Good news.
Cleared all 3 cut offs comfortably.
0918 was a good paper.
Maths was skewed towards Permutation and Comnbination.
Many questions from that topic.
DI was tough.
There was one set on pipes that was very easy.
No other set was easy.2-3 sets had 1 or 2 easy questions though.
That saved the day.
Verbal was a lengthy paper.
But IMO verbal was pretty easy.
Sunday, June 29, 2008
Wednesday, June 18, 2008
Progressions
Most of the points that I have listed below are very basic.
In a Arithmetic Progression a,b,c.
The arithmetic mean b = (a+c)/2.
If we insert 'n' arithmetic means between 2 terms a and b,
we will get an AP with n+2 terms.
Take 2 numbers 10 and 20.
If we are inserting 1 Arithmetic mean between these 2 terms,
it will be 15.
Now 15 is the Arithmetic mean of 10 and 20.
If we are inserting 3 Arithmetic means between 10 and 20.
The AP becomes
10,12.5,15,17.5,20
Now the Arithmetic mean of the 3 inserted numbers 12.5,15,17.5 is
15 which is the arithmetic mean of the original 2 numbers.
Here in the AP 10,12.5,15,17.5,20 we have 5 terms,
To determine the arithmetic means to be inserted,
we can use the formula
(Difference between the 2 terms)/Number of means to be inserted+1
Lets see the case explained above.
We need to insert 1 AM between 10 and 20.
So 20-10/(1+1)
We get 5.
We need add this to the smaller number.
We get 15.
We see that in an AP with odd number of terms,
first term+last term = second term + second last term = 2*middle term
Some properties that will of help while solving problems.
If Sum of 'P' terms = Sum of 'Q' terms then
Sum of 'P+Q' terms = 0
Consider the AP 2,1,0,-1,-2
Here Sum upto 1 term is 2
Sum of 4 terms is also 2.
Sum of the 5 terms is 0.
If 'P'th term=Q and 'Q'th term=P
'P+Q'th term = 0
Consider 4,3,2,1,0
4th term is 1
1st term is 5
and 5th term is 0
In a Arithmetic Progression a,b,c.
The arithmetic mean b = (a+c)/2.
If we insert 'n' arithmetic means between 2 terms a and b,
we will get an AP with n+2 terms.
Take 2 numbers 10 and 20.
If we are inserting 1 Arithmetic mean between these 2 terms,
it will be 15.
Now 15 is the Arithmetic mean of 10 and 20.
If we are inserting 3 Arithmetic means between 10 and 20.
The AP becomes
10,12.5,15,17.5,20
Now the Arithmetic mean of the 3 inserted numbers 12.5,15,17.5 is
15 which is the arithmetic mean of the original 2 numbers.
Here in the AP 10,12.5,15,17.5,20 we have 5 terms,
To determine the arithmetic means to be inserted,
we can use the formula
(Difference between the 2 terms)/Number of means to be inserted+1
Lets see the case explained above.
We need to insert 1 AM between 10 and 20.
So 20-10/(1+1)
We get 5.
We need add this to the smaller number.
We get 15.
We see that in an AP with odd number of terms,
first term+last term = second term + second last term = 2*middle term
Some properties that will of help while solving problems.
If Sum of 'P' terms = Sum of 'Q' terms then
Sum of 'P+Q' terms = 0
Consider the AP 2,1,0,-1,-2
Here Sum upto 1 term is 2
Sum of 4 terms is also 2.
Sum of the 5 terms is 0.
If 'P'th term=Q and 'Q'th term=P
'P+Q'th term = 0
Consider 4,3,2,1,0
4th term is 1
1st term is 5
and 5th term is 0
Speed Maths 3
Yet another method to find Squares faster.
I will start with an example.
Say We need to find the square of 16.
The nearest 10 is 20.
To get 20, we add 4 to 16,
i.e 16+4=20
Since we have added 4, to get the nearest ten.
We subtract 4 from 16..
16-4=12
To get the square of 16, 20 * 12 + (4 ^ 2)
=240 + 16 = 256.
For a 2 digit number we go for the nearest 10.
If it is a 3 digit number we should go for the nearest 100.
Few more examples
1) 24
Nearest 10 is 20.
24- 4=20
So to get the square, 20 * 28 + (4 ^ 2 )
The answer is 576.
2) 111
Nearest 100 is 100
111-11 = 100
So to get the square 100 * 122 + (11^ 2)
=12200+121
=12321
Will post the next one on progressions.
But new concepts in progressions are limited.
So that will be a small post .
I will start with an example.
Say We need to find the square of 16.
The nearest 10 is 20.
