Base System : Laying down the foundation

It's been a while but we are back with a new article today that takes up basic principles of base system and discovers its applications in various kinds of problems. This should set you up for other variants in this domain and a special new variant is present in our Number System E-book if you have bought a copy of that. Let's get started then..

General principle
The number of digits used in a particular system of writing numbers is known as the base. Our well known decimal system (or base 10) has 10 digits (0 to 9).

In general, a base-n system uses digits 0 to (n – 1). Like our count in decimal system runs in cycles of 10, another base n runs in cycles of n and moves to the next significant digit to the left when its digits get exhausted for the count.

Some basic principles
1) A number in base system n, say abc, is written as c + b*n + a*n^2. This would be familiar to writing a number, say 145, in decimal system as 5 + 4*10 + 1*100. Same principle applies for a general base n.
2) A base system N uses digits 0 to (N – 1) only for its representation.
3) A number is base N is divisible by N-1, when the sum of digits in base N is divisible by N-1
4) When digits of a number N1 in base N are rearranged to form a number N2, then N2-N1 is always divisible by N-1.
5) If a number in base N has even number of digits and that number is a palindrome, then the number is divisible by N+1


Example problems
Q1 A number 2342a121 is in base 8 and it is divisible by 7. Find the value of a.
Using principle 3,
(2 + 3 + 4 + 2 + a + 1 + 2 + 1) is divisible by 7.
(15 + a) is divisible by 7
a= 6

Q2 The value of (222) in base ‘x’ when converted to base 10 is ‘P’. The value of (222) in base ‘y’ when
converted to base 10 is Q. If (P – Q) in base 10 = 28, then what is the value of (Q – x) in base 10?
Use principle 1,
P = 2 + 2x + 2x^2
Q = 2 + 2y + 2y^2
(P – Q) = 28
2(x^2 – y^2) + 2(x – y) = 28
(x – y)(x + y + 1) = 14 = 1*14 = 2*7
[Note: Second bracket is always greater than the first hence we need not check for reverse combinations of 14*1 and 7*2.]
(x – y) = 1 and (x + y + 1) = 14
(x – y) = 1 and (x + y) = 13
x = (13 + 1)/2 = 7, y = (13 – 1)/2 = 6
Q = 2 + 2(6) + 2(36) = 86
(Q – x) = 79

(x – y) = 2 and (x + y + 1) = 7
(x – y) = 2 and (x + y) = 6
x = (6 + 2)/2 = 4, y = (6 – 2)/2 = 2
Not possible as if base y is 2 then it cannot have a number 222 existing in it since a base n uses only digits 0 to (n – 1).

Hence our answer will be distinct and equal to 79. If more than one valid values appear, you get your answer as cannot be determined.

Q3) Find the 1000th term of the sequence : 1,3,4,7,8,9,10,11,13,14,... in which there is no number which contain digit 2,5 or 6.

Standard Approach
_ : 6 numbers ; _ _ : 6*7 = 42 numbers ; _ _ _ : 6*7*7 = 294 numbers
342 numbers added till 999
1 _ _ _ : 7*7*7 = 343 added = 685 numbers till now. Remaining 1000 - 685 = 315
3 0 _ _ : 49 added
3 1 _ _ : 49 added
3 3 _ _ : 49 added
3 4 _ _ , 3 7 _ _ , 3 8 _ _ = 49*3 = 147
Total 294 more. Remaining 21
3 9 0 _ : 7
3 9 1 _ : 7
3 9 3 _ : 7
3939

Smart Approach[Using base system]
Analogous mapping of digits of this new system where 2,5,6 are absent unlike the known base 7 where 7,8,9 are not there. Remember base system is just a count. You can always map it to different digits and count differently. Only problem people face here is because we are too used to counting from 1 to 9 and further.
Mapping the available digits to the normal count, we have,


0 -> 0
1 -> 1
2 -> 3
3 -> 4
4 -> 7
5 -> 8
6 -> 9

1000th term of our series will thus be 1000[Read as one zero zero zero] in base 7 = 2626
Mapping it back to original digits <-> 3939.
[1000 = 2*343 + 6*49 + 2*7 + 6*1. Hence 1000 = 2626 in base 7.]


