Divisibility Tests for 7,13,17 & 19

Divisibility test of 7
1).Double the last digit (digit at the rightmost place) and subtract it from the number left (excluding the last digit). If this number is divisible with 7 then the original number is divisible by 7.
This procedure can be followed as many times as required (until the number is reduced to 2 digit number). Then the number so obtained can be checked whether it is divisible by 7 or not. If the number so obtained is divisible by 7 then the original number is divisible by 7 and if not then original number is not divisible by 7.
E.g.-
Consider the number 1057.
Now the last digit is 7. On doubling it we get 14.
On subtracting it from 105 we get 91.
Now it can be seen that 91 is divisible by 7 so the original number is divisible by 7.
(It can further be simplified by doubling 1 and subtracting it from 9 and thus we get 7 which is divisible by7.)
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Divisibility test of 13
To check whether a number is divisible by 13 we follow the procedure as follows:
1). Multiply the last digit with 4 and add it to the number left (after removing the last digit).
2). Follow this method again and again and reduce the number to 2-digit number form.
3). Now check whether the number is divisible by 13 or not.
If the 2-digit number so obtained is divisible by 13 then the original number is divisible by 13 otherwise not.
E.g.-
Let us consider the number 195.
Now the last digit is 5 and on multiplying it with 4 we get 20.
Now on adding this with the remaining number (i.e. 19) we get 39. Now as 39 is divisible by 13 therefore the original
number id divisible by 13.
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Divisibility test of 17
To check whether a number is divisible by 17 we follow the procedure as follows:
1). Multiply the last digit with 5 and subtract it from the number left (after removing the last digit).
2). Follow this method again and again and reduce the number to 2-digit number form.
3). Now check whether the number is divisible by 17 or not.
If the 2-digit number so obtained is divisible by 17 then the original number is divisible by 17 otherwise not.
E.g.-
Let us consider the number 221.
Now the last digit is 1 and on multiplying it with 5 we get 5.
Now on subtracting 5 from the remaining number (i.e. 22) we get 17. Now as 17 is divisible by 17 therefore the original number id divisible by 17.
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Divisibility test of 19
To check whether a number is divisible by 19 we follow the procedure as follows:
1). Multiply the last digit with 2 and add it to the number left (after removing the last digit).
2). Follow this method again and again and reduce the number to 2-digit number form.
3). Now check whether the number is divisible by 19 or not.
If the 2-digit number so obtained is divisible by 19 then the original number is divisible by 19 otherwise not.
E.g.-
Let us consider the number 209.
Now the last digit is 9 and on multiplying it with 2 we get 18.
Now on adding th18 with the remaining number(i.e. 20) we get 38. Now as 38 is divisible by 19 therefore the original number id divisible by 19.
Number Multiply last digit -/+
7 2 -
13 4 +
17 5 -
19 2 +

Quant… Basic Formulae

Consolidated some of the basic formula.
ALGEBRA :
1. Sum of first n natural numbers = n(n+1)/2
2. Sum of the squares of first n natural numbers = n(n+1)(2n+1)/6
3. Sum of the cubes of first n natural numbers = [n(n+1)/2]2
4. Sum of first n natural odd numbers = n2
5. Average = (Sum of items)/Number of items
Arithmetic Progression (A.P.):
An A.P. is of the form a, a+d, a+2d, a+3d, …
where a is called the ‘first term’ and d is called the ‘common difference’
1. nth term of an A.P. tn = a + (n-1)d
2. Sum of the first n terms of an A.P. Sn = n/2[2a+(n-1)d] or Sn = n/2(first term + last term)
Geometrical Progression (G.P.):
A G.P. is of the form a, ar, ar2, ar3, …
where a is called the ‘first term’ and r is called the ‘common ratio’.
1. nth term of a G.P. tn = arn-1
2. Sum of the first n terms in a G.P. Sn = a|1-rn|/|1-r|
Permutations and Combinations :
1. nPr = n!/(n-r)!
2. nPn = n!
3. nP1 = n
1. nCr = n!/(r! (n-r)!)
2. nC1 = n
3. nC0 = 1 = nCn
4. nCr = nCn-r
5. nCr = nPr/r!
Number of diagonals in a geometric figure of n sides = nC2-n