To get 20, we add 4 to 16,
i.e 16+4=20
Since we have added 4, to get the nearest ten.
We subtract 4 from 16..
16-4=12
To get the square of 16, 20 * 12 + (4 ^ 2)
=240 + 16 = 256.
For a 2 digit number we go for the nearest 10.
If it is a 3 digit number we should go for the nearest 100.
Few more examples
1) 24
Nearest 10 is 20.
24- 4=20
So to get the square, 20 * 28 + (4 ^ 2 )
The answer is 576.
2) 111
Nearest 100 is 100
111-11 = 100
So to get the square 100 * 122 + (11^ 2)
=12200+121
=12321
Will post the next one on progressions.
But new concepts in progressions are limited.
So that will be a small post .
Friday, June 13, 2008
Is High reading speed necessary to do well in the Verbal Section
This is something that goes through everybody's mind.
There are all kinds of views on this.
In my opinion "A high reading speed is not necessary to do well
in the verbal section". It is certainly not sufficient to do well
in the verbal section.
Off late , CAT has moved towards smaller passages.
So Numbe of words in a RC Passage is typically 800-900
as opposed to 1200+.Also passages are very abstract and difficult.
So you may have to read the passage , and then come back to the passage
after taking a look at the questions.Considering the difficulty level of the CAT
passages these days, I think regression is unavoidable.
CAT has increased the time available and simultaneously reduced the
number of questions.Even if you have a very high reading speed , assuming
that you get 3 RC passages in CAT , the time you save will be 2-3 minutes.
Now IMHO this is negligible.
What is required is a moderate reading speed and good comprehension.
Also an ability to read vague and abstract passages.There are times when I
attempt a very difficult passage but the effort of reading drains me
completely and I have no enthusiasm to attend the next section.
One must ideally read a lot and read from a cross section of subjects.
Increased reading is the only way to do well in RCs.If you do have a
great reading speed, feel happy even though it may not give you a
great advantage in CAT.Everybody who talks about reading speed,
talks about the case studies and the amount of reading required while
doing MBA.So maybe your reading speed will keep you in good stead there.
What is a high reading speed ? 500+
Moderate- 300+
Even with a speed of 250+, one should be able to RCs well.
But a speed less than that means no regular reading or more likely
an aversion to reading.Now that's not a good thing.Read a few books,
read 2 papers regularly,read editorials regularly.Read atleast 1 weekly
magazine.6 months of this should definitely bring the reading speed to
250+. That should be enough to keep your heads above the water.
There are all kinds of views on this.
In my opinion "A high reading speed is not necessary to do well
in the verbal section". It is certainly not sufficient to do well
in the verbal section.
Off late , CAT has moved towards smaller passages.
So Numbe of words in a RC Passage is typically 800-900
as opposed to 1200+.Also passages are very abstract and difficult.
So you may have to read the passage , and then come back to the passage
after taking a look at the questions.Considering the difficulty level of the CAT
passages these days, I think regression is unavoidable.
CAT has increased the time available and simultaneously reduced the
number of questions.Even if you have a very high reading speed , assuming
that you get 3 RC passages in CAT , the time you save will be 2-3 minutes.
Now IMHO this is negligible.
What is required is a moderate reading speed and good comprehension.
Also an ability to read vague and abstract passages.There are times when I
attempt a very difficult passage but the effort of reading drains me
completely and I have no enthusiasm to attend the next section.
One must ideally read a lot and read from a cross section of subjects.
Increased reading is the only way to do well in RCs.If you do have a
great reading speed, feel happy even though it may not give you a
great advantage in CAT.Everybody who talks about reading speed,
talks about the case studies and the amount of reading required while
doing MBA.So maybe your reading speed will keep you in good stead there.
What is a high reading speed ? 500+
Moderate- 300+
Even with a speed of 250+, one should be able to RCs well.
But a speed less than that means no regular reading or more likely
an aversion to reading.Now that's not a good thing.Read a few books,
read 2 papers regularly,read editorials regularly.Read atleast 1 weekly
magazine.6 months of this should definitely bring the reading speed to
250+. That should be enough to keep your heads above the water.
Thursday, June 12, 2008
TIME AIMCAT 0919
The results of the next AIMCAT is out.
Compared to 920, This went a lot better for me and that
shows in the result.I have 2 ver good sections -DI and Verbal.
Quant was horrible.
Made many stupid mistakes in Quant.Need to improve the accuracy
and be careful while doing Quant.
Positives :
.For the first time, very good accuracy rate in Verbal and DI
.A good DI section after a long time.
