (2×3=6), add five (+5=11), add half the

neighbor (+4=15), and add the carry (+1=16). Now we

double a zero off to the left of our 3852

(Trachtenberg wrote the zero out there: 03852) and add

half the neighbor (0+1=1), and add the carry (+1=2).

And we have our answer.

Notice that the carries are smaller than they were in

normal multiplication by 7. The above rule is not

simple, but once mastered, it is easy to use. It

should be about as fast as multiplying normally (which

requires memorizing the multiplication table).

Multiplication by other small numbers (3 through 12)

uses similar rules.

————————————-

Square 2-digit

Rule:

1. Square the second digit

2. Multiply the two digits and double

3. Square the first digit

example

49

81 L.D. 1 cy=8

4*9*2=72 +8 = 80 cy=8

4*4=16 + 8=24

ans=2401

——————————————-

Multiply just-under numbers

Example:

98

x 99

100 – 98 = 2 picture 98 2

100 – 99 = 1 picture 99 1

97 98 – 1 (upper left – lower right)

02 2 x 1 (multiply two right numbers)

9702

98 x 99 = 9702

99*94

99 1

94 6

(99-6)1*6=9306

————————————————————

Multiply just-over numbers

Example:

104

x106

104 -100 = 4 picture 104 4

106 -100 = 6 picture 106 6

110 104+6 (upper left + lower right)

24 4 x 6 (multiply two right numbers)

11024

104 x 106 = 11024

————————————————————–

Multiply just over and just under numbers

97*104

97 -3

104 4

101 -12

100 88 (subtract 1 from the result of cross addition. take the complement of the right multiplication)

10088

103*96

103 3

96 -4

99 -12

9888

————————————————————

38*32=1216.ie,3*4=12 and 8*2=16.only criteria is that the units digit number should add to 10 and that the ten’s digit number shud b the same

————————————————————

3)multiply 57 by 63

method :

5 7

| \ / |

| / \ |

6 3

__________

35 9 1

There are 3 steps:

a)Multiply vertically on the right: 7 x 3 = 21

This gives the last figure of the answer 1 and 2 as carry.

b) Multiply crosswise and add: (5 x 3 + 6 x 7) + 2(carry) = 59

This gives the middle figure 9 and 5 as carry.

c) Multiply vertically on the left: (5 x 6) + 5(carry) = 35.

This gives the first figure of the answer.

for 3 digits

127*93

12 7

9 3

118 1 1

——————————————————

Square just-under numbers

Example:

96^2

100 – 96 is 4

92 96 – 4 is 92

16 4 squared is 16

9216

962 = 9216

—————————————

Square just-over numbers

Example:

106^2

106 – 100 is 6

112 106 + 6

36 6^2

Ans: 11236

—————————————-

Square of number between 50 and 60

Example:

56^2

31 25 + last digit (25+6)

36 sqare of last digit (62)

3136

562 = 3136

———————–

Find remainder when 2222^5555 + 5555^2222 is divided by 7

2222 mod 7=3

5555 mod 7 = 4

so prob reduces to 3^5555 +4^2222 remainder 7

now this can b written as 243^1111 + 16^1111 remainder 7

now use direct formula tht x^n + y^n is divisible by x+y if n is odd

given expression is divisible by 243+16=259

so it is also divisible by 7

and hence remainder=0

anoter method:

Well fermat’s rule states that

a^(p-1) whenever divided by p where p is a prime no. and a is coprime to p, then the remainder would be 1.

as the divisor is 7 and both 2222 and 5555 are coprime to 7 thus any1 of them raised to the power of 6 or a multiple of it would give remainder as 1

2222^5555 + 5555^2222 is equivqlent to

3^5555 + 4^2222 which is

3^(6k+5) + 4^(6m+2)

which is 3^5 + 4^2 (since 3^6 and 4^6 will give 1 as remainder)

——————————————————–

Some pythagorean triplets:

3,4,5 (3^2=4+5)

5,12,13 (5^2=12+13)

7,24,25 (7^2=24+25)

8,15,17 (8^2 / 2 = 15+17 )

