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Discussion

  • 2016-11-04 Bhushan Badgujjar

    72

  • 2016-11-04 Bhushan Badgujjar

    4 ways

  • 2016-11-04 Bhushan Badgujjar

    10

  • 2016-11-04 Bhushan Badgujjar

    180^9

  • 2016-11-04 Bhushan Badgujjar

    124416

  • 2016-11-04 Bhushan Badgujjar

    1000

  • 2016-11-04 Bhushan Badgujjar

    49

  • 2016-11-04 Bhushan Badgujjar

    30

  • 2016-11-04 Mahesh Bharadwaj

    C) 16

  • 2016-11-04 Mahesh Bharadwaj

    C) 10 Min

  • 2016-11-04 Mahesh Bharadwaj

    A 30

  • 2016-11-04 Shashank Prabhu

    3. A sprinter starts running on a circular path of radius r metres. Her average speed (in metres/minute) is πr during the first 30 seconds, πr/2 during next one minute, πr/4 during next 2 minutes, πr/8 during next 4 minutes, and so on. What is the ratio of the time taken for the nth round to that for the previous round? (CAT 2004)

    (a) 4 (b) 8 (c) 16 (d) 32

  • 2016-11-04 Shashank Prabhu

    2. In a 4000 meter race around a circular stadium having a circumference of 1000 meters, the fastest runner and the slowest runner reach the same point at the end of the 5th minute, for the first time after the start of the race. All the runners have the same staring point and each runner maintains a uniform speed throughout the race. If the fastest runner runs at twice the speed of the slowest runner, what is the time taken by the fastest runner to finish the race? (CAT 2003 leaked)

    (a) 20 min (b) 15 min (c) 10 min (d) 5 min

  • 2016-11-04 Shashank Prabhu

    1. Three students Arun, Barun and Kiranmala start moving around a circular track of length 60m from the same point simultaneously in the same direction at speeds of 3 m/s, 5 m/s and 9 m/s respectively. When will they meet for the first time after they started moving?

    (a) 30 seconds (b) 60 seconds (c) 15 seconds (d) 10 seconds

  • 2016-11-04 Shashank Prabhu

    Circular motion is a topic feared by many when it comes to CAT and other entrance tests. While the concepts are few, it takes a lot of imagination to solve the questions. Also, the questions are easier compared to what we see in mocks and so, the people who give up on the concept end up losing out on easy marks most of the time. Let's look at the basic concepts now:

    1. Number of points of meeting for two objects A and B moving in the same direction along a circular path will be |a-b| where a:b is the ratio of their speeds in the simplest form
    2. Number of points of meeting for two objects A and B moving in the opposite direction along a circular path will be |a+b| where a:b is the ratio of their speeds in the simplest form
    3. If the objects are moving in the same direction, the faster object will have to cover a circumference more than the slower object when they meet and they will meet after a period C/|a-b| where a and b are the speeds of the two objects
    4. If the objects are moving in the opposite direction, both of them would have covered a distance equal to the circumference of the track and they will meet after a period C/|a+b| where a and b are the speeds of the two objects
    5. The time after which, the objects moving along a circular track will at the origin will be given by LCM(times taken by individual objects to complete a single round along the track)

    For more than two objects, you can take pairs and then plot the meeting points and then deal with the common points. Also remember that the meeting points will be equally spaced and they will be covered in a fixed order before the objects meet again at the starting point.

  • 2016-11-03 Mahesh Bharadwaj

    Sir I have question, What if the Power is less than the Euler Number ? How to approach such questions ?
    Ex : 12^107 mod 37
    Using Euler's I am ending up with 12^35, next step ?
    Thanks

  • 2016-10-28 Shashank Prabhu

    3. The sum of 3 positive integers a, b, and c is equal to 30. What is the maximum possible value of a*b^2*c^2

  • 2016-10-28 Shashank Prabhu

    2. The sum of 3 positive integers is 30. What would be the maximum value of their product?

  • 2016-10-28 Shashank Prabhu

    1. The sum of two positive numbers is 14. What would be the maximum value of their product?

  • 2016-10-28 Shashank Prabhu

    Happy Diwali guys! Today I will be covering the AM, GM, HM concept. This is completely theoretical and there would be a few questions which you simply won't be able to solve if you don't know the concept. To start with,