.After a poor verbal in CAT 07, Good verbal scores in the mocks
give me some confidence
Negatives:
No improvement in Quant despite the effort put in.
Need to do practice more sums ....
Compared to 920, This went a lot better for me and that
shows in the result.I have 2 ver good sections -DI and Verbal.
Quant was horrible.
Made many stupid mistakes in Quant.Need to improve the accuracy
and be careful while doing Quant.
Positives :
.For the first time, very good accuracy rate in Verbal and DI
.A good DI section after a long time.
.After a poor verbal in CAT 07, Good verbal scores in the mocks
give me some confidence
Negatives:
No improvement in Quant despite the effort put in.
Need to do practice more sums ....
Wednesday, June 11, 2008
Last 2 digits of an Expression
Last 2 digits is the remainder when Divided by 100.
This can be found usind Binomial theorem.
The method described below is quick and is
an extension of the Binomial Theorem.
For expressions ending in '5'
Last 2 digits will always be 25.
For '1'
I will first so some examples and then follow them with an
explanation.
Take 41 ^ 73
The last digit/ the right most digit will always be 1
For the last but one digit take the ten's digit of the base i.e 4 in the above
example and multiply it will the unit's digit of the power i.e 3 in the example.
So 2.
The last 2 digits will be 21.
Some more examples.
31 ^ 123 =
Right most is one.
Last but one will be 3 (ten's digit of the base) * 3 (unit's digit of the power)=9
So last 2 digits will be 91
81 ^ 1423 = 41
Last but one here will be 8*3
For Numbers ending in 3,7,9
From the power cycles,
3 ^ 4k/4k+4 and 7 ^ 4k/4k+4 , 9 ^ 2k will end in 1.
So using this we can convert them into expressions ending in 1.
233 ^ 56
The last 2 digits will depend on 33 ^ 56
This becomes (33 ^ 4 ) ^ 14
33 ^ 4 is (33 ^ 2 ) * (33 ^ 2)
I have already written about quicker methods to get squares.
33 ^ 2 will end in 89
Finally we get 21 ^ 14
This can be found using the method explained for 1.
For expressions ending in even numbers
Some pointers to remember
1024 ^ even number will always end in 76
1024 ^ odd number will always end in 24
1024 is 2 ^ 10
For most expressions ending in an even number,
we can take 2 ^ x as common and express that as 10k+
Say we have 74 ^ 288
74 is 2 * 37
so we get (2 ^ 288)*(37^288)
2 ^ 288 will be (2 ^ 10)^28 * (2 ^ 8)
We can proceed with this
and use earlier methods to get 37 ^ 288.
Agreed it's not an easy method by any means.
This can be found usind Binomial theorem.
The method described below is quick and is
an extension of the Binomial Theorem.
For expressions ending in '5'
Last 2 digits will always be 25.
For '1'
I will first so some examples and then follow them with an
explanation.
Take 41 ^ 73
The last digit/ the right most digit will always be 1
For the last but one digit take the ten's digit of the base i.e 4 in the above
example and multiply it will the unit's digit of the power i.e 3 in the example.
So 2.
The last 2 digits will be 21.
Some more examples.
31 ^ 123 =
Right most is one.
Last but one will be 3 (ten's digit of the base) * 3 (unit's digit of the power)=9
So last 2 digits will be 91
81 ^ 1423 = 41
Last but one here will be 8*3
For Numbers ending in 3,7,9
From the power cycles,
3 ^ 4k/4k+4 and 7 ^ 4k/4k+4 , 9 ^ 2k will end in 1.
So using this we can convert them into expressions ending in 1.
233 ^ 56
The last 2 digits will depend on 33 ^ 56
This becomes (33 ^ 4 ) ^ 14
33 ^ 4 is (33 ^ 2 ) * (33 ^ 2)
I have already written about quicker methods to get squares.
33 ^ 2 will end in 89
Finally we get 21 ^ 14
This can be found using the method explained for 1.
For expressions ending in even numbers
Some pointers to remember
1024 ^ even number will always end in 76
1024 ^ odd number will always end in 24
1024 is 2 ^ 10
For most expressions ending in an even number,
we can take 2 ^ x as common and express that as 10k+
Say we have 74 ^ 288
74 is 2 * 37
so we get (2 ^ 288)*(37^288)
2 ^ 288 will be (2 ^ 10)^28 * (2 ^ 8)
We can proceed with this
and use earlier methods to get 37 ^ 288.
Agreed it's not an easy method by any means.
Friday, June 6, 2008
Speed Maths 2
Squares of Numbers from 70-130
Let the Number be X such that X is between 70 and 130.
Step 1: X-100.
