# Maths shortcuts for competitives

**find out if a number is divisible by seven:**

Take the last digit, double it, and subtract it from the rest of the

number.

If the answer is more than a 2 digit number perform the above

again.

If the result is 0 or is divisible by 7 the original number is also

divisible by 7.

Example 1 ) 259

9*2= 18.

25-18 = 7 which is divisible by 7 so 259 is also divisible by 7.

Example 2 ) 2793

3*2= 6

279-6= 273

now 3*2=6

27-6= 21 which is divisible by 7 so 2793 is also divisible by 7 .

Now find out if following are divisible by 7

1) 2841

2) 3873

3) 1393

4) 2877

**TO FIND SQUARE OF A NUMBER BETWEEN 40 to 50**

Sq (44) .

1) Subtract the number from 50 getting result A.

2) Square A getting result X.

3) Subtract A from 25 getting result Y

4) Answer is xy

EXAMPLE 1 : 44

50-44=6

Sq of 6 =36

25-6 = 19

So answer 1936

EXAMPLE 2 : 47

50-47=3

Sq 0f 3 = 09

25-3= 22

So answer = 2209

NOW TRY To Find Sq of 48 ,26 and 49

**TO FIND SQUARE OF A 3 DIGIT NUMBER :**

LET THE NUMBER BE XYZ

SQ (XYZ) is calculated like this

STEP 1. Last digit = last digit of SQ(Z)

STEP 2. Second Last Digit = 2*Y*Z + any carryover from STEP 1.

STEP 3. Third Last Digit 2*X*Z+ Sq(Y) + any carryover from STEP

2.

STEP 4. Fourth last digit is 2*X*Y + any carryover from STEP 3.

STEP 5 . In the beginning of result will be Sq(X) + any carryover

from Step 4.

EXAMPLE :

SQ (431)

STEP 1. Last digit = last digit of SQ(1) =1

STEP 2. Second Last Digit = 2*3*1 + any carryover from STEP

1.= 6

STEP 3. Third Last Digit 2*4*1+ Sq(3) + any carryover from STEP

2.= 2*4*1 +9= 17. so 7 and 1 carryover

STEP 4. Fourth last digit is 2*4*3 + any carryover (which is 1) . =

24+1=25. So 5 and carry over 2.

STEP 5 . In the beginning of result will be Sq(4) + any carryover

from Step 4. So 16+2 =18.

So the result will be 185761.

If the option provided to you are such that the last two digits are

different, then you need to carry out first two steps only , thus

saving time. You may save up to 30 seconds on each

calculations and if there are 4 such questions you save 2

minutes which may really affect UR Percentile score.

**PYTHAGORAS THEROEM :**

In any given exam there are about 2 to 3 questions based on pythagoras theorem. Wouldn’t it be nice that you remember some of the pythagoras triplets thus saving up to 30 seconds in each question. This saved time may be used to attempt other questions. Remember one more right question may make a lot of difference in UR PERCENTILE score.

The unique set of pythagoras triplets with the Hypotenuse less than 100 or one of the side less than 20 are as follows :

(3,4,5), (5, 12, 13), (8, 15, 17), (7, 24, 25), (20, 21, 29), (12, 35, 37), (9, 40, 41), (28, 45, 53), (11, 60, 61), (33, 56, 65), (16, 63, 65), (48, 55, 73), (36, 77, 85), (13, 84, 85), (39, 80, 89), and (65, 72, 97).

(15,112,113), (17,144,145), (19,180,181), (20,99,101)

If you multiply the digits of the above mentioned sets by any constant you will again get a pythagoras triplet .

Example : Take the set (3,4,5).

Multiply it by 2 you get (6,8,10) which is also a pythagoras triplet.

Multiply it by 3 you get ( 9,12,15) which is also a pythagoras triplet.

Multiply it by 4 you get (12,16,20) which is also a pythagoras triplet.

You may multiply by any constant you will get a pythagoras triplet

Take another example (5,12,13)

Multiply it by 5,6 and 7 and check if you get a pythagoras triplet.

**TIPS FOR SMART GUESSING :**

You will notice that in any case, whether it is a unique triplet or it is a derived triplet (derived by multiplying a constant to a unique triplet), all the three numbers cannot be odd.