    1. AM is the arithmetic mean or average of n numbers. So, for a series a, b, c, ... n, AM is given by (a+b+c...n)/n
    2. GM is the geometric mean of n numbers. So, for a series a, b, c, ... n, GM is given by (a*b*c...n)^(1/n)
    3. HM is the harmonic mean of n numbers. So, for a series a, b, c, ... n, HM is given by n/(1/a+1/b+1/c...1/n)

    The basic identity that we need to know is that, for any set of n 'positive' numbers, AM >= GM >= HM

    Most of the questions are easy and involve a direct application of AM >= GM. Also, we simply need to understand how to take the variables and we are almost done. For example:

    Given that 5x+2y=50. What is the maximum value of x^2 * y^3?

    Now, we have to split the variables in the sum into equal parts using the number in the powers. So, 5x has to be split into 2 equal parts (because the power of x is 2) which will be 5x/2 and 5x/2. Similarly, 2y has to be split into 3 parts which will be 2y/3, 2y/3 and 2y/3.

    Now we have 5 elements the sum of which will give us something in the form of 5x+2y and the product of which will give the variable as x^2 * y^3

    So, using the AM >= GM inequality,

    (5x/2 + 5x/2 + 2y/3 + 2y/3 + 2y/3)/5 >= (5x/2 * 5x/2 * 2y/3 * 2y/3 * 2y/3)^(1/5)
    (5x+2y)/5 >= (50/27*x^2*y^3)^1/5
    10 >= (50/27*x^2*y^3)^1/5
    10^5 >= (50/27*x^2*y^3)
    x^2 * y^3 has to be less 54000.

    Let's solve a few more questions from this topic.

  • 2016-10-21 Mahesh Bharadwaj

    Hello Sir,
    164 = 41*4
    39! mod 41 gives 1
    39! mod 4 gives 0
    41a+1=4b
    a=3, b=31
    Hence, Remainder will be 124

  • 2016-10-21 Shashank Prabhu

    4. How many natural numbers exist such that they are coprime to 216 and are less than 216?

  • 2016-10-21 Shashank Prabhu

    3. Find the number of ways in which you can express 36800 as a product of two positive integers coprime to each other.

  • 2016-10-21 Shashank Prabhu

    2. Find the number of ways in which you can write 21680 as a product of 2 of its factors.

  • 2016-10-21 Shashank Prabhu

    1. Find the product of all the factors of 180.

  • 2016-10-21 Shashank Prabhu

    This week, I will be covering the remaining concepts from factors. There are direct applications of the same and you simply need to remember the formulas and answer the questions. The direct formulas for each type are as follows:

    1. Product of factors of a number N that has f factors is given by:

    P(N) = [N]^(f/2)

    For example: 6 has 4 factors 1, 2, 3, 6 and so, the product is 1*2*3*6=36 which is the same as 6^(4/2) = 6^2 = 36

    2. Number of ways in which a number N that has f factors can be expressed as a product of two factors is simply given by [f/2] where [x] is a number greater than or equal to x. In simple words,

    number of ways in which 6 can be expressed as a product of 2 factors is given by [4/2] i.e. 2
    number of ways in which 16 can be expressed as a product of 2 factors is given by [5/2] = [2.5] = 3 viz. (1, 16)(2, 8)(4, 4)

    3. Number of ways in which a number N = a^x * b^y * c^x ... n^k where a, b, c are the prime factors can be expressed as a product of two factors co prime to each other in 2^(n-1) ways. So, a number say 180 which can be expressed as 2^2*3^2*5 can be expressed as a product of 2 coprimes in 2^(3-1) = 4 ways viz. (1, 180)(4, 45)(9, 20)(5, 36)

    4. Number of positive integers that are less than a number N = a^m * b^n * c^p and are coprime to it is simply the Euler's totient function given by N * (1-1/a)(1-1/b)(1-1/c)

    Let's solve a few examples and get better at these concepts.