Add this value to X.
The value will be the left most 2 or 3(As the case maybe) digits of the answer.
Step 2: Square (X-100)
This will be the right most 2 digits.
In case you get 3 digits, carry over to the left digit.
Example:
1) Number is 82.
82-100 = -18
82-18=64
Now 18 ^ 2 is 324.
3 will be carried over to 4 and that becomes 67.
Answer is 6724
2) 112
112-100 = 12
112 + 12 =124
124 will be the 3 left most digits.
12 ^ 2 is 144.
1 will be carried over to the 4 and
the Answer is 12544
Let the Number be X such that X is between 70 and 130.
Step 1: X-100.
Add this value to X.
The value will be the left most 2 or 3(As the case maybe) digits of the answer.
Step 2: Square (X-100)
This will be the right most 2 digits.
In case you get 3 digits, carry over to the left digit.
Example:
1) Number is 82.
82-100 = -18
82-18=64
Now 18 ^ 2 is 324.
3 will be carried over to 4 and that becomes 67.
Answer is 6724
2) 112
112-100 = 12
112 + 12 =124
124 will be the 3 left most digits.
12 ^ 2 is 144.
1 will be carried over to the 4 and
the Answer is 12544
Thursday, June 5, 2008
Speed Maths 1
A small post.
For Finding squares of numbers between 30-70
Say the number is 'X' and X is between 30 and 70.
Step 1: Find x-25.
X-25 will be the first digit/first 2 digits of the required answer.
Step 2:Find (50-X)^2
This will be the right most 2 digits.In case the square is a 3 digit number,
carry it over .
Example: 43
43-25
=18
50-43
=7
7 ^ 2 = 49
43 ^ 2 is 1849
Example 2:35
35-25=10
50-35=15
15 ^ 2 =225
Answer: 10, from 225 25 will be the 2 right most digits and carry over 2 to the 0 on the left side
So it becomes 1225.
Answer: 1225
For Finding squares of numbers between 30-70
Say the number is 'X' and X is between 30 and 70.
Step 1: Find x-25.
X-25 will be the first digit/first 2 digits of the required answer.
Step 2:Find (50-X)^2
This will be the right most 2 digits.In case the square is a 3 digit number,
carry it over .
Example: 43
43-25
=18
50-43
=7
7 ^ 2 = 49
43 ^ 2 is 1849
Example 2:35
35-25=10
50-35=15
15 ^ 2 =225
Answer: 10, from 225 25 will be the 2 right most digits and carry over 2 to the 0 on the left side
So it becomes 1225.
Answer: 1225
Tuesday, June 3, 2008
CoPrimes, Euler's Number and Fermat's theorem
Co-Primes are very important , when it comes to finding remainders.
First things first, What is a Co-Prime ?
2 numbers are said to be co-primes/relatively prime if their HCF is 1.
That is the 2 numbers have no other common factor apart from the number.
Co-Primes by themselves serve no purpose.You need to use them with
Fermat's theorem / Chinese Remainder Theorem etc.
Fermat's theorem states that
Remainder (x ^ ( Euler's Number of N) )/N ) = 1
Here X and N have to Co-Primes.
Now , what is Euler's Number?
Euler's number of a particular number is the number of co-primes less than
the number. The total number of co-primes that a certain number can have
is infinite. So we need to determine the number of coprimes less than the number.
Euler's number for a prime number P is P-1. As all numbers less than that
prime number will coprimes of the prime number. As in 1,2,3,4,5,6 will all be
coprimes of 7.
For other numbers :
Euler's Number of a number X = X(1-1/a)(1-1/b)....
Here a and b are the prime factors of the number.
We look at the number of the prime numbers that are factors of the given number.
We ignore the number of times the prime number occurs.
So say the number is 18.
18 has 2 and 3 as factors.
So Eulers No of 18 = 18(1-1/2)(1-1/3)
= 6
So if we get a sum like ,
Find the remainder when (11 ^ 97 )/7
1)See if 11 and 7 are relatively prime to each other
2)If they are relatively prime, Find the Euler's number of the denominator.
3)Express power in terms of the euler's number.
For eg, in this case the Euler's number for 7 is 6.
So 97 = 6k+1.
We know that if we can factorize the numerator in a division, the end remainder
will be the product of the remainders of each individual factor.
So this becomes (11 ^ 6)/7.....(16 times)* (11 ^ 1)/7
Now as per Fermat's theorem,
(11 ^ 6) /7 is 1.
So the answer is the remainder when (11^1) is divided by 7.
This can be used when the numerator can be factorized.
This can be used when the numerator and the denominator are co-primes/relatively prime.
