Quant Boosters  Hemant Malhotra  Set 9

div by 7 ,11,13 so div by 1001
1001=10^3+1
so make group of 3
and add with alternate sign from right side
000  abc + 674  307 + 1 = abc+368 so abc=368

Q3) 34! = 295232799cd96041408476186096435ab000000, determine the digits a, b, c & d.

34! has 7 trailing zeroes so b = 0
first non zero digits from right side
34! = 5 * 6+4
so 2^6 * 6! * 4!
4 * 2 * 4 = 2
so a = 2
now number is div by 7,11,13 so div by 1001 so make group 3 and add with alternate sign from right side
000000+5ab643+609618+847140+604cd9+799232+295=2041cd9
so 2041cd9 mod 1001=0
so 39cd9 mod 1001=0
so c=0 and d=3

Q4) Find the number of ordered pair of digits (A,B ) such that A3640548981270644B is divisible by 99.

4b + 64 + 12 + 98 + 48 + 05 + 64 + a3 =
291 + 70 + 40 + b + 10a + 3
404 + ab/99
8 + ab/99
where ab is a 2 digit number
clearly ab can only be 91 , so only 1 pair

Q5) Let 123abc231bca312cab be a 18 digit number in base 7. How many ordered pairs (a,b,c) exist such that the given 18 digit number is divisible by 4

Any number in base k is divisible by k+1 or its divisors if and only if the sum of the digits(alternating and sign changed) is divisible by k+1 or its divisors respectively.
like in base 10, if a number is div. by 11 then sum of its digits(sign changed alternatively) is div. by 11 and viceversa.
In base 7 k = 7, k+1 = 8. 4 is the divisor of 8. The alternating sum of the digits of 123abc231bca312cab is 12+3a+bc+23+1b+ca+31+2c+ab = 6 (a +b +c) is divisible by 4
Now (a+b+c) can be 2,6,10,14, or 18
Solutions in each case is 6,28,36,15,1
total solution's = 6+28+36+15+1 = 86

Q6) Let N = 44444444......4444444 (2045 times). find remainder when N is divided by 103 ?

44444 mod 103 = 51
any number written p  1 times div by p where p is prime
so only last 5 digits will be dere so 44444 mod 103=51

Q7) What is remainder when 2014^2015 is divided by 121 ?

Use Eulers theorm, 121 = 11 * 11 = > E (121)=121 * 10/11 = 110
now 2014^2015 mod 121 = (43)^35 mod121
= 43 * (43 * 43)^17 mod 121
= 43 * (34)^17 mod 121
= 43 * 34 * (34 * 34)^8 mod 121
= 43 * 34 * (54 * 54)^4
= 43 * 34 * (12)^4 mod 121
= 43 * 34 * 144 * 23 mod 121
= 43 * 34 * 23^2 mod 121
= 43 * 34 * 45 Mod 121
= 87 mod 121 = 34.OR
78^2015 mod 121
= (1+77)^2015 mod 121
= 77C0 + 2015C1 * 77 + rest of the terms divisible by 11
= (1 + 2015 * 77 ) mod 121
= 34

Q8) If m is a natural no. and N = 2^m, what is the remainder when N! is divided by 2^N ?

highest power of 2 in N! = (2^m)1 = N 1
N! = 2^(N1)*k
N! mod 2^N = 2^(N1) { k mod 2} =2^(N1)

Q9) P is a prime number greater than 30. When P is divided by 30, the remainder is x. How many different values of x are possible ?
a. 9
b. 8
c. 10
d. 11

P = 30n + x, P has to be 6k+1 or 6k1. Hence x also has to be of the form 6k+1 or 6k1. Possible values of x are 1,5,7,11,13,17,19,23,25,29. out of these x cannot be 5 or 25. hence 8 values
OR
Just find number of coprime less than 30 = 30 * 1/2 * 2/3 * 4/5 = 8

Q10) If n = 1 + x, where x is the product of 4 consecutive natural numbers, then n is:
I. an odd number.
II. an even number.
III. a prime number.
IV. a perfect square.a) II only
b) III and IV only
c) I and IV only
d) I only
e) None of these

For 4 consecutive natural number there will be atleast 2 even numbers so product will be even
= > x is even and x+1 is odd
so 1 is true and 2nd is false
now to decide where n is prime or not
try for first four natural numbers which give a product of 24 and therefore n =25, therefore n is not prime but could be prefect squareOR
n= (a1) * (a) * (a+1)(a+2)
n= (a^2+a1)(a^2+a1)
so n is perfect squareOR
x = 2 * 3 * 4 * 5.
then 120 + 1 = 121 = 11^2so option 3.

Q11) If 2n + 1 and 3n + 1 ( n # 0 ) are perfect squares, find remainder when n is divide by 40 ?

n = 40 so 2n+1 = 81 = 9^2
3n+1 = 121 = 11^2
40 mod 40 = 0

Q12) The sum of 6 natural numbers (not necessarily distinct) is 23. Let M denote the positive difference of the maximum and minimum of the LCM of these 6 numbers. Then total number of divisors that divide M^2 but doesn't divide M are
(a) 22
(b) 33
(c) 18
(d) 26
(e) none