  • 2016-10-14 Shashank Prabhu

    5. Three of the four sentences form a sequence when arranged in the logically correct order. One sentence does not belong to this. Identify the incorrect sentence:

    A. I had six thousand acres of land, and had thus got much spare land besides the coffee plantation. Part of the farm was native forest, and about one thousand acres were squatters’ land, what [the Kikuyu] called their shambas.
    B. The squatters’ land was more intensely alive than the rest of the farm, and was changing with the seasons the year round. The maize grew up higher than your head as you walked on the narrow hard-trampled footpaths in between the tall green rustling regiments.
    C. The squatters are Natives, who with their families hold a few acres on a white man’s farm, and in return have to work for him a certain number of days in the year. My squatters, I think, saw the relationship in a different light, for many of them were born on the farm, and their fathers before them, and they very likely regarded me as a sort of superior squatter on their estates.
    D. The Kikuyu also grew the sweet potatoes that have a vine like leaf and spread over the ground like a dense entangled mat, and many varieties of big yellow and green speckled pumpkins.

  • 2016-10-14 Shashank Prabhu

    4. The first sentence is fixed. Arrange the other statements in the logically correct order:

    A. In America, highly educated women, who are in stronger position in the labour market than less qualified ones, have higher rates of marriage than other groups.
    B. Some work supports the Becker thesis, and some appears to contradict it.
    C. And, as with crime, it is equally inconclusive.
    D. But regardless of the conclusion of any particular piece of work, it is hard to establish convincing connections between family changes and economic factors using conventional approaches.
    E. Indeed, just as with crime, an enormous academic literature exists on the validity of the pure economic approach to the evolution of family structures.

  • 2016-10-14 Shashank Prabhu

    3. The first sentence is fixed. Arrange the other statements in the logically correct order:

    A. Personal experience of mothering and motherhood are largely framed in relation to two discernible or “official” discourses: the “medical discourse and natural childbirth discourse”. Both of these tend to focus on the “optimistic stories” of birth and mothering and underpin stereotypes of the “good mother”.
    B. At the same time, the need for medical expert guidance is also a feature for contemporary reproduction and motherhood. But constructions of good mothering have not always been so conceived - and in different contexts may exist in parallel to other equally dominant discourses.
    C. Similarly, historical work has shown how what are now taken-for-granted aspects of reproduction and mothering practices result from contemporary “pseudoscientific directives” and “managed constructs”. These changes have led to a reframing of modern discourses that pattern pregnancy and motherhood leading to an acceptance of the need for greater expert management.
    D. The contrasting, overlapping, and ambiguous strands within these frameworks focus to varying degrees on a woman’s biological tie to her child and predisposition to instinctively know and be able to care for her child.
    E. In addition, a third, “unofficial popular discourse” comprising “old wives” tales and based on maternal experiences of childbirth has also been noted. These discourses have also been acknowledged in work exploring the experiences of those who apparently do not “conform” to conventional stereotypes of the “good mother”.

  • 2016-10-14 Shashank Prabhu

    2.

    A. He felt justified in bypassing Congress altogether on a variety of moves.
    B. At times he was fighting the entire Congress.
    C. Bush felt he had a mission to restore power to the presidency.
    D. Bush was not fighting just the democrats.
    E. Representatives democracy is a messy business, and a CEO of the white House does not like a legislature of second guessers and time wasters.

  • 2016-10-14 Shashank Prabhu

    1.

    A. The two neighbours never fought each other.
    B. Fights involving three male fiddler crabs have been recorded, but the status of the participants was unknown
    C. They pushed or grappled only with the intruder.
    D. We recorded 17 cases in which a resident that was fighting an intruder was joined by an immediate neighbour, an ally.
    E. We therefore tracked 268 intruder males until we saw them fighting a resident male.

  • 2016-10-14 Shashank Prabhu

    Hi all! Today's topic is going to be Parajumbles. It goes without saying that Parajumbles are tough. Removing the options made it tougher. If last year was any indication, parajumbles could be the trap that could decide how your performance pans out at the end. Even if you do not lose marks in case you get them wrong, they do suck into the available time thereby making things a bit tense. So, my first suggestion is to tackle them like any other question (and not as a non-negative bonus). Also, if you are not very sure of the correct sequence and are stuck between say an ACBDE or ACDBE, it makes perfect sense to enter your answer, mark the question for review and move ahead. Spending more time is not really recommended and chances are high that you would end up being more confused and eventually let go.

    I would be sharing a few tips that you can look at when it comes to tackling these questions. But at the end of the day, nothing is carved in stone and there will always be exceptions. If there is nothing you can do about, just mark the first sequence that comes to your mind and forget about it.

