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Theory Of Equations - Finding Number Of Integral Solutions ( 60 Questions )

Finding the integral solutions of a given equation is a question type that commonly appear in CAT and similar topics. Most of these questions requires a fundamental understanding of Algebra and PnC. Some times this could get bit tricky and we will end up missing cases or wasting time to find the cases. There are many useful tricks and direct formulas in this area which comes handy during exams. In this article we have collated 60 questions from this topic which were shared in various forums. Solve them all and the concepts and tricks would be clear to you. Happy learning.

Cauchy Schwarz Inequality - Hemant Malhotra

x2 + y2 - 14x - 6y - 6 = 0
(x-7)2 + (y-3)2 -64 = 0
now let 3x + 4y = k now this line will touch circle then only it will be maximum or minimum
perpendicular from center to line will be equal to radius
The length of perpendicular from a point (x1, y1) to a line ax + by + c = 0 is |(ax1 + by1 + c)/ root(a2+b2)|

Quant Marathon by Gaurav Sharma - Set 10

Two digit numbers with no digit repeated = 9C1 x 9C1 = 81
Three digit numbers with no digit repeated = 9C1 x 9C1 x 8C1 = 9 x 9 x 8 = 648
Four digit numbers less than 2000, with no digit repeated = 9C1 x 8C1 x 7C1 x 1 = 9 x 8 x 7 = 504
Total numbers with no digit repeated = 9 + 81 + 684 + 504 = 1242
Numbers with at least two digits same = 2000 – Numbers with no repetition
2000 – 1242 = 758

Formula To Find Ordered & Unordered Pairs Possible For A Given LCM - Hemant Malhotra

let two numbers x and y and their lcm is 22 * 33 * 53
now atleast one of x and y , highest power of 2 is 2. highest power of 3 is 3 and highest power of 5 is 3
now lets take 22 so x can be 20 or 21 or 22.. same in case of y
so 3 * 3 = 32 ways for them
but there will be a case when neither x nor y have the power 22 so we have to remove that case because lcm won't be 22
so we will remove cases when x and y have only 20 or 2^1 so 2*2 = 2^2 cases we have to remove

Remainder theorem - Atreya Roy

Example :
1) 15! Mod 17 : Can be written as (17-2)! Mod 17 where p = 17. Hence the remainder in this case = 1
2) 11! Mod 13 = 1. Where p = 13
These two concepts of Wilson theorem are very important and will help you solve sums at a faster rate. Applying Euler’s Theorem to co-prime numbers may also work, but with the help of Wilson’s Theorem, you will save a lot of time while solving. Hope the Theory of Wilson’s Theorem for Remainders is clear now. In our Next Concept explanation, we will talk about Fermat’s Theorem, which is another very important theorem used in the field of remainder calculations.

Numbers and digits - Atreya Roy

Cyclicity of 3 : 3^1 =3; 3^2 = 9; 3^3 = 27 ; 3^4 = 81 ; 3^5= 243 . Hence again, after 4 terms starting from 1, the unit digit of powers of 3 get repeated. Hence the cycle order for 3 = 4
Cyclicity of 4 : 4^ odd always ends with 4 / 4^even always ends with 6 => Hence the cycle order of 4 = 2
For 5 and 6, 5^a and 6^b : where a and b are positive integers, both will end in 5 and 6 respectively.
Similarly we can find the cyclicity of other numbers

Factorial - Atreya Roy

Highest Power of a prime in a number
Let us take an example for the following.
What is the highest power of 3 such that the number will divide 12!
If we write 12! , we get 12! = 1*2*3*4*5*6*7*8*9*10*11*12.
So from 3,6,9 and 12 we will get 3s.
Total 3s = 1 from 3,

Factors - Atreya Roy

Hence, with these factors if we multiply the 2 we took out, we will get the total even factors.
Total even factors = (3+1)*(2+1)*(2+1) = 4*3*3 = 36
From this we can also find out the odd factors.
Total Factors = Even Factors + Odd factors.
Hence if the total number of factors = 45 and even factors are 36 then 45-36 = 9 odd factors are present in the number.
Note : To find the number of odd factors present in a number we can also calculate them by removing all the 2s present in the number.

Factors - Hemant Malhotra

First we will understand some basic stuffs: Let say you have 10 mango of same kind, 20 banana of same kind and 30 apple of same kind. Now you have options to select none or 1 or more than 1 items (any number of items)
CASE 1 - (mango) we can select 0 mango or 1 mango or 2 mango or 3 mango and so on , so we have 11 ways to select mangoes (10+1)=11
CASE 2 - Banana, same (20+1) ways
CASE 3 - Apple same (30+1) = 31 ways
So total number of ways = (10+1) (20+1) (30+1)
Same idea we will apply in case of finding number of divisors
We break number in to their prime factors and then apply this theory

Quant Boosters by Sabyasachi Mishra, CAT 100 percentiler.

Now there are 4 ways to choose a, 4 ways for b and 4 ways for c => There are 4 *4 * 4 = 64 three-digit numbers that can be made using 1,2,3 and / or 4.
Let’s write all these numbers one below another to add them up. Now each digit would appear 16 times in ones place.
Their sum = 16 *(1 + 2+ 3 + 4) = 160
Similarly the tens place would add up to 160 and the hundreds place too.
So the actual sum = 160 (1 + 10 + 100) = 160 * 111 = 17760