Tests of Divisibility :
1. A number is divisible by 2 if it is an even number.
2. A number is divisible by 3 if the sum of the digits is divisible by 3.
3. A number is divisible by 4 if the number formed by the last two digits is divisible by 4.
4. A number is divisible by 5 if the units digit is either 5 or 0.
5. A number is divisible by 6 if the number is divisible by both 2 and 3.
6. A number is divisible by 8 if the number formed by the last three digits is divisible by 8.
7. A number is divisible by 9 if the sum of the digits is divisible by 9.
8. A number is divisible by 10 if the units digit is 0.
9. A number is divisible by 11 if the difference of the sum of its digits at odd places and the sum of its digits at even places, is divisible by 11.
H.C.F and L.C.M :
H.C.F stands for Highest Common Factor. The other names for H.C.F are Greatest Common Divisor (G.C.D) and Greatest Common Measure (G.C.M).
The H.C.F. of two or more numbers is the greatest number that divides each one of them exactly.
The least number which is exactly divisible by each one of the given numbers is called their L.C.M.
Two numbers are said to be co-prime if their H.C.F. is 1.
H.C.F. of fractions = H.C.F. of numerators/L.C.M of denominators
L.C.M. of fractions = G.C.D. of numerators/H.C.F of denominators
Product of two numbers = Product of their H.C.F. and L.C.M.

PERCENTAGES :
1. If A is R% more than B, then B is less than A by R / (100+R) * 100
2. If A is R% less than B, then B is more than A by R / (100-R) * 100
3. If the price of a commodity increases by R%, then reduction in consumption, not to increase the expenditure is : R/(100+R)*100
4. If the price of a commodity decreases by R%, then the increase in consumption, not to decrease the expenditure is : R/(100-R)*100
PROFIT & LOSS :
1. Gain = Selling Price(S.P.) – Cost Price(C.P)
2. Loss = C.P. – S.P.
3. Gain % = Gain * 100 / C.P.
4. Loss % = Loss * 100 / C.P.
5. S.P. = (100+Gain%)/100*C.P.
6. S.P. = (100-Loss%)/100*C.P.
Short cut Methods:
1. By selling an article for Rs. X, a man loses l%. At what price should he sell it to gain y%? (or)
A man lost l% by selling an article for Rs. X. What percent shall he gain or lose by selling it for Rs. Y?
(100 – loss%) : 1st S.P. = (100 + gain%) : 2nd S.P.
2. A man sold two articles for Rs. X each. On one he gains y% while on the other he loses y%. How much does he gain or lose in the whole transaction?
In such a question, there is always a lose. The selling price is immaterial.
Formula: Loss % =
3. A discount dealer professes to sell his goods at cost price but uses a weight of 960 gms. For a kg weight. Find his gain percent.
Formula: Gain % =
RATIO & PROPORTIONS:
1. The ratio a : b represents a fraction a/b. a is called antecedent and b is called consequent.
2. The equality of two different ratios is called proportion.
3. If a : b = c : d then a, b, c, d are in proportion. This is represented by a : b :: c : d.
4. In a : b = c : d, then we have a* d = b * c.
5. If a/b = c/d then ( a + b ) / ( a – b ) = ( d + c ) / ( d – c ).
TIME & WORK :
1. If A can do a piece of work in n days, then A’s 1 day’s work = 1/n
2. If A and B work together for n days, then (A+B)’s 1 days’s work = 1/n
3. If A is twice as good workman as B, then ratio of work done by A and B = 2:1
PIPES & CISTERNS :
1. If a pipe can fill a tank in x hours, then part of tank filled in one hour = 1/x
2. If a pipe can empty a full tank in y hours, then part emptied in one hour = 1/y
3. If a pipe can fill a tank in x hours, and another pipe can empty the full tank in y hours, then on opening both the pipes,
the net part filled in 1 hour = (1/x-1/y) if y>x
the net part emptied in 1 hour = (1/y-1/x) if x>y
TIME & DISTANCE :
1. Distance = Speed * Time
2. 1 km/hr = 5/18 m/sec
3. 1 m/sec = 18/5 km/hr
4. Suppose a man covers a certain distance at x kmph and an equal distance at y kmph. Then, the average speed during the whole journey is 2xy/(x+y) kmph.
PROBLEMS ON TRAINS :
1. Time taken by a train x metres long in passing a signal post or a pole or a standing man is equal to the time taken by the train to cover x metres.
2. Time taken by a train x metres long in passing a stationary object of length y metres is equal to the time taken by the train to cover x+y metres.
3. Suppose two trains are moving in the same direction at u kmph and v kmph such that u>v, then their relative speed = u-v kmph.
4. If two trains of length x km and y km are moving in the same direction at u kmph and v kmph, where u>v, then time taken by the faster train to cross the slower train = (x+y)/(u-v) hours.
5. Suppose two trains are moving in opposite directions at u kmph and v kmph. Then, their relative speed = (u+v) kmph.
6. If two trains of length x km and y km are moving in the opposite directions at u kmph and v kmph, then time taken by the trains to cross each other = (x+y)/(u+v)hours.
7. If two trains start at the same time from two points A and B towards each other and after crossing they take a and b hours in reaching B and A respectively, then A’s speed : B’s speed = (√b : √
SIMPLE & COMPOUND INTERESTS :
Let P be the principal, R be the interest rate percent per annum, and N be the time period.
1. Simple Interest = (P*N*R)/100
2. Compound Interest = P(1 + R/100)N – P
3. Amount = Principal + Interest
LOGORITHMS :
If am = x , then m = logax.
Properties :
1. log xx = 1
2. log x1 = 0
3. log a(xy) = log ax + log ay
4. log a(x/y) = log ax – log ay
5. log ax = 1/log xa
6. log a(xp) = p(log ax)
7. log ax = log bx/log ba
Note : Logarithms for base 1 does not exist.
AREA & PERIMETER :
Shape Area Perimeter
Circle ∏ (Radius)2 2∏(Radius)
Square (side)2 4(side)
Rectangle length*breadth 2(length+breadth)
1. Area of a triangle = 1/2*Base*Height or
2. Area of a triangle = √ (s(s-(s-b)(s-c)) where a,b,c are the lengths of the sides and s = (a+b+c)/2
3. Area of a parallelogram = Base * Height
4. Area of a rhombus = 1/2(Product of diagonals)
5. Area of a trapezium = 1/2(Sum of parallel sides)(distance between the parallel sides)
6. Area of a quadrilateral = 1/2(diagonal)(Sum of sides)
7. Area of a regular hexagon = 6(√3/4)(side)2
8. Area of a ring = ∏(R2-r2) where R and r are the outer and inner radii of the ring.
VOLUME & SURFACE AREA :
Cube :
Let a be the length of each edge. Then,
1. Volume of the cube = a3 cubic units
2. Surface Area = 6a2 square units
3. Diagonal = √ 3 a units
Cuboid :
Let l be the length, b be the breadth and h be the height of a cuboid. Then
1. Volume = lbh cu units
2. Surface Area = 2(lb+bh+lh) sq units
3. Diagonal = √ (l2+b2+h2)
Cylinder :
Let radius of the base be r and height of the cylinder be h. Then,
1. Volume = ∏r2h cu units
2. Curved Surface Area = 2∏rh sq units
3. Total Surface Area = 2∏rh + 2∏r2 sq units
Cone :
Let r be the radius of base, h be the height, and l be the slant height of the cone. Then,
1. l2 = h2 + r2
2. Volume = 1/3(∏r2h) cu units
3. Curved Surface Area = ∏rl sq units
4. Total Surface Area = ∏rl + ∏r2 sq units
Sphere :
Let r be the radius of the sphere. Then,
1. Volume = (4/3)∏r3 cu units
2. Surface Area = 4∏r2 sq units
Hemi-sphere :
Let r be the radius of the hemi-sphere. Then,
1. Volume = (2/3)∏r3 cu units
2. Curved Surface Area = 2∏r2 sq units
3. Total Surface Area = 3∏r2 sq units
Prism :
Volume = (Area of base)(Height

