Number System Practice Gym by Anubhav Sehgal  Part 4

How many numbers from 1 to 100 are NOT divisible by 2 or 3 or 5 or 7 ?
105 * 1/2 * 2/3 * 4/5 * 6/7  {101,103} = 22
Why 105 ?
Step 1 : Find numbers not divisible by 2,3,5 or 7 by simple Euler number's application.
Step 2 : Remove numbers like 103 which being not divisible by 2,3,5 or 7 were removed in euler but do not belong to our range of 1100 for which answer is to be calculated.
105 Chosen as it is the closest(to100) multiple of LCM of uncancelled denominators 5,7HCF of 2 numbers is 12 and their sum if 144. Find the maximum value of their product
12(a + b) = 144
a + b = 12
As near to each other as possible with they being coprime to each other.
5, 7
Max product = (12 * 5) * (12 * 7) = 60 * 84 = 5040.Highest power of 8 in 17! + 18! + 19! + . . . + 100!?
17!(1 + 18 + 18 * 19 + ... 18 * 19 * 20 * ..100)
17!(1 + even)
17! * odd
Find power of 2 and subsequently 8 in 17!
Power of 2 : 8 + 4 + 2 + 1 = 15
Power of 8 : [15/3] = 5What is the highest power of 3 available in the expression 58!  38!
38!(58 * 57 * .. * 39  1)
Power of 3 in 38! = 12 + 4 + 1 = 17
Power of 3 inside bracket = 0
Highest power to divide 58!  38! = 17Remainder when 26^57 is divided by 29?
E(29) = 28
26^57 mod 29
26^56 * 26 mod 29
26How many scalene triangles are there for which the lengths of all sides are integers & the perimeter is 24 cm?
For this particular case, [(n  6)^2/48] where [.] nearest integer function.
[(24  6)^2 / 48 ] = 7No of trailing zeroes in (100!)!
Two different concepts :
Double factorial : n!!
n!! = n.(n  2).(n  4)...6.4.2 for n even
n!! = n.(n  2).(n  4)...5.3.1 for n odd
n!! = 1 for n = 0,1 by definitionSimple factorial nested twice
(n!)!
Case 1 : 100!!
100!! = 100.98.96.....6.4.2
=> Number of zeroes = Number of 5s = 12
Case 2 : (100!)!
Use stirling approximation or C(2n,n) maybe.
The number of 4letter words that can be formed out of the letters of the word FIITJEE is
F,I2,T,J,E2
All different : 5c4 * 4! = 120
2 same 2 different : 2c1 * 4c2 * 4!/2! = 144
2 same 2 same : 2c2 * 4!/2!2! = 6
Total : 270Find the sum of the first 125 terms of sequence 1, 2, 1, 3, 2, 1, 4, 3, 2, 1, 5, 4, 3, 2 ...
1 ; 1 + 2 ; 1 + 2 + 3 ; ... 125 terms
120 terms if we have 1 ; 1,2 ; 1,2,3 ; .... ; 1,2,3,4,..,15
So,sum till 120 terms = n(n + 1)(2n + 1)/12 + n(n + 1)/4
where n = 15 => Sum = 620 + 60 = 680
After this,121st to 125th term : 16,15,14,13,12 = 70
Total : 750Useful Tip :
If N is a natural number, number of natural numbers in the range :
Case 1 : A < = N < = B
Number of natural numbers in the range : (B  A) + 1
Case 2 : A < = N < B OR A < N < = B
Number of natural numbers in the range : (B  A)
Case 3 : A < N < B
Number of natural numbers in the range : (B  A)  1Practice examples :
Q1) How many numbers,N, in the range :
a) 100 < = N < = 200 b) 100 < = N < 200 c) 100 < N < 200
Q2) How many even numbers,N, in the range :
a) 100 < = N < = 250 b) 120 < = N < 360
Q3) How many numbers,N, of the form 3k,where k is any positive integer in the range :
a) 50 < N < 300 b) 100 < = N < 500 c) 100 < = N < = 1000Should help at a lot of places. So practice well if you are starting your preparations.
When a natural number is divided by 4 and 7, it gives remainder 3 and 2 respectively. What is the remainder obtained when same natural number is divided by 11?
4a + 3 = 7b + 2
4a = 7b  1
a = b + (3b  1)/4
b = 3, a = 5
23 is the first number.
General form : 23 + LCM(4,7) * k = 23 + 28k
Hence remainder cannot be uniquely determined