9,40,41 (9^2=40+41)

11,60,61 (11^2=60+61)

12,35,37 (12^2 / 2 = 35+37)

16,63,65 (16^2 /2 = 63+65)

20,21,29(EXCEPTION)

—————————————————-

For a set of fractions ( num<den), if the difference between the numerator and the denominator is the same, then the fraction with the largest numerator is the largest and the one with the smallest numerator is the smallest

e.g. 21/27, 23/29, 19/26

largest: 23/29

smallest: 19/26

———————————————————–

For a set of fractions ( num>den), if the difference between the numerator and the denominator is the same, then the fraction with the smallest numerator is the largest and the one with the largest numerator is the smallest

e.g. 27/21, 29/23, 26/19

smallest: 29/23

largest: 26/19

——————————————————————-

i guess we could find sqrt for numbers whose nearby sqrt we already know.

steps are :-

1. take the difference from the nearest smaller perfect square. for eg. sqrt (39) nearest perfect square is 36(whose sqrt we know as 6). and the difference is 39-36=3.

2. divide the difference with product of 2 and sq root of the number which we know.

ie difference 3/(2* sqrt(36) )

3/(2*6)=.25

3.sum the above with the sqrt which we know

ie .25+6=6.25

also works for subtraction

sqrt(33) = 6-.25 = 5.75 (aprox)

4. there is a great chance of getting the answer right with even the first decimal

is correct in most of the cases.

——————————————————————

Finding square root:

Splitting the difference:

Sqrt(3125) : group the digits taking two at a time from the left (like in the usual way of finding sq roots)

First pair: 31

1)We know 5*5 =25

2)rough estimate sqrt is 50

3)divide 3125by 50 =62.5

4)subtracton: 62.5-50 =12.5

5)divide 12.5 by 2 =6.25

round it off downwards :6

rough estimate 50+6=56

try this one:

sqrt(93560)

9,35,60

roughly :300

div 93560 by 300

311.8667-300=11.87

11.87/2=5.935

approx 6

ans 306

sqrt(3847214

38 47 21 48

approx 6000

38472148/6000

6412.014-6000=412.014

412.114/2=206.05

6206.05…(has considerable error)

In most cases error is just 1%

——————————————————————

——————————————

We all know the traditional formula to compute interest…

CI = P*(1+R/100)^N – P

The calculation get very tedious when N>2 (more than 2 years). The method suggested below is elegant way to get CI/Amount after ‘N’ years.

You need to recall the good ol’ Pascal’s Triange in following way:

Code:

Number of Years (N)

——————-

1 1

2 1 2 1

3 1 3 3 1

4 1 4 6 4 1

. 1 …. …. … … 1

Example: P = 1000, R=10 %, and N=3 years. What is CI & Amount?

Step 1: 10% of 1000 = 100, Again 10% of 100 = 10 and 10% of 10 = 1

We did this three times b’cos N=3.

Step 2:

Now Amount after 3 years = 1 * 1000 + 3 * 100 + 3 * 10 + 1 * 1 = Rs.1331/-

The coefficents – 1,3,3,1 are lifted from the pascal’s triangle above.

Step 3:

CI after 3 years = 3*100 + 3*10 + 3*1 = Rs.331/- (leaving out first term in step 2)

If N =2, we would have had, Amt = 1 * 1000 + 2 * 100 + 1 * 10 = Rs. 1210/-

CI = 2 * 100 + 1* 10 = Rs. 210/-

This method is extendable for any ‘N’ and it avoids calculations involving higher powers on ‘N’ altogether!

A variant to this short cut can be applied to find depreciating value of some property. (Example, A property worth 100,000 depreciates by 10% every year, find its value after ‘N’ years).