Divisibility tests

A number is divisible by 2 if its last digit is also (i.e. 0,2,4,6 or 8).

A number is divisible by 3 if the sum of its digits is also. Example: 534: 5+3+4=12 and 1+2=3 so 534 is divisible by 3.

A number is divisible by 5 if the last digit is 5 or 0.
 .
Test for divisibility by 7. Double the last digit and subtract it from the remaining leading truncated number. If the result is divisible by 7, then so was the original number. Apply this rule over and over again as necessary. Example: 826. Twice 6 is 12. So take 12 from the truncated 82. Now 82-12=70. This is divisible by 7, so 826 is divisible by 7 also.

There are similar rules for the remaining primes under 40, i.e. 11,13, 17,19,23,29,31,37,41,43 and 47.
Test for divisibility by 11. Subtract the last digit from the remaining leading truncated number. If the result is divisible by 11, then so was the first number. Apply this rule over and over again as necessary.
Example: 19151--> 1915-1 =1914 -->191-4=187 -->18-7=11, so yes, 19151 is divisible by 11.


Test for divisibility by 13. Add four times the last digit to the remaining leading truncated number. If the result is divisible by 13, then so was the first number. Apply this rule over and over again as necessary.
Example: 50661-->5066+4=5070-->507+0=507-->50+28=78 and 78 is 6*13, so 50661 is divisible by 13.

Test for divisibility by 17. Subtract five times the last digit from the remaining leading truncated number. If the result is divisible by 17, then so was the first number. Apply this rule over and over again as necessary.
Example: 3978-->397-5*8=357-->35-5*7=0. So 3978 is divisible by 17.

