# 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 1-100 for which answer is to be calculated.
105 Chosen as it is the closest(to100) multiple of LCM of uncancelled denominators 5,7

HCF 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] = 5

What 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! = 17

Remainder when 26^57 is divided by 29?

E(29) = 28
26^57 mod 29
26^56 * 26 mod 29
26

How 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 ] = 7

No of trailing zeroes in (100!)!

Two different concepts :

1. 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 definition

2. Simple 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 4-letter words that can be formed out of the letters of the word FIITJEE is

F,I-2,T,J,E-2
All different : 5c4 * 4! = 120
2 same 2 different : 2c1 * 4c2 * 4!/2! = 144
2 same 2 same : 2c2 * 4!/2!2! = 6
Total : 270

Find 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 : 750

Useful 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) - 1

Practice 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 < = 1000

Should 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

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