In case of unique triplet , the hypotenuse is always odd and one of the remaining side is odd the other one is even.

Below are the first few unique triplets with first number as Odd.

3 4 5

5 12 13

7 24 25

9 40 41

11 60 61

You will notice following trend for unique triplets with first side as odd.

Hypotenuse = (Sq(first side) +1) / 2

Other side = Hypotenuse -1

Example : First side = 3 ,

so hypotenuse = (3*3+1)/2= 5 and other side = 5-1=4

Example 2: First side = 11

so hypotenuse = (9*9+1)/2= 41 and other side = 41-1=40

Please note that the above is not true for a derived triplet for example 9,12 and 15, which has been obtained from multiplying 3 to the triplet of 3,4,5. You may check for other derived triplets.

Below are the first few unique triplets with first number as Even .

4 3 5

8 15 17

12 35 37

16 63 65

20 99 101

You will notice following trend for unique triplets with first side as Even.

Hypotenuse = Sq( first side/ 2)+1

Other side = Hypotenuse-2

Example 1. First side =8

So hypotenuse = sq(8/2) +1= 17

Other side = 17-2=15

Example 2. First side = 16

So hypotenuse = Sq(16/2) +1 =65

Other side = 65-2= 63

**PROFIT AND LOSS** : In every exam there are from one to three

questions on profit and loss, stating that the cost was first

increased by certain % and then decreased by certain %. How

nice it would be if there was an easy way to calculate the final

change in % of the cost with just one formula. It would really help

you in saving time and improving UR Percentile. Here is the

formula for the same :

Suppose the price is first increase by X% and then decreased

by Y% , the final change % in the price is given by the following

formula

Final Difference % = X- Y – XY/100.

EXAMPLE 1. : The price of T.V set is increased by 40 % of the

cost price and then decreased by 25% of the new price . On

selling, the profit for the dealer was Rs.1,000 . At what price was

the T.V sold.

From the above mentioned formula you get :

Final difference % = 40-25-(40*25/100)= 5 %.

So if 5 % = 1,000

then 100 % = 20,000.

C.P = 20,000

S.P = 20,000+ 1000= 21,000.

EXAMPLE 2 : The price of T.V set is increased by 25 % of cost

price and then decreased by 40% of the new price . On selling,

the loss for the dealer was Rs.5,000 . At what price was the T.V

sold.

From the above mentioned formula you get :

Final difference % = 25-40-(25*45/100)= -25 %.

So if 25 % = 5,000

then 100 % = 20,000.

C.P = 20,000

S.P = 20,000 – 5,000= 15,000.

Now find out the difference in % of a product which was :

First increased by 20 % and then decreased by 10 %.

First Increased by 25 % and then decrease by 20 %.

First Increased by 20 % and then decrease by 25 %.

First Increased by 10 % and then decrease by 10 %.

First Increased by 20 % and then decrease by 15 %.

**TIPS TO IMPROVE UR PERCENTILE :**

HOW ABOUT SOLVING THE FOLLOWING QUESTION IN JUST

10 SECONDS

Ajay can finish work in 21 days and Blake in 42 days. If Ajay,

Blake and Chandana work together they finish the work in 12

days. In how many days Blake and Chandana can finish the

work together ?

(21*12 )/(24-12) = (21*12)/9= 7*4= 28 days.

NOW CAREFULLY READ THE FOLLOWING TO SOLVE THE

TIME AND WORK PROBLEMS IN FEW SECONDS.

**TIME AND WORK :**

1. If A can finish work in X time and B can finish work in Y time

then both together can finish work in (X*Y)/ (X+Y) time.

2. If A can finish work in X time and A and B together can finish

work in S time then B can finish work in (XS)/(X-S) time.

3. If A can finish work in X time and B in Y time and C in Z time

then they all working together will finish the work in

(XYZ)/ (XY +YZ +XZ) time

4. If A can finish work in X time and B in Y time and A,B and C

together in S time then :

C can finish work alone in (XYS)/ (XY-SX-SY)

B+C can finish in (SX)/(X-S)

and A+ C can finish in (SY)/(Y-S)

Here is another shortcut

TYPE 1 : Price of a commodity is increased by 60 %. By how

much % should the consumption be reduced so that the

expense remain the same.