Pascal’s triangle works in all cases but the problem being that it’s lot easier if P(Principal) and R(Rate of interest) are multiples of 10.The problem arises when either the principal or rate of interest or both are not multiples of 10.The calculation becomes a li’l bit complicated …or at least that’s what i think…in the example below R is not 10 and still it works…

For example:P=700,R=8%,N=3

(8/100)*700=56

(8/100)*56=4.48

(8/100)*4.48=0.3584

So CI=(1*700+3*56+3*4.48+1*0.3584)-700=181.7984

and if you would hav calculated in the original method

C.I=700(1+8/100)^3 – 700=700(1.08*1.08*1.0-700=181.7984

——————————————

When interest is calculated as CI, the number of years for the Amount to double (two times the principal) can be found with this following formula:

R * N ~ 72 (approximately equal to).

Exampe, if R=6% p.a. then it takes roughly 12 years for the Principal to double itself.

Note: This is just a approximate formula (when R takes large values, the error % in formula increases).

When interest is calculated as SI, number of years for amt to double can be found as:

N * R = 100 . BTW this formula is exact!

———————————————–

Let’s say we want to find 14.25% of 3267.

What will 10% of 3267 be… 326.7

And 1%… obviously 32.67. Then what would 4% be… 128 for 32 and 2.68 for 0.67…, i.e. 130.6

Thus, 14% will be 326.7 + 130.6, i.e. 457.3

If I want an even more accurate answer, if 1% is 32.6, then 0.25% will be 1/4 of 32.6, i.e 8.15

14.25% of 3267 will be 465.4

Knowing reciprocal percentage equivalent, I should have thought of an even better factorisation as 14.28% – 0.03% and since 14.28% is nothing but 1/7, the answer can directly be found by dividing 3267 by 7, i.e. 466.7

————————————————

a% of b = b% of a

so, sometimes if 4% of 25 is to be found, finding 25% of 4 would be a more sensible way

—————————————————

Percentage Table

SUBTRACTED ADDED

4 4.16(1/24) 3.86(1/26)

5 5.26(1/19) 4.76(1/21)

10 11.11(1/9) 9.09(1/11)

12.5 14.28(1/7) 11.11(1/9)

15 17.55 13

20 25 16.67(1/6)

25 33.33 20(1/5)

30 42.8 23

40 66.67(2/3) 28

50 100 33.33(1/3)

60 150 37.5(3/

when subtracted then add this% when added the subtarct this%

neha’s table of percentages.

first its uses: the info is xtremely important when dealing with problems of type..

say price increased by x% how much shud consumption decrease so that expenditure remains same.

also since most of the time speed dist..problems are in the end ratio proportion kind

this info will be xtremely useful..

so the info given is if increase is 25% decrease reqd to negate it is 20%.

In fact the table need not be memorized.the whole work can be done mentally without much problem..

the trick is to use 100 as a base and work on it..

for eg ..imagine price and consumption both as 100

now price increases by 25%..therefore ratio of new price/old price = 125/100

=> ratio of new consumption/old consumption = inverse of prev ratio = 100/125

=>change (decrease)= (125-100)/125 = 25/125= 20%

ive unnecessarily gone to such detail with the calc ..with xperience this is very simple..

for eg: 4% decrease

here’s how one shud work it out.

4% decrease =>96/100 invert it=>100/96=>change 4/96 = > 1/24

this is as given in the table.

nother one 5% increase => 105/100=>100/105=>5/105=>1/21

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Problem : 2485/179379 = ?

1. Take the denominator(179379)

2. See what percentage would give you same no of digits as numerator. In this case 1% of 179379 = 1793.79.

3. Now you need another~ 600 to get to 2400 which is 30% (.3% w.r.t original denominator) of 1793.79 ~ 1800.

4. This gives us all together 1.3% of denominator or .013 as quotient.

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The sum of terms from x to y where x and y are non negative intergers and y > x is

(sqr(y) – sqr(x) + x + y )/2

d/dx of (x+a)(x+b)(x+c)(x+d) = (x+b)(x+c)(x+d)d/dx(x+a) + (x+a)(x+c)(x+d)d/dx(x+b) + (x+b)(x+a)(x+d)d/dx(x+c) + (x+b)(x+c)(x+a)d/dx(x+d)

Cheers

VIVEK

references

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great blog

great collection

really impresive

By:

parvezon July 18, 2011at 5:16 pm