This brings us to the end of the helicopter view of the base system domain in number system. Someday if we find time we will take some more special application problems to enhance our knowledge or you could get started with them on your own using this as a foundation.
Until next time,


Cheers
AS

0

Number System Ebook

I will be launching soon the e-book on number systems for which pre-registrations have already begun.
The e-book will contain :
+ All prevalent concepts covered with explanation and examples wherever required
+ Special individual concepts that frequent in MBA Examinations
+ Practice exercise with solutions to extend your learning into application of basic concepts
+ 30 days promised support through email for any doubts that you may have for the content of the e-book cleared directly by the author to the best of his knowledge.
Price : Rs 200
Let us know if you are interested by filling this form with your valid information.
Expected release : End of April or first week of May.

You may do your registrations on the following link if interested :
Registration Link : Click here


Cheers
AS

1

Calculation Tricks : Multiplication

I write this article today amidst the excitement of release of my first ever compiled ebook on parajumbles. Visit the FB page : https://www.facebook.com/mbadecoded for more on that. Today we look at the calculation tricks for multiplication. First we start with the specific tricks that pertain to multiplication by a specified number. Then eventually we look at how multiplication can be done faster in general.

Multiplication by 11

Step 1 : Put the right most digit of number being multiplied as the right most digit of your answer.
Step 2 : Each successive digit of number being multiplied is added to the digit next to it and put as the next digit moving right to left. If the sum is of two digits the ten's place digit is carried over to the left.


example : 5892 * 11
Step 1 : Answer : -----2 which is nothing but 2 + 0(as no right hand side neighbour)
Step 2 : 9 + 2 = 11. Carry over 1. Put 1 beside 2. Answer : ----12
Step 2(Contd) : 8 + 9 + 1(carry) = 18. Carry over 1. Put 8 beside 12. Answer : ----812
Step 2(Contd) : 5 + 8 + 1(carry) = 14. Carry over 1. Put 4 beside 812. Answer : ---4812
Step 2(Final) : 5 + 1(carry) = 6. Put 6 beside 4812 to get your final answer : 64812

How about multiplication by 12 or 13 or ..?
Before adding the right hand neighbour you double each digit first in each step.
5324 * 12 = Written as : 05324 * 12
4*2 = 8. Answer : ----8
2*2 + 4 = 8. Answer : ----88
3*2 + 2 = 8. Answer : ----888
5*2 + 3 = 13.Add 3 to the left and carry over 1. Answer : --3888
0*2 + 5 + 1(carry) = 6. Final answer : 63888
Try figuring out the logic and you will be able to generalize multiplication for more numbers..For now moving onto other special cases.


Multiplication by 25

As simple as it can get. 25 = 100/4. You add two zeroes to your number and divide it by 4. Just an alternative to your usual calculation.

example : 1761253 * 25 = 176125300/4 = 44031325

Multiplication by 111

To multiply a two-digit number by 111, add the two digits and if the sum is a single digit, write this digit twice  in between the original digits of the number.

example : 35x111= 3885 ; 63x111= 6993

The same idea works if the sum of the two digits is not a single digit, but you need to take the carry over to the left.

example : 57x111= 6(5 + 1) | 3(=2 + 1) | 2 | 7 = 6327
because 5+7=12

For 3 digit numbers,
Carry if any of these sums is more than one digit.

example :
123x111 = 1 | 3 (=1+2) | 6 (=1+2+3) | 5 (=2+3) | 3 = 13653
241x111 = 2 | 6(=2 + 4)| 7(=2 + 4 + 1) | 5(=4 + 1) | 1 = 26751

352x111=3 | 9 (=3+5 + 1) | 10 (=3+5+2)| 7 (=5+2)| 2 = 39072



Multiplication of numbers close to 100(called the "base")

Case 1 : 102*106 i.e both numbers greater than 100.