Now The question arises what is they are not coprimes.
If 2 numbers are not co-primes,it means they have a common factor.
See if you can remove the common factor from both the numerator and denominator.
In case that can't be done , we will have to use other methods to find the remainder.
First things first, What is a Co-Prime ?
2 numbers are said to be co-primes/relatively prime if their HCF is 1.
That is the 2 numbers have no other common factor apart from the number.
Co-Primes by themselves serve no purpose.You need to use them with
Fermat's theorem / Chinese Remainder Theorem etc.
Fermat's theorem states that
Remainder (x ^ ( Euler's Number of N) )/N ) = 1
Here X and N have to Co-Primes.
Now , what is Euler's Number?
Euler's number of a particular number is the number of co-primes less than
the number. The total number of co-primes that a certain number can have
is infinite. So we need to determine the number of coprimes less than the number.
Euler's number for a prime number P is P-1. As all numbers less than that
prime number will coprimes of the prime number. As in 1,2,3,4,5,6 will all be
coprimes of 7.
For other numbers :
Euler's Number of a number X = X(1-1/a)(1-1/b)....
Here a and b are the prime factors of the number.
We look at the number of the prime numbers that are factors of the given number.
We ignore the number of times the prime number occurs.
So say the number is 18.
18 has 2 and 3 as factors.
So Eulers No of 18 = 18(1-1/2)(1-1/3)
= 6
So if we get a sum like ,
Find the remainder when (11 ^ 97 )/7
1)See if 11 and 7 are relatively prime to each other
2)If they are relatively prime, Find the Euler's number of the denominator.
3)Express power in terms of the euler's number.
For eg, in this case the Euler's number for 7 is 6.
So 97 = 6k+1.
We know that if we can factorize the numerator in a division, the end remainder
will be the product of the remainders of each individual factor.
So this becomes (11 ^ 6)/7.....(16 times)* (11 ^ 1)/7
Now as per Fermat's theorem,
(11 ^ 6) /7 is 1.
So the answer is the remainder when (11^1) is divided by 7.
This can be used when the numerator can be factorized.
This can be used when the numerator and the denominator are co-primes/relatively prime.
Now The question arises what is they are not coprimes.
If 2 numbers are not co-primes,it means they have a common factor.
See if you can remove the common factor from both the numerator and denominator.
In case that can't be done , we will have to use other methods to find the remainder.
Monday, June 2, 2008
Tests for Divisibility
Test for Divisibility:
This again is a very basic topic.
Most of u know the tests to check divisibility.
But how many of us know why those particular methods are employed ?
After the class, the logic behind the tests look so simple and I feel dumb
for not having realised this earlier.I guess most of you know the reason
as well. But I am sure there will be a fair number who don't know why.
For their sake, I am making this post.
For 3,9
Everybody knows that for 3 and 9 we find the digitsum and see if that
digit sum is divisible by 3 or 9 , whatever the case maybe.
Let us take a 2 digit number.
A two digit can be written as 10A+B
10A+B=9A+A+B
Now, 9A is always divisible by 3/9.So whether the number is divisible by 3 or 9
depends on A+B.
Any number will be of the form 10 ^ 3 A + 10 ^ 2 B + 10 C+10^0 D .....................
If we divide each co-efficient (i.e 10 ^ 0 , 10 ^ 1 , 10 ^ 2 ............) by 9, the remainder is +1.
That is why we add all the digits.
For 11,7..
Taking the previous example.
10 ^ 3 A + 10 ^ 2 B + 10 C+10^0 D
If we divide 10^0 by 11 remainder is +1.
if we divide 10 ^ 1 by 11 remainder is -1 or (10)
If we divide 10 ^ 2 by 11 remainder is + 1
If we divide 10 ^ 3 by 11 remainder is -1.
We see that for 11, it alternates between + and -.
That is why we use the difference of the sum of odd terms and sum of even terms and see if that difference
is divisible by 11.
For 7.
10 ^ 3 A + 10 ^ 2 B + 10 C+10^0 D
If we divide 10^0 by 7 remainder is +1.
if we divide 10 ^ 1 by 7 remainder is 3
If we divide 10 ^ 2 by 7 remainder is + 2
If we divide 10 ^ 3 by 7 remainder is -1 or (6)
Here we see the same pattern as in 11, but for every 3 digits.
Thats why for 7 we use difference of sum of alternate triplets.
The same holds good for 13 .
For 5,25,125.......
The first power of ten that exactly divides 5 is 1.
That is why we use the last digit of a number to see if it is
divisible by 5.
The first power of ten that exactly divides 25 is 2.