    1. Central idea: Just read the jumbled sentences and try to understand what exactly is the purpose of the paragraph. You may want to write the central idea down and then think to yourself as to how would you have gone about it.

    2. Identify the leads/nouns: By now it would be pretty obvious that the paragraph talks about an individual or an event or a principle or some theory. The introduction should come right at the start and so, most of the time, it would be your starting sentence.

    3. More information: If there is a sentence that explains the central idea, it should ideally follow the introductory sentence.

    4. Pronouns: The sentences with pronouns will come after the introductory sentence. However, it is very important to identify what the pronoun is and the subject that it refers to. If there are multiple subjects, be very careful when arranging these sentences.

    5. Connectors: And, but, although, even though, finally, eventually, etc. are good connectors and it will be easy for you to form pairs of sentences that should come one after the other.

    6. Flow: Generally the flow is intro -> information -> conclusion. If your concluding statement is a positive one, you will rearrange the sentences from negative to positive. If it ends on a negative note, you will arrange the in between sentences from positive to negative.

    Odd sentence out parajumbles:

    The question will ask you to find out the sentence that does not fit into a paragraph made by the remaining 3 sentences. Invariably, the subject will be present in at least 3 of the option statements which will definitely add to the confusion. The trick is to move beyond the subject and figure out the gist of the overall passage that would be formed.

    To crack these, you have to find two statements that connect each other and then try plugging in each of the remaining two statements, at the start or at the end. The basics remain the same: search for the keywords and central idea, get the flow, use the pronouns and find a connecting pair.

    Finally, remember that not every parajumble can be cracked using technical methods. Any editor/author does not write paragraphs visualising that one day it would make it to an aptitude test in the form of a parajumble and so, there would be times when you would not be able to digest the solutions to a few questions. Simple suggestion: let go of these questions. The investment in terms of time and the returns gained in terms of knowledge/marks are disproportionate and as a budding manager, you are supposed to disregard such choices.

    Let's attempt a few questions and check if we have got a bit better at arranging jumbled sentences.

  • 2016-10-12 Shashank Prabhu

    The doors will be open after an odd number of iterations. The only numbers that have an odd number of factors are perfect squares. So, if we look at say 36, we get that it has 2^2*3^2 -> 9 factors. Number of perfect squares in the first 100 natural numbers are 10 and so, these 10 lockers will be open.

  • 2016-10-12 Shashank Prabhu

    176400=2^4*3^2*5^2*7^2
    (2^1+2^2)(3^0+3^1+3^2)(5^0+5^1+5^2)(7^0+7^1+7^2)
    6*13*131*57
    582426

  • 2016-10-12 Shashank Prabhu

    176400=2^4*3^2*5^2*7^2
    2*3*3*3=54

  • 2016-10-12 Shashank Prabhu

    14400=2^6*3^2*5^2
    (2^7-1)/(2-1)*(3^3-1)/(3-1)*(5^3-1)/(5-1)
    127*13*31
    51181

  • 2016-10-12 Shashank Prabhu

    240=2^4*3*5
    So, 1*2*2=4 factors (1, 3, 5, 15)

  • 2016-10-07 Shashank Prabhu

    5. Mr. Potter is standing in the locker room of Hogwarts, which is lined with 100 closed lockers. Using a magic spell 'Alohomora', Mr. Potter can alter the condition of each and every locker that he intends to. First, he opens all the lockers. In the second instance, he alters the condition of every second locker (so the 2nd, 4th, 6th…98th and 100th are all closed). Then, he goes to every third locker and opens it if it is closed or closes it if it is open (let’s call this toggling the locker for our discussion). He proceeds to toggle every nth locker on instance number n. So, for example, on pass number 20 – he will toggle every 20th locker. After his hundredth instance of this activity, in which he toggles only locker number 100, how many lockers will now be open?