Left part of the answer will be : (102 + (106 - 100)) = 108 OR (106 + (102 - 100) = 108

Right part of the answer will be : A two digit multiplication of surplus from both numbers = 2*6 = 12. Carry the left most digits of this right part if the answer is more than two digits.

Answer : 10812


Case 2 : 95*97 i.e both numbers are less than 100
Similar approach. Just surplus replaced with deficits. Left part : (95 - 3) = (97 - 5) = 92
Right part : 3*5 = 15. Carry if required.

Answer : 9215


Case 3 : 104*95 i.e one number is less and other greater than 100

Again the similar approach. Left part will be : (104 + (-5)) = (95 + 4) = 99

Right part = 4*(-5) = -20.

Answer : 99 | -20 <- Does that look right to you? I hope so no. Let's look at how to make it look right.

Removing the negative sign
You borrow a 1 from the neighbouring 9 which by value is a 100. So your answer will become :
99 - 1 | 100 - 20 = 9880

Another example for such a kind where you need to borrow again but this time something different. Idea is to bring down the right hand side to a positive two digit number.

91*112 = (91 + 12) | -9*12 = 103|-108 = 103 - 2 | 200 - 108 = 101|92 = 10192

Let's end this one here today. Practice a lot on these techniques(Squares one earlier as well) if you want them to actually save your time in calculations. Solve at least 15-20 questions on each type by constructing various number combination. Like and share if you like the content on Lagom.
Next time we ll discuss the general multiplication techniques as well as when the base is changed from 100 i.e to say the numbers are close to say 50 or 400 or 1000 or anything like that. Keep following..Till then..



Cheers!
AS

0

Mass Point Geometry : Mastering the art

This article will take your through some of the beautiful geometry problems that will be solved using mass point geometry(MPG) and unveil its extensions & applications in the geometry domain.
You can read the article on MPG basics here : http://mbadecoded.blogspot.in/2015/03/mass-point-geometry-explored.html

So moving on to the agenda of this article, let's see our first problem :

Q1 In triangle ABC, medians AD and CE intersect at P, PE = 1.5 , PD = 2, and DE = 2.5. What is the area of AEDC?
a) 13  b) 13.5  c) 14  d) 14.5  e) 15
                             

Assign B mass m. Thus, because E is the midpoint of AB, A also has a mass of m.

Similarly, C has a mass of m. D and E each have a mass of 2m because they are between B and C and A and B respectively. Note that the mass of D is twice the mass of A, so AP must be twice as long as PD.

PD has length 2, so AP has length 4 and AD has length 6. Similarly, CP is twice PE and PE = 1.5, so CP = 3 and CE = 4.5.

Now note that triangle PED is a 3-4-5 right triangle with the right angle DPE. This means that the quadrilateral AEDC is a kite.

The area of a kite is half the product of the diagonals, AD and CE. Recall that they are 6 and 4.5 respectively, so the area of AEDC is 6*4.5/2 = 13.5


You can refer the alternate solutions to this problem here : http://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_16


Q2 Triangle ABC has AB = 21, AC = 22 and BC = 20. Points D and E are located on AB and
AC, respectively, such that DE is parallel to BC and contains the centre of inscribed circle of triangle ABC. Then, DE = m/n, where m and n are relatively prime positive integers. Find m + n.

You can refer the alternate solutions to this problem here : http://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_7


Q3 In the given figure, DE || BC and AD : DB = 5 : 4 , find the ratio of areas of triangles BDE and BCF.
a) 70 : 81  b) 16 : 25  c) 56 : 81  d) 52 : 81

Let the area of triangle ABC be A. Then Area of triangle ADE = 5*5*A/[(5 + 4)(5 + 4)] = 25A/81
Applying MPG as following and getting the shown ratios.

We can say area of triangle BDE = (4/5)(25A/81) [Areas divided in the ratio of their shared base]
Then we find the area of CEF similarly and subtract these areas from total to get area of BCF and our answer.



Reference : You can try more interesting problems which haven't been discussed here in detail and try reaching the answer yourself using MPG.
http://artofproblemsolving.com/wiki/index.php/Mass_points




This brings us to the close of article series on Mass point geometry. I hope after going through the two articles here on blog and practicing problems on the given link you will start reaching a good confidence level while dealing with problems in this area.