That is why we use the last 2 digits of a number to see if it is
divisible by 25
The first power of ten that exactly divides 125 is 3.
That is why we use the last 3 digits of a number to see if it is
divisible by 125
The first power of ten that exactly divides 4 is 2.
That is why we use the last 2 digits of a number to see if it is
divisible by 4
The first power of ten that exactly divides 8 is 3.
That is why we use the last 3 digits of a number to see if it is
divisible by 8
This again is a very basic topic.
Most of u know the tests to check divisibility.
But how many of us know why those particular methods are employed ?
After the class, the logic behind the tests look so simple and I feel dumb
for not having realised this earlier.I guess most of you know the reason
as well. But I am sure there will be a fair number who don't know why.
For their sake, I am making this post.
For 3,9
Everybody knows that for 3 and 9 we find the digitsum and see if that
digit sum is divisible by 3 or 9 , whatever the case maybe.
Let us take a 2 digit number.
A two digit can be written as 10A+B
10A+B=9A+A+B
Now, 9A is always divisible by 3/9.So whether the number is divisible by 3 or 9
depends on A+B.
Any number will be of the form 10 ^ 3 A + 10 ^ 2 B + 10 C+10^0 D .....................
If we divide each co-efficient (i.e 10 ^ 0 , 10 ^ 1 , 10 ^ 2 ............) by 9, the remainder is +1.
That is why we add all the digits.
For 11,7..
Taking the previous example.
10 ^ 3 A + 10 ^ 2 B + 10 C+10^0 D
If we divide 10^0 by 11 remainder is +1.
if we divide 10 ^ 1 by 11 remainder is -1 or (10)
If we divide 10 ^ 2 by 11 remainder is + 1
If we divide 10 ^ 3 by 11 remainder is -1.
We see that for 11, it alternates between + and -.
That is why we use the difference of the sum of odd terms and sum of even terms and see if that difference
is divisible by 11.
For 7.
10 ^ 3 A + 10 ^ 2 B + 10 C+10^0 D
If we divide 10^0 by 7 remainder is +1.
if we divide 10 ^ 1 by 7 remainder is 3
If we divide 10 ^ 2 by 7 remainder is + 2
If we divide 10 ^ 3 by 7 remainder is -1 or (6)
Here we see the same pattern as in 11, but for every 3 digits.
Thats why for 7 we use difference of sum of alternate triplets.
The same holds good for 13 .
For 5,25,125.......
The first power of ten that exactly divides 5 is 1.
That is why we use the last digit of a number to see if it is
divisible by 5.
The first power of ten that exactly divides 25 is 2.
That is why we use the last 2 digits of a number to see if it is
divisible by 25
The first power of ten that exactly divides 125 is 3.
That is why we use the last 3 digits of a number to see if it is
divisible by 125
The first power of ten that exactly divides 4 is 2.
That is why we use the last 2 digits of a number to see if it is
divisible by 4
The first power of ten that exactly divides 8 is 3.
That is why we use the last 3 digits of a number to see if it is
divisible by 8
Power cycles
I will make a series of posts detailing Power cycles,Divisibility Tests,Remainders
using Basic Remainder Theorem,Fermat's Theorem,Chinese Remainder theorem ,
Constant remainders in an A P and successive remainders.
Power cycles is an easy and fairly well known concept.
However for the benefit of those who donot know what it means
I will provide some explanation .
Power Cycles can be used to find the last digit of a number raised to some
power. Its limited in it's use as we can only get the last digit.
Now the idea is 1 raised to any power will end in 1.
So the power cycle for 1 is 1 with a frequency of 1.
For 2,it can be 2, 4, 8 and then 6 .After 6 we get 2 and the cycle repeats.
As in 2 ^ 1 = 2 , 2 ^ 2 =4,2^3 =8 ,2 ^ 4 =16, 2^ 5 =32 and so on.
Similarly for 3, It is 3,9,7 and 1
For 4, it is 4 and 6
For 5 it is 5
For 6 it is 6
And for 7 it is 7,9,3 and 1
For 8 it is 8,4,2,6
For 9 it is 9 and 1
For 0 it is 0.
So 2,3,7 and 8 have power cycles with frequency 4.
0,1,5,6 have power cycles with frequency 1.
4 and 9 have a frequency 2.
Now if we have to find the last digit of 2 ^ 73,
We need to express 73 as 4k+1/4k+2/4k+3/4k (4 because the
Frequency of 2’s power cycle is 4).
Here in this case it is in the format 4k+1 , so the last digit will be 2.
If it is of the form 4k+2, the last digit will be 4 and if 4k+3 then 8 and so on.