  • 2016-10-07 Shashank Prabhu

    4. Find the sum of all the factors of 176400 that are multiples of 2 but not multiples of 8.

  • 2016-10-07 Shashank Prabhu

    3. Find the number of factors of 176400 that are multiples of 2 but not multiples of 8.

  • 2016-10-07 Shashank Prabhu

    2. Find the sum of all the factors of 14400.

  • 2016-10-07 Shashank Prabhu

    1. Find the number of factors of 240 that are odd.

  • 2016-10-07 Shashank Prabhu

    Hello all! Hope your CAT prep is in full swing. This week, I will be telling you a few things about factors. For CAT, you should ideally be well versed with these question types:

    1. Number of factors with and without conditions
    2. Sum of factors with and without conditions
    3. Product of factors
    4. Number of ways in which a number can be expressed as a product of two factors
    5. Number of ways in which a number can be expressed as a product of two factors co prime to each other
    6. Number of positive integers that are less than a number and are coprime to it

    I will be covering the first two types in this post and will be continuing it next week.

    Number of factors:

    If N=a^m* b^n* c^p* d^q…where a,b,c,d… are the prime factors of N

    Number of factors of N=(m+1)(n+1)(p+1)(q+1)…

    A lot of students know the formula but are not aware of the intricacies of the same that leads to half knowledge while attempting questions leading to wastage of precious time during the test. To understand the formula, we have to know why we add 1 to each power. When we say that am is a factor of a number N, we understand that each of a(m), a(m-1), a(m-2)… a(2), a(1), a(0) will also be a factor of N. Simply speaking, we have to count all the factors from 0 till m and that’s why we add 1 to each of the factors. Now, say for example that we require all the factors that do not have a(3) in them, then we will ignore all the factors starting from a(3) till am and consider only a(0), a(1) and a(2). A lot of questions are formed around this concept and if one isn’t careful, it can lead to loss of easy marks.

    A few important things about the number of factors:
    -) The highest prime factor of a number will always be less than or equal to the square root of the number
    -) Squares of prime numbers have exactly 3 factors
    -) If a number has an odd number of factors, it has to be a perfect square, conversely, if a number does not have an odd number of factors, it cannot be a perfect square
    -) The number can be formed by multiplying two factors on either side of the square root of the number
    -) A prime number will have exactly two factors, 1 and the number itself
    -) In special cases and question types, you might have to consider negative factors as well; generally though, the questions would involve positive factors of a number only
    -) Every number greater than 1 will have at least two factors: 1 and the number itself
    -) Any prime number greater than 3 can be expressed in the form on (6k + 1)

    Sum of factors:

    It is a slightly more complicated concept and involves finding the sum of factors. Again, it is an extension of the number of factors and the properties of a geometric progression and extremely derivable.

    If N=a^m* b^n* c^p* d^q…where a,b,c,d… are the prime factors of N

    Sum of all the factors will be nothing but S=(a^0+a^1+a^2+⋯+a^m )(b^0+b^1+b^2…+b^n )…

    Using sum of geometric progression,S= ((a^(m+1)-1))/((a-1))*((b^(n+1)-1))/((b-1) )*((c^(p+1)-1))/((c-1) )*…

    which is essentially the sum of the factors. Again, using a bit of logic, we can figure out that if a particular factor is not required, we can simply eliminate it from the required bracket and leave the rest as it is and calculate the answer.

    Lets try out a few questions based on the concept.

    Also, you may suggest the topics that you would like me to cover as a part of this series by dropping a comment below.

  • 2016-10-01 Shashank Prabhu

    164 can be split as 41*3
    39! mod 41 = 1 (using Wilson's theorem)
    39! mod 3 = 0
    41a+1 = 3b
    a=1, b=14
    Remainder will be 42.

  • 2016-10-01 Shashank Prabhu

    485 can be split as 97*5
    95! mod 97 = 1 (using Wilson's theorem)
    95! mod 5 = 0 (as it would be divisible)
    97a+1 = 5b
    a=2, b=39
    Remainder will be 195.

  • 2016-10-01 Shashank Prabhu

    12^123 mod 100
    100 can be split as 25*4
    12^123 mod 25 = 12^3 mod 25 (as Euler's function of 25 is 20) = 1728 mod 25 = 3
    12^123 mod 4 = 0 (as 12^123 would be divisible by 4)
    25a+3=4b
    a=1, b=7
    Remainder=28 which would be the last two digits.

  • 2016-10-01 Shashank Prabhu

    77 can be split as 7*11
    12^176 mod 7 = 5^2 mod 7 = 4
    12^176 mod 11 = 1
    7a+4=11b+1
    7a=11b-3
    b=6, a=9
    Remainder is 67.