TYPE 2 : Price of a commodity is decreased by 60 %. By how

much % can the consumption be increased so that the expense

remain the same.

Solution :

TYPE1 : (100* 60 ) / (100+60) = 37.5 %

TYPE 2 : (100* 60 ) / (100-60) = 150 %

## Maths formulae List for CAT

THEORY OF EQUATIONS:

—————————–

(1) If an equation (i:e f(x)=0 ) contains all positive co-efficients of any powers of x , it has no positive roots then.

eg: x^4+3x^2+2x+6=0 has no positive roots .

(2) For an equation , if all the even powers of x have some sign coefficients and all the odd powers of x have the opposite sign coefficients , then it has no negative roots .

(3)Summarising DESCARTES RULE OF SIGNS:

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) .

Hence the remaining are the minimum number of imaginary roots of the equation(Since we also know that the index of the maximum power of x is the number of roots of an equation.)

(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 roots could be found out by finding the sum of the roots of the equation and subtracting (2+3i)+(2-3i)=4 from that sum. (More about finding sum and products of roots next time )

——————————————

THEORY OF EQUATIONS

——————————————

(1) 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

(2) For a biquadratic equation ax^4+bx^3+cx^2+dx+e = 0

sum of the roots = – b/a

sum of the product of the roots taken three at a time = c/a

sum of the product of the roots taken two at a time = -d/a

product of the roots = e/a

(3) 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.

(4) Besides Complex roots , even irrational roots occur in pairs. Hence if 2+root(3) is a root , then even 2-root(3) is a root .

(All these are very useful in finding number of positive , negative , real ,complex etc roots of an equation )

—————-

—————-

(1) If for two numbers x+y=k(=constant), then their PRODUCT is MAXIMUM if

x=y(=k/2). The maximum product is then (k^2)/4 .

(2) If for two numbers x*y=k(=constant), then their SUM is MINIMUM if

x=y(=root(k)). The minimum sum is then 2*root(k) .

(3) |x| + |y| >= |x+y| (|| stands for absolute value or modulus )

(Useful in solving some inequations)

(4) 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 .

1) For any regular polygon , the sum of the exterior angles is equal to 360 degrees

hence measure of any external angle is equal to 360/n. ( where n is the number of sides)

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

(3) If a trapezium can be inscribed in a circle it must be an isosceles trapezium (i:e oblique sies equal).

(4) For an isosceles trapezium , sum of a pair of opposite sides is equal in length to the sum of the other pair of opposite sides .(i:e AB+CD = AD+BC , taken in order) .

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

1) For any 2 numbers a>b

a>AM>GM>HM>b (where AM, GM ,HM stand for arithmetic, geometric , harmonic menasa respectively)

(2) (GM)^2 = AM * HM

(3) For three positive numbers a, b ,c

(a+b+c) * (1/a+1/b+1/c)>=9

(4) For any positive integer n

2<= (1+1/n)^n <=3

(5) a^2+b^2+c^2 >= ab+bc+ca

If a=b=c , then the equality holds in the above.

(6) a^4+b^4+c^4+d^4 >=4abcd

(7) (n!)^2 > n^n (! for factorial)

————————–

(1) 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 .

(2)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 for these equations.(<> means not equal to )

If a1/a2 <> b1/b2 , then we have a unique solutions for these equations..

(3) For any quadrilateral whose diagonals intersect at right angles , the area of the quadrilateral is

0.5*d1*d2, where d1,d2 are the lenghts of the diagonals.

(4) Problems on clocks can be tackled as assuming two runners going round a circle , one 12 times as fast as the other . That is ,

the minute hand describes 6 degrees /minute

the hour hand describes 1/2 degrees /minute .

Thus the minute hand describes 5(1/2) degrees more than the hour hand per minute .

(5) The hour and the minute hand meet each other after every 65(5/11) minutes after being together at midnight.

(This can be derived from the above) .

1)If n is even , n(n+1)(n+2) is divisible by 24

(2)If n is any integer , n^2 + 4 is not divisible by 4

(3)Given the coordinates (a,b) (c,d) (e,f) (g,h) of a parallelogram , the coordinates of the meeting point of the diagonals can be found out by solving for

[(a+e)/2,(b+f)/2] =[ (c+g)/2 , (d+h)/2]

(4)Area of a triangle

1/2*base*altitude = 1/2*a*b*sinC = 1/2*b*c*sinA = 1/2*c*a*sinB = root(s*(s-a)*(s-b)*(s-c)) where s=a+b+c/2

=a*b*c/(4*R) where R is the CIRCUMRADIUS of the triangle = r*s ,where r is the inradius of the triangle .