Similarly we can use it for all other numbers.
The post is getting bigger and bigger,
Will elaborate on other concepts in a separate post.
using Basic Remainder Theorem,Fermat's Theorem,Chinese Remainder theorem ,
Constant remainders in an A P and successive remainders.
Power cycles is an easy and fairly well known concept.
However for the benefit of those who donot know what it means
I will provide some explanation .
Power Cycles can be used to find the last digit of a number raised to some
power. Its limited in it's use as we can only get the last digit.
Now the idea is 1 raised to any power will end in 1.
So the power cycle for 1 is 1 with a frequency of 1.
For 2,it can be 2, 4, 8 and then 6 .After 6 we get 2 and the cycle repeats.
As in 2 ^ 1 = 2 , 2 ^ 2 =4,2^3 =8 ,2 ^ 4 =16, 2^ 5 =32 and so on.
Similarly for 3, It is 3,9,7 and 1
For 4, it is 4 and 6
For 5 it is 5
For 6 it is 6
And for 7 it is 7,9,3 and 1
For 8 it is 8,4,2,6
For 9 it is 9 and 1
For 0 it is 0.
So 2,3,7 and 8 have power cycles with frequency 4.
0,1,5,6 have power cycles with frequency 1.
4 and 9 have a frequency 2.
Now if we have to find the last digit of 2 ^ 73,
We need to express 73 as 4k+1/4k+2/4k+3/4k (4 because the
Frequency of 2’s power cycle is 4).
Here in this case it is in the format 4k+1 , so the last digit will be 2.
If it is of the form 4k+2, the last digit will be 4 and if 4k+3 then 8 and so on.
Similarly we can use it for all other numbers.
The post is getting bigger and bigger,
Will elaborate on other concepts in a separate post.
Sunday, May 25, 2008
Byju's free quant workshop
Byju did a free quant workshop yesterday at the JyothiNivas College,koramangala.
Nearly 250 people attended the workshop.
Byju started with Speed Calculation techniques.The first thing was how to convert all divisions to single digit divisions.Then he moved on to squares and square roots.He spent a good 20 minutes on these techniques.
Then he moved on to Algebra.He emphasized on converting a standard sum to a specific sum using
simple sample values.He showed some examples where using this strategy simplifies the problem.
He spent some time on geometry and Permutation and Combination.
A major part of the workshop was on NUMBERS.He showed the use of power cycles.He also explained how Basic remainder theorem,Fermat's theorem,Chinese remainder theorem can be used to find the remainders.
Merely stating the content of the workshop is of no use to anybody.I do realize that.But my grasp on the techniques that Byju explained yesterday is very limited.In a few days , I hope to practice some of the techniques/Methods and gain a much better understanding.At that point in time I will make detailed posts on the techniques and how they can be used to solve CAT problems.
Coming to Byju himself, He is enormously talented.There is no doubt there.
But he certainly doesnot suffer from humility.He brags and boasts about
his capabilities every couple of minutes.We have to sit through that.But if
we can tolerate his arrogance , which stem from his superior capabilities ,
there is a lot to gain from his classes.His classes last continuosly for 4+ hours and
he teaches to a class of 200+ people.Inspite of these huge numbers , the takeaway
is huge and can make a difference.
Nearly 250 people attended the workshop.
Byju started with Speed Calculation techniques.The first thing was how to convert all divisions to single digit divisions.Then he moved on to squares and square roots.He spent a good 20 minutes on these techniques.
Then he moved on to Algebra.He emphasized on converting a standard sum to a specific sum using
simple sample values.He showed some examples where using this strategy simplifies the problem.
He spent some time on geometry and Permutation and Combination.
A major part of the workshop was on NUMBERS.He showed the use of power cycles.He also explained how Basic remainder theorem,Fermat's theorem,Chinese remainder theorem can be used to find the remainders.
Merely stating the content of the workshop is of no use to anybody.I do realize that.But my grasp on the techniques that Byju explained yesterday is very limited.In a few days , I hope to practice some of the techniques/Methods and gain a much better understanding.At that point in time I will make detailed posts on the techniques and how they can be used to solve CAT problems.
Coming to Byju himself, He is enormously talented.There is no doubt there.
But he certainly doesnot suffer from humility.He brags and boasts about
his capabilities every couple of minutes.We have to sit through that.But if
we can tolerate his arrogance , which stem from his superior capabilities ,
there is a lot to gain from his classes.His classes last continuosly for 4+ hours and
he teaches to a class of 200+ people.Inspite of these huge numbers , the takeaway
is huge and can make a difference.
Thursday, May 22, 2008
AIMCAT 920 results
The results are out.