  • 2016-10-01 Shashank Prabhu

    65 can be split as 13*5
    7^6666 mod 13 is 7^6 mod 13 (as Euler's function of 13 is 12)
    7^6 mod 13 = 12
    7^6666 mod 5 is 2^2 mod 5 = 4
    So, using CRT
    13a+12=5b+4
    5b=13a+8
    a=4, b=12
    Remainder will be 64.

  • 2016-09-30 Shashank Prabhu

    5. Find the remainder when 39! is divided by 164.

  • 2016-09-30 Shashank Prabhu

    4. Find the remainder when 95! is divided by 485.

  • 2016-09-30 Shashank Prabhu

    3. Find the last two digits of 12^123.

  • 2016-09-30 Shashank Prabhu

    2. Find the remainder when 12^176 is divided by 77.

  • 2016-09-30 Shashank Prabhu

    1. Find the remainder when 7^6666 is divided by 65.

  • 2016-09-30 Shashank Prabhu

    Ahoy mateys! Today we will be dealing with the Chinese remainder theorem - a favourite of the mocks, a stranger when it comes to CAT. Anyway the concept is pretty useful and it won't hurt to know a bit about it.

    As we saw in the previous posts, to find the remainders of the type a^b mod n, Euler's theorem/Fermat's little theorem is fairly useful. However, if you have something in the form of a^b mod n where n is a composite number or a and n are not co-prime, it becomes a bit tedious to solve the expression. The Chinese remainder theorem comes handy in these cases. I will not get into the theoretical definition of CRT but will demonstrate it's use through a few examples.

    What is the remainder when 199 is divided by 36?

    Duh! Fairly simple, right? The answer as we know should be 19. But this can be used to demonstrate CRT as well.
    Step 1: Split the composite divisor into two co-prime factors. So, 36 should be split as 4*9 (and not 3*12 or 2*18 or 6*6. Extremely important and often forgotten by a lot of know-it-alls)
    Step 2: Find the individual remainders when divided by these two co-prime factors.
    199 mod 4 -> 3
    199 mod 9 -> 1
    Step 3: Express the two expressions in the form of Dividend = (Divisor * Quotient) + Remainder
    199 = 4a + 3
    199 = 9b + 1
    Step 4: Equate the two expressions to get the original equation
    4a + 3 = 9b + 1
    Step 5: Get an integral solution for both (a, b)
    Tip: Shift everything to the side that has the larger co-efficient to make sure you are dealing with fewer cases
    4a = 9b - 2
    b = 0, there is no integral solution to a
    b = 1, there is no integral solution to a
    b = 2, a = 4.... this is the smallest solution that we can get
    Step 6: Substitute either of the values back into our original equation
    4(4) + 3 = 19 OR
    9(2) + 1 = 19
    That's the required remainder.

    Let's try a bigger number:

    Find the remainder when 9^2013 is divided by 77.

    77 split as 11*7
    9^2013 mod 11 will be 3
    9^2013 mod 7 will be 1
    11a+3 = 7b+1
    7b = 11a+2
    Smallest solution will be when a=3, b=5
    Remainder is 36.

    Hope it's a bit clear now. Let's try out a few examples!

  • 2016-09-26 Shashank Prabhu

    3*9 will be 27 and so the last two digits will be 71.

  • 2016-09-26 Shashank Prabhu

    35 raised to an odd power will end in 75.

  • 2016-09-26 Shashank Prabhu

    57^560
    01^140
    01
    Basically, any number that ends in 7 and is raised to a power in the form of 4k will give the last two digits as 01.