(5)In any triangle

a=b*CosC + c*CosB

b=c*CosA + a*CosC

c=a*CosB + b*CosA

(6)If a1/b1 = a2/b2 = a3/b3 = ………….. , then each ratio is equal to

(k1*a1+ k2*a2+k3*a3+…………..) / (k1*b1+ k2*b2+k3*b3+…………..) , which is also equal to

(a1+a2+a3+…………./b1+b2+b3+……….)

(7)In any triangle

a/SinA = b/SinB =c/SinC=2R , where R is the circumradius

cosC = (a^2 + b^2 – c^2)/2ab

sin2A = 2 sinA * cosA

cos2A = cos^2(A) – sin^2 (A)

1) 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)

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

(2a) 2 < e < 3

(3) log(1+x) = x – (x^2)/2 + (x^3)/3 – (x^4)/4 ………to infinity [ Note the alternating sign . .Also note that the ogarithm is with respect to base e ]

(4) In a GP the product of any two terms equidistant from a term is always constant .

(5) For a cyclic quadrilateral , area = root( (s-a) * (s-b) * (s-c) * (s-d) ) , where s=(a+b+c+d)/2

(6) For a cyclic quadrilateral , the measure of an external angle is equal to the measure of the internal opposite angle.

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

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

(2)The sum of an infinite GP = a/(1-r) , where a and r are resp. the first term and common ratio of the GP .

(3)The equation whose roots are the reciprocal of the roots of the equation

ax^2+bx+c is cx^2+bx+a

(4) The coordinates of the centroid of a triangle with vertices (a,b) (c,d) (e,f)

is((a+c+e)/3 , (b+d+f)/3) .

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

(6) Area of a parallelogram = base * height

(7)APPOLLONIUS THEOREM:

In a triangle , if AD be the median to the side BC , then

AB^2 + AC^2 = 2(AD^2 + BD^2) or 2(AD^2 + DC^2) .

_________________

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

(2) The HCF and LCM of two nos. are equal when they are equal .

(3) Volume of a pyramid = 1/3 * base area * height

(4) In an isosceles triangle , the perpendicular from the vertex to the base or the angular bisector from vertex to base bisects the base.

(5) In any triangle the angular bisector of an angle bisects the base in the ratio of the

other two sides.

(6) the quadrilateral formed by joining the angular bisectors of another quadrilateral is

always a rectangle.

(7) Roots of x^2+x+1=0 are 1,w,w^2 where 1+w+w^2=0 and w^3=1

( |a|+|b| = |a+b| if a*b>=0

else |a|+|b| >= |a+b|

(9) 2<= (1+1/n)^n <=3

(10) WINE and WATER formula:

If Q be the volume of a vessel

q qty of a mixture of water and wine be removed each time from a mixture

n be the number of times this operation be done

and A be the final qty of wine in the mixture

then ,

A/Q = (1-q/Q)^n

(11) Area of a hexagon = root(3) * 3 * (side)^2

(12) (1+x)^n ~ (1+nx) if x<<<1

(13) 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)

(14) Appolonius theorem could be applied to the 4 triangles formed in a parallelogram.

(15) Area of a trapezium = 1/2 * (sum of parallel sids) * height = median * height

where median is the line joining the midpoints of the oblique sides.

(16) 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 .

(17) ANy function of the type y=f(x)=(ax-b)/(bx-a) is always of the form x=f(y) .

(1 Let W be any point inside a rectangle ABCD .

Then

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

(19) 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))

(20) Let ‘x’ be certain base in which the representation of a number is ‘abcd’ , then the decimal value of this number is a*x^3 + b*x^2 + c*x + d

5) For a cyclic quadrilateral , area = root( s* (s-a) * (s-b) * (s-c) * (s-d) ) , where s=(a+b+c+d)/2

Here are some neat shortcuts on Simple/Compound Interest.

Shortcut #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).

