Number System Practice Gym by Anubhav Sehgal  Part 6

Let X be the largest positive number less than 10^6 such that when written in base 2, the binary representation consists of only 1s. Find the remainder when (X+2) is divided by 3.
2^10 = 1024 > 10^3
(2^10)^2 > 10^6
=> 2^19 is the largest power of 2 less than 10^6
For a binary representation with all 1s, X must be
2^19  1
So (X + 2) mod 3 = (2^19 + 1) mod 3 = 0Find the LARGEST 5digitinteger N so that 2N is also a 5digitinteger and all digits 0, 1, 2, 3, ..., 9 are contained in both N and 2N
Idea is that we have to maximize the digits.
4 has to be first digit of N. Giving us 9 as first digit of 2N ( 2 * 4 + 1 carryover)
Second digit of N cannot be 9, Since 9 is already first digit in 2N.
So let 8 be second digit of N. We have 2 * 8 = 16 +1 carryover = 17 giving 7 as the second digit of 2N.
Now let 6 be the 3rd digit of N. we have 2 * 6 = 12 + 1 carryover = 13 giving us 3 as the 3rd digit of 2N.
Now let 5 be the 4th digit of N. We have 2 * 5 = 10 giving us 0 as the 4th digit of 2N.
Finally we have 1 as the fifth digit of N giving us 2 as the last digit of 2N.
So the required numbers are : 48651 and 97302.Smallest whole number which when divided by 5,7,9 and 11 leaves the remainder 1,2,3 and 4 respectively.
N = 5a + 1 = 7b + 2 = 9c + 3 = 11d + 4
2N = 5(2a) + 2 = 7(2b) + 4 = 9(2c) + 6 = 11(2d) + 8
2N + 3 = 5(2a + 1) = 7(2b + 1) = 9(2c + 1) = 11(2d + 1)
2N + 3 is the LCM of (5,7,9,11)
N = [LCM(5,7,9,11)  3]/2 = 1731A positive integer p is called almost prime, when it has only 1 divisor aside from 1 and p. Find the sum of the 6 smallest almost primes.
Only squares of primes have exactly 3 factors.
Hence your required sum
= 2^2 + 3^2 + 5^2 + 7^2 + 11^2 + 13^2 = 377Find n if sqrt(17^2 + 17^2 + ... n times) = 3 * 17^2
3 * 17^2 = sqrt(9 * 17^4) = sqrt(9 * 17^2 * 17^2)
i.e. 9 * 17^2 times 17^2 inside the square root.
i.e 51^2 times
i.e 2601a!b! = a! + b! and find (a + b). (Positive integers)
General approach to solve such equations
a!b!  a!  b! + 1 = 1
(a!  1)(b!  1) = 1
a!  1 = b!  1 = 1
a! = b! = 2
a = b = 2
a + b = 4Which is the third smallest number when subtracted from 6300 results in a perfect square ?
6300  n = k^2
80^2 = 6400
79^2 will give us the first smallest number on subtracting..
77^2 will give us the third
6300  5929 = 371.How many different marks are possible in CAT if there are 100 questions with marking scheme of +3 , 1 & 0 for correct, wrong & non attempt resp
If no of marks are +n for correct and 1 for incorrect, then number of scores which are not possible is given by :
(n  1) + (n  2) + .. + 1 = n(n  1)/2
Take +4 1 scheme
30 correct gives : 120
29 correct gives : 116
With 30 correct you can have 0 wrong
So 1,2,3 wrong are not possible with 30 correct
With 29 correct you can have 1 wrong
But 2,3 Qs wrong is not..
..
(n  1) + (n  2) + .. + 1 = n(n  1)/2
Which is nothing but C(n,2)How many positive integer solution for the equation x1 + x2 + x3 + … + xk =n, where 1 < = x1 < = x2 < = x3 … < = xk?
a + b + c = 10
1 < = a < = b < = c
a = p + 1
b = p + q + 1
c = p + q + r + 1
where p,q,r are non negative integers.
3p + 2q + r = 7
p = 0, 2q + r = 7 => 4 solutions
p = 1, 2q + r = 4 => 3 solutions
p = 2, 2q + r = 1 => 1 solution
Total : 8 solutionsFind the sum of all positive integral value a for which (2a + 124)/(a + 3) is an integer.
2a + 124)/(a + 3)
2(a + 62)/(a + 3)
2[(a + 3 + 59)]/(a + 3)
2 + 118/(a +3)
a + 3 is a factor of 118
118 = 2 * 59
1,2,59,118
a = 2,1,56,115
Sum of positive integral values = 171