  • 2016-09-26 Shashank Prabhu

    93^5437
    93^5436 * 93
    01^1359 * 93
    93

  • 2016-09-26 Shashank Prabhu

    79^5643
    79^5640 * 79^3
    81^1410 * 41 * 79
    41 * 79
    39

  • 2016-09-23 Sakshi Sindhwani

    01

  • 2016-09-23 Sakshi Sindhwani

    71

  • 2016-09-23 Sakshi Sindhwani

    75

  • 2016-09-23 shibashish chatterjee

    41*79 = 39(Last two digit)

  • 2016-09-23 shibashish chatterjee

    93

  • 2016-09-23 shibashish chatterjee

    01

  • 2016-09-23 shibashish chatterjee

    75

  • 2016-09-23 Shashank Prabhu

    5. What are the last two digits of 179^5643 ?

  • 2016-09-23 Shashank Prabhu

    4. What are the last two digits of 9893^5437 ?

  • 2016-09-23 Shashank Prabhu

    3. What are the last two digits of 13957^560 ?

  • 2016-09-23 Shashank Prabhu

    2. What are the last two digits of 135^567 ?

  • 2016-09-23 Shashank Prabhu

    1. What are the last two digits of 31^89 ?

  • 2016-09-23 Shashank Prabhu

    It is important to remember that the last two digits of perfect squares will show a cyclicity around the multiples of 50. So, 12^2 will have the same last two digits as (50-12)^2=38^2, (50+12)^2=62^2, (100-12)^2=88^2 and so on.

    For cases when the last digit of the base ends in 1, 3, 7 or 9

    Step I: Bring it down to the form of (a^4)k * (a smaller manageable number)
    Step II: Now that the unit’s place digit will be 1, the only job left is to find the tens place digit. This can be found by multiplying the tens place digit of the base with the unit’s place digit of the power. Sounds confusing? Let’s see a couple of examples:

    Find the last two digits of 7^2008 (CAT 2008)
    The power is already in the form of 4k and so, we will simply write it as (7^4)502
    We will consider only the last two digits of 7^4 and then we would be almost done with the result. 7^4 can be written as 49^2 which would end in 01 and so, we would get (01)^502

    The unit’s place digit will be 1 while the tens place digit will be equal to the product of the tens place digit of the base (0) and the units place digit of the power (2) ie. 0*2=0
    So, the last two digits are 01.

    Find the last two digits of 223^513

    This becomes a bit longer than the previous one. First of all, we consider it as 23^513 as the digits preceding 23 have no effect whatsoever on the last two digits (consequently, it would be important to note that the result would be the same even for 523513, 298723513 and so on)

    Again, we split the original term into the form of a4k * (smaller term)
    23^512 * 23
    (23^4)128 * 23
    Last two digits of 23^4 can be found out by writing it as (23^2)^2. Now, we know that the last two digits of 23^2 will be same as those of 27^2 i.e. 29. So, 29^2 is what we are left the last two digits of which are again the same as that of 21^2 i.e. 41.

    (41)^128 * 23

    41^128 will have unit’s place digit = 1 and the tens place digit will be given by the product of the tens place digit of the base and the units place digit of the power i.e. (4*8=32) which would be 2. So, the last two digits would be 21. Now we have to do 21*23 which will give us 83 as the last two digits.

    For cases when The base ends in 5

    If the tens place digit is even, all powers greater than 1 will end in 25
    If the tens place digit is odd, all odd powers greater than 1 will end in 75
    If the tens place digit is odd, all even powers will end in 25

    Hope you have got the gist. Let's solve a few examples now:

  • 2016-09-22 Shashank Prabhu

    This week's session will be on finding the last two digits of an odd number raised to any power! Also, do let me know in case you want me to cover any specific topic. Will try to cover it in the coming weeks :)

  • 2016-09-17 Shashank Prabhu

    Easy question. Remainder will be 1.

  • 2016-09-17 Shashank Prabhu

    Easy one. The answer will be 1.

  • 2016-09-17 Shashank Prabhu

    40! mod 43
    41! mod 43 = 1
    41*40! mod 43 = 1
    -2r mod 43 = 1
    -2r = 43k + 1
    k = 1, r = -22 or 21

  • 2016-09-17 Shashank Prabhu

    We know that 95! mod 97 is 1
    So, 95*94! mod 97 is 1
    Remainder (95 mod 97) * Remainder (94! mod 97) = 1
    -2 * Remainder (94! mod 97) = 1
    -2r mod 97 = 1
    -2r = 97k + 1
    We have to find the smallest non-negative integral value of r and k
    If k = 1, r = -49
    So, remainder will be either -49 or 97-49 = 48.

  • 2016-09-16 Sakshi Sindhwani

    197

  • 2016-09-16 Bhushan Badgujjar

    how to solve this?