Shortcut #2:

————-

(i) 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:

P * 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).

(ii) 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!

Adding to what ‘Peebs’ said, this shortcut does work for any P/N/R.

Basically if you look closely at this method, what I had posted is actually derived from the Binomial expansion of the polynomial — (1+r/100)^n but in a more “edible” format digestable by us!

BTW herez one shortcut on recurring decimals to fractions …Its more easier to explain with an example..

Eg: 2.384384384 ….

Step 1: since the 3 digits ’384′ is recurring part, multiply 2.384 by 1000 = so we get 2384.

Next ’2′ is the non recurring part in the recurring decimal so subtract 2 from 2384 = 2382.

If it had been 2.3848484.., we would have had 2384 – 23 = 2361. Had it been 2.384444.. NR would be 2384 – 238 = 2146 and so on.

We now find denominator part …….

Step 3: In step 1 we multiplied 2.384384… by 1000 to get 2384, so put that first.

Step 4: next since all digits of the decimal part of recurring decimal is recurring, subtract 1 from step 3. Had the recurring decimal been 2.3848484, we need to subtract 10. If it had been 2.3844444, we needed to have subtracted 100 ..and so on…

Hence here, DR = 1000 – 1 = 999

Hence fraction of the Recurring decimal is 2382/999!!

Some more examples ….

1.56787878 … = (15678 – 156) / (10000 – 100) = 15522/9900

23.67898989… = (236789 – 2367) / (10000 – 100) = 234422/9900

124.454545… = (12445 – 124) / (100 – 1) = 12321/99

Numbers

1) 2^2n-1 is always divisible by 3

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

= 3M +1 -1

= 3M, thus divisible by 3

2) What is the sum of the divisors of 2^5.3^7.5^3.7^2?

ANS : (2^6-1)(3^8-1)(5^4-1)(7^3-1)/2.4.6

Funda : if a number ‘n’ is represented as

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*…)

3) 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

4) x + y + z = 7 and xy + yz + zx = 10, then what is the maximum value of x? ( CAT 2002 has similar question )

ANS: 49-20 = 29, now if one of the y,z is zero, then the sum of other 2 squares shud be equal to 29, which means, x can take a max value at 5

5) 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

6) In 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.

7) How many times the digit 0 will appear from 1 to 10000

ANS: In 2 digit numbers : 9,

In 3 digit numbers : 18 + 162 = 180,

In 4 digit numbers : 2187 + 486 + 27 = 2700,

total = 9 + 180 + 2700 + 4 = 2893

8 ) What is the sum of all irreducible factors between 10 and 20 with denominator as 3?

ANS :

sum = 10.33 + 10.66 + 11.33 + 11.66 + 12.33 + 12.66 + 13.33 + 13.66…….

= 21 + 23 + ……

= 300

9) if n = 1+x where x is the product of 4 consecutive number then n is,

1) an odd number,

2) is a perfect square

SOLN : (1) is clearly evident

(2) let the 4 numbers be n-2,n-1,n and n+1 then by multing the whole thing and adding 1 we will have a perfect square

10) When 987 and 643 are divided by same number ‘n’ the reminder is also same, what is that number if the number is a odd prime number?

ANS : since both leave the same reminder, let the reminder be ‘r’,

then, 987 = an + r

and 643 = bn + r and thus

987 – 643 is divisible by ‘r’ and

987 – 643 = 344 = 86 * 4 = 43 * 8 and thus the prime is 43

hence ‘r’ is 43

11) 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

12) a^n – b^n is always divisible by a-b

13) if a+b+c = 0 then a^3 + b^3 + c^3 = 3abc

EXAMPLE: 40^3-17^3-23^3 is divisble by

since 40-23-17 = 0, 40^3-17^3-23^3 = 3*40*23*17 and thus, the number is divisible by 3,5,8,17,23 etc.

14) There is a seller of cigerette and match boxes who sits in the narrow lanes of cochin. He prices the cigerattes at 85 p, but found that there are no takers. So he reduced the price of cigarette and managed to sell all the cigerattes, realising Rs. 77.28 in all. What is the number of cigerattes?

a) 49

b) 81

c) 84

d) 92

ANS : (d)

since 77.28 = 92 * 84, and since price of cigarette is less than 85, we have (d) as answer

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