If you like the content here on Lagom, please like,share and spread the posts and our name to your peers. Happy sharing :)


Cheers!
AS

0

Percentages - Absolute Basics

Today's article will deal with some basics of percentages that can be implemented in DI section as well its application areas of Profit, Loss & Discount(PLD) ; Simple and Compound Interest(SI_CI) and the like which I will be presenting soon as a follow up to this article. So let's start!
The above presented table would prove really helpful to you in several sections of Quantitative aptitude and Data Interpretation. Basically it would help you wherever you deal with numbers and percentages. PLD,SI_CI as well as in Time and Work problems if you are accustomed to the efficiency approach. Moreover, DI problems involving %ages would definitely deal with these numbers. So first things first, learn and keep this handy with you to keep revising it till they are on your fingertips.


Concept 1:

Then, I would like to talk about the data interpretation calculation tips in the context of using percentages as a tool. Let's call this a 10% rule for DI.
Let's see what this is all about..

Simple data table :
5 students : A,B,C,D,E give 3 exams P,Q,R which have total marks as 120,200 and 250 respectively.
Let's consider student A. His scores are 34%, 68% and 45.4% in P,Q,R respectively and similar data is provided for all 5 students. You are asked to find the subject wise and overall topper and second place student of the class assuming there are only 5 students in the class [else answer would cannot be determined]

How do you think you'll find 5*3 = 15 individual %age calculations and save time?

Let's unveil the 10% rule now.

10% of a number pq.xyz means moving the decimal one place to the left : p.qxyz
1% means moving it two places to the left : .pqxyz
and 0.1% means moving it three places : .0pqxyz

Using this for our task :

A scores 34% marks in P(total : 120)
34% = (30 + 4)% = 3*(12) + 4*(1.2) = 36 + 4.8 = 40.8 <= Did you see what happened there? 

Let's do this thing one more time..
 
A scores 68% marks in Q(total : 200)
68% = (70 - 2)% = 7*20 - 2*2 = 140 - 4 = 136 <= How about now? Are you getting how it's done.

3 times' the charm..

A scores 45.4% marks in R(total : 250)
45.4% = (40 + 5 + 0.4)% = 4*25 + 5*2.5 + 4*0.25 = 100 + 12.5 + 1 = 113.5
 
I hope next time you see such figures you'll not start with seeing them as fractions and dividing strange figures(well it might be useful at times actually. But that's your observation)
 

Concept 2:

Another place to use this is while comparing ratios..
123/140 , 115/213 , 110/124..

Find what % of 140 is 123 coming out to be and same for rest.
Smart thing to do here would be to put 115/213 as the smallest[try to figure out why..simple ratio rules..Leave comments if you don't get it still]

140 : 10% = 14 , 1% = 1.4 , 0.1% = 0.14
You try going as near as possible depending on the closeness of ratios.
123 ~ 112 + 9.8 + 1.12 = (80 + 7 + 0.8)% = 87.8%

124 : 10% = 12.4 , 1% = 1.24 , 0.1% = 0.124
110 ~ 99.2 + 9.92 = Already 88. something hence greater than 123/140.


Final order : 115/213 < 123/140 < 110/124

Concept 3:
 
Let's see one final important concept in percentages that has widespread applications in PLD,SI_CI,etc..successive percentage change.


A quantity is increased by 10% and then the changed value is dropped by 4% [Sensing something of the profit loss discount scenarios? ]. What is the net % change from original cost?

Say, quantity is X. You increase it by 10% => X' = X + X*10/100 = X(1 + 10/100)

Now this quantity X' is decreased by 4% => X" = X' - X' * 4/100 = X'(1 - 4/100)
 
= X(1 + 10/100)(1 - 4/100) and some things start to look like our compound interest..
= X(1 - 4/100 + 10/100 - 10*4/10000) = X[1 + ((a - b - ab/100)/100)]
= A net change of (a - b - ab/100) %

So the general formula for two successive % change = a + b + ab/100 as the net % change over the original quantity with appropriate signs for a rise(+ve) and drop(-ve).