My Verbal scores are right up there.
Quant and DI- the less said, the better.
The very poor showing in 2 very important sections even before the battle has begun,
is frightening to say the least.
Need to pick up momentum.
Long Nights ahead..............................
My Verbal scores are right up there.
Quant and DI- the less said, the better.
The very poor showing in 2 very important sections even before the battle has begun,
is frightening to say the least.
Need to pick up momentum.
Long Nights ahead..............................
Wednesday, May 21, 2008
Byju CAT Classes
When I went to Jyothi Nivas College to take the mock,a few people were
standing outside the college issuing a notice for a free quant workshop by
Byju.The notice said that Byju was a serial CAT topper .
Since the workshop was supposed to start at 1 PM only , I thought I will
take the mock and then attend the workshop.But the mock got over only
at 2 PM and I had to miss the workshop.
I returned home and googled Byju. There were lots of positive reviews for
the classes that he conducts.Some folks had reviewed his classes in a
Pagalguy thread for Bangalore coaching centers.
Looks like Byju is doing another workshop this coming Saturday at 4 PM.
Venue:Jyothi Nivas College.
If you were one of those guys who missed the last workshop, go for this.
Best of Luck..............................
standing outside the college issuing a notice for a free quant workshop by
Byju.The notice said that Byju was a serial CAT topper .
Since the workshop was supposed to start at 1 PM only , I thought I will
take the mock and then attend the workshop.But the mock got over only
at 2 PM and I had to miss the workshop.
I returned home and googled Byju. There were lots of positive reviews for
the classes that he conducts.Some folks had reviewed his classes in a
Pagalguy thread for Bangalore coaching centers.
Looks like Byju is doing another workshop this coming Saturday at 4 PM.
Venue:Jyothi Nivas College.
If you were one of those guys who missed the last workshop, go for this.
Best of Luck..............................
Tuesday, May 20, 2008
TIME AIMCAT 920
TIME conducted it's first AIMCAT for CAT 2008 on May 18th. TIME AIMCAT series was originally supposed to start on the 25th. IMS which usually has it's mocks from July decided to start early and increase the number of mocks forcing TIME to change the dates
and reduce the prices as well.A good bargain for CAT aspirants in the end.
Since both the IMS SIMCAT and TIME AIMCAT were held at the same time,I decided
to go with TIME as I have done these last couple of years.TIME did a bad job in terms of organizing the first AIMCAT,at least that's how it was in my center. There were lots of people walking in after ten without prior registration.TIME had very few people to manage the crowd.
The test started 20 minutes late.
The mock itself was pretty okay and was in the lines of last year's CAT.No
surprises there. I started with the verbal section(again no surprises there).
Did a few questions. Then moved to quant.Did the first 2 questions and
then came back to verbal. Completed it.Went to quant and then DI.
Not a great strategy, I admit.I was just experimenting.
As has been the case these last few years, I did well in verbal but got
badly screwed in Quant and DI.I have always started the AIMCAT
series well.I managed a sub 150 rank last time and I had 98+ percentile
the year before that. I lose momentum after the first 7-8 mocks.I
started missing sectional cut-offs regularly.Then the overall ranking and
the percentile drop to alarmingly low levels.
By all indications , It's a bad start this year.
I hope things get better from here
TAKEAWAYS:
and reduce the prices as well.A good bargain for CAT aspirants in the end.
Since both the IMS SIMCAT and TIME AIMCAT were held at the same time,I decided
to go with TIME as I have done these last couple of years.TIME did a bad job in terms of organizing the first AIMCAT,at least that's how it was in my center. There were lots of people walking in after ten without prior registration.TIME had very few people to manage the crowd.
The test started 20 minutes late.
The mock itself was pretty okay and was in the lines of last year's CAT.No
surprises there. I started with the verbal section(again no surprises there).
Did a few questions. Then moved to quant.Did the first 2 questions and
then came back to verbal. Completed it.Went to quant and then DI.
Not a great strategy, I admit.I was just experimenting.
As has been the case these last few years, I did well in verbal but got
badly screwed in Quant and DI.I have always started the AIMCAT
series well.I managed a sub 150 rank last time and I had 98+ percentile
the year before that. I lose momentum after the first 7-8 mocks.I
started missing sectional cut-offs regularly.Then the overall ranking and
the percentile drop to alarmingly low levels.
By all indications , It's a bad start this year.
I hope things get better from here
TAKEAWAYS:
Need to stick to a proper time plan.
Should not keep DI for last
Need to practice DI regularly
I missed a few sitters in Quant.Should read all quant questions in the future.
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