Test for divisibility by 19. Add two times the last digit to the remaining leading truncated number. If the result is divisible by 19, then so was the first number. Apply this rule over and over again as necessary.
EG: 101156-->10115+2*6=10127-->1012+2*7=1026-->102+2*6=114 and 114=6*19, so 101156 is divisible by 19.

If you want you can stop here.However, I have extended the list up to 50.

Test for divisibility by 23. Add seven times the last digit to the remaining leading truncated number. If the result is divisible by 23, then so was the first number. Apply this rule over and over again as necessary.
Example: 17043-->1704+7*3=1725-->172+7*5=207-->20+7*7=69 which is 3*23, so 17043 is also divisible by 23.

Test for divisibility by 29. Add three times the last digit to the remaining leading truncated number. If the result is divisible by 29, then so was the first number. Apply this rule over and over again as necessary.
Example: 15689-->1568+3*9=1595-->159+3*5=174-->17+3*4=29, so 15689 is also divisible by 29.

Test for divisibility by 31. Subtract three times the last digit from the remaining leading truncated number. If the result is divisible by 31, then so was the first number. Apply this rule over and over again as necessary.
Example: 7998-->799-3*8=775-->77-3*5=62 which is twice 31, so 7998 is also divisible by 31.

Test for divisibility by 37. This is (slightly) more difficult, since it perforce uses a double-digit multiplier, namely eleven. People can usually do single digit multiples of 11, so we can use the same technique still. Subtract eleven times the last digit from the remaining leading truncated number. If the result is divisible by 37, then so was the first number. Apply this rule over and over again as necessary.
Example: 23384-->2338-11*4=2294-->229-11*4=185 which is five times 37, so 23384 is also divisible by 37.

Test for divisibility by 41. Subtract four times the last digit from the remaining leading truncated number. If the result is divisible by 41, then so was the first number. Apply this rule over and over again as necessary.
Example: 30873-->3087-4*3=3075-->307-4*5=287-->28-4*7=0, remainder is zero and so 30873 is also divisible by 41.

Test for divisibility by 43. Now it starts to get really difficult for most people, because the multiplier to be used is 13, and most people cannot recognise even single digit multiples of 13 at sight. You may want to make a little list of 13*N first. Nevertheless, for the sake of completeness, we will use the same method. Add thirteen times the last digit to the remaining leading truncated number. If the result is divisible by 43, then so was the first number. Apply this rule over and over again as necessary.
Example: 3182-->318+13*2=344-->34+13*4=86 which is recognisably twice 43, and so 3182 is also divisible by 43.
We can also do following:, since we are working to modulo43, instead of adding factor 13 times the last digit, we can subtract 30 times it, because 13+30=43.

Finally, the Test for divisibility by 47. This too is difficult for most people, because the multiplier to be used is 14, and most people cannot recognise even single digit multiples of 14 at sight. You may want to make a little list of 14*N first. Nevertheless, for the sake of completeness, we will use the same method. Subtract fourteen times the last digit from the remaining leading truncated number. If the result is divisible by 47, then so was the first number. Apply this rule over and over again as necessary.
Example: 34827-->3482-14*7=3384-->338-14*4=282-->28-14*2=0 , remainder is zero and so 34827 is divisible by 47.

Why d we sometimes say ADD and for other primes say SUBTRACT, and where from the apparently arbitrary factors come from. So let us do some algebra to show the method in my madness.
We know that recursive divisibility test of number N as f-M*r where f are the front digits of N, r is the rear digit of N and M is some multiplier. And we want to see if N is divisible by some prime P. We need a method to work out the values of M. What you do is to calculate (mentally) the smallest multiple of P which ends in a 9 or a 1. If it's a 9 we are going to ADD, if it's a 1 we are going to SUBTRACT later. Then we will use the leading digit(s) of the multiple as our multiplier M.
Example for P=17 : three times 17 is 51 which is the smallest multiple of 17 that ends in a 1 or 9. Since it's a 1 we are going to SUBTRACT later. The leading digit is a 5, so we are going to SUBTRACT five times the remainder r. The algorithm was stated above. Now let's do the algebraic proof. Writing N=10f+r, we can multiply by -5 (as shown in the example for 17), getting -5N=-50f-5r. Now we add 51f to both sides (because 51 was the smallest multiple of P=17 to end in a 1 or a 9), giving one f (which we want), so 51f-5N=f-5r. Now if N is divisible by P (here P=17), we can substitute to get 51f-5*17*x=f-5r and rearrange the left side as 17*(3f-5x)=f-5r and therefore f-5r is a multiple of P=17 also. Q.E.D.