  • 2016-09-16 Bhushan Badgujjar

    [1500/13]+[1500/13^2] =123

  • 2016-09-16 Vaibhav Panchal

    198

  • 2016-09-16 Vaibhav Panchal

    -1/2=48

  • 2016-09-16 Vaibhav Panchal

    -1/2=21

  • 2016-09-16 Vaibhav Panchal

    (p-2)!mod p=1

  • 2016-09-16 partha prateek

    48

  • 2016-09-16 partha prateek

    21

  • 2016-09-16 shibashish chatterjee

    49

  • 2016-09-16 shibashish chatterjee

    again 01

  • 2016-09-16 shibashish chatterjee

    should be 01

  • 2016-09-16 shibashish chatterjee

    01

  • 2016-09-16 Shashank Prabhu

    5. 195! mod 394 will be equal to ?

  • 2016-09-16 Shashank Prabhu

    4. What is the value of 94! mod 97?

  • 2016-09-16 Shashank Prabhu

    3. What is the value of 40! mod 43?

  • 2016-09-16 Shashank Prabhu

    2. What is the value of 87! mod 89?

  • 2016-09-16 Shashank Prabhu

    1. What is the value of 71! mod 73?

  • 2016-09-16 Shashank Prabhu

    Hello all! Hope your prep is coming along nicely. In continuation to last week's topic (factorials), I will be simplifying an obscure yet interesting concept: Wilson's theorem.

    It simply states that for a prime number ‘p’, (p-1)! will give a remainder of (p – 1) when it is divided by p. In other words, let’s say we consider the prime number 5. Then, 4! when divided by 5 will give a remainder of 24 mod 5 or 4. Similarly, when 6! is divided by 7, we get the remainder to be 6.

    A few other examples

    40! mod 41 will be 40
    42! mod 43 will be 42
    96! mod 97 will be 96
    and so on.

    Remember that the denominator (or the divisor) should be a prime number and so, it would not completely divide the numerator.

    Extending the Wilson’s theorem further, we can see that for a prime number ‘p’, (p – 2)! will give a remainder of 1 when it is divided by p. This can be proved by simply using the concept of negative remainders.

    Let us say that the remainder of (p – 2)! mod p is r.
    We know that, using Wilson’s theorem,
    (p – 1)! mod p is (p – 1)
    So, (p – 1) * (p – 2)! mod p is (p – 1)
    Using the property of remainders of a product, we can say that
    [(p – 1) mod p] * [(p – 2)! mod p] is (p – 1)
    Now, if p is a prime number, (p – 1) mod p will always be (p – 1)
    So, for LHS = RHS to happen, the second part of the expression should give the remainder equal to 1.
    (p – 2)! mod p will be equal to 1.

    Let's solve a few examples that focus on application of this concept.

  • 2016-09-10 Shashank Prabhu

    100=2*5 and as the powers of the prime factors are the same, we go for the higher prime factor i.e. 5. So, number of 5s in 100! is given by 20+4=24 which will be the highest power of 10 in 100! and consequently, the number of zeros in which, 100! ends.

  • 2016-09-10 Shashank Prabhu

    24 = 2^3 * 3
    Number of 2s in 96! = 48+24+12+6+3+1 = 94
    Number of 3s in 96! = 32+10+3+1 = 46
    Number of buckets of 2^3 that is possible is 31
    As 31 is the limiting number, the highest power of 24 present in 96! is 31.

  • 2016-09-10 Shashank Prabhu

    This is a classic case of jumping 5s. If you observe carefully, 20! will end in 4 zeros. Similarly 21!, 22!, 23! and 24! will end in 4 zeros. However, when it comes to 25!, there is a jump by two 5s and so, there will be no n such that n! ends in 5 zeros. The best way to tackle questions such as these is by trying to find out the number of zeros around powers of 5. In this case, as the options are near 150, we can check for an n less than 750 (as 150*5 will be 750 and there will surely be a few overlapping factors). So, we check for 624 and 625. 624! will have 124+24+4=152 zeros at the end whereas, 625! will have 125+25+5+1=156 zeros at the end. So, the correct answer is option d.

    PS: There was an error in the question. It has been rectified now. Apologies.

  • 2016-09-10 Shashank Prabhu

    Number of 3s in 150! will be 50+16+5+1=72. So, 24 packets in total. Perfectly done :)


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