 
Since you know the concept now..try discovering what it would have been for three successive % change :)

Like,Rate and Share if you like the content on Lagom. Do leave comments for any doubts/queries. Do not shy away from it until cleared. Responsibility is yours and yours only to clear your doubts. 

Until next time..
Cheers!
AS

2

PnC : Sum of numbers formed by digits provided

This article deals with various possibilities that exist when it comes to questions of finding the sum of all the numbers that can be formed by using some digits provided in the question with some constraints. Let's discuss today how and why these problems are dealt the way they are.
Case 1 : When digits are not repeated and doesn't include zero

Let's take a small example to understand the generic concept. Say, we have, digits 1,2 and 3 and we form all possible three digit numbers from it using each digit only once. We'll get 123,132,213,231,312,321 as the possible 6(=3!) numbers.
But, what about their sum?
Look at it this way : Choose a digit out of these and keep it fixed at unit's place and arrange the other.Following the same for all provided digits we'll get the sum as :
123
213
132
312
231
321
So what eventually happens is we get each of the n digits at each of the places(unit,ten's,etc) for (n - 1)!
times since we fixed one of the digits while forming the arrangement. Assigning face value of each place(1 to unit's, 10 to ten's , 100 to hundredth's , etc) we get the final result as :

(Sum of the digits provided) * (10^(n - 1) + 10^(n - 2) + ... + 1) * (n - 1)!
for sum of all n-digit numbers formed by the given n digits.


So here we get : (1 + 2 + 3) * (100 + 10 + 1) * (3 - 1)! = 6 * 111 * 2 =  1332.


Case 2 : When digits can be repeated and zero is not included.
Going by the same concept here, the only difference here is that when you arrange the digits after fixing one of the digits you still can take the selected digit again. Hence instead of (n - 1)*(n - 2) *... = (n - 1)! options, you will have n*n*n...(n - 1) times = n^(n - 1) permutations.


Hence the formula becomes :(Sum of the digits provided) * (10^(n - 1) + 10^(n - 2) + ... + 1) * n^(n - 1)


Case 3 : When digits include 0 as well.
We know that if we are considering n-digit numbers, zero cannot be placed at the MSB position(or in simple terms the leftmost digit of number). So what we do for this type of a case is we calculate the sum of numbers as we would for Case 1 or Case 2 assuming 0 as part of the digit set and find sum for n-digit numbers(including n).

But subtract from this the sum of all (n - 1) digit numbers that can be formed using other digits using Case 1 or Case 2 again. This is done as a 5 digit number 01234 is same as a 4 digit number 1234.
Hence calculating sum of all 5 digit numbers from 0,1,2,3,4 and subtracting from it sum of 4 digit numbers using 1,2,3,4 will remove the unnecessary sum of 5 digit numbers that start from 0 at MSB.

Hence, say digits : 0,2,3,4 (no repetition)
Then,
Sum of all 4 digit numbers = (0 + 2 + 3 + 4)*(1111)*3! - (2 + 3 + 4)*111*2!


Case 4 : When digit set itself contains repeated digits.
If you have understood the Case 1 and Case 2 by logic you would understand that change would occur only in the permutation part of n! by dividing for repetition that occurs to compensate for repeated permutations.
It's same like : Arrange 1,2,3,4 = 4! but if it's arrange 1,2,3,3, it will be 4!/2! and for that matter to eliminate any confusion arranging 1,2,2,3,3 would be 5!/2!2!. Same goes for this case.

example : Find sum of all 5 digit numbers formed from 1,3,5,5,7 with no digit repeated other than repetition mentioned.

Solution : (1 + 3 + 5 + 5 + 7)*(11111)*[(5 - 1)!/2!] = Answer.


You should be able to understand a question with repetition allowed and a repeated digit set wouldn't make sense.

Hope this article resolves your conflicts with this domain of problems in PnC that appear in MBA examinations. Rate, Like and Share if you like the content. Leave comments here in case of any query.



Cheers!
AS

6