# Quant Boosters - Hemant Malhotra - Set 9

• Number of Questions - 30
Topic - Quant Mixed Bag
Solved ? - Yes
Source - Elite's Grid Prep Forum

• Q1) 17! = 355687abc096000, where a,b and c are digits, find the value of a + b + c
a) 23
b) 14
c) 5
d) 9

• divisible by 7 ,11,13 so divisible by 1001
1001 = 10^3+1
so make group of 3 and add with alternate sign from right side
so 000-096+abc-687+355
(-428 + abc) mod 1001 = 0
so abc = 428
so sum of digit = 4 + 2 + 8 = 14

• Q2) 15! = 1307674abc000, find digital sum of abc

• 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
000-000+5ab-643+609-618+847-140+604-cd9+799-232+295=2041-cd9
so 2041-cd9 mod 1001=0
so 39-cd9 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 vice-versa.
In base 7 k = 7, k+1 = 8. 4 is the divisor of 8. The alternating sum of the digits of 123abc231bca312cab is 1-2+3-a+b-c+2-3+1-b+c-a+3-1+2-c+a-b = 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^(N-1)*k
N! mod 2^N = 2^(N-1) { k mod 2} =2^(N-1)

• 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 6k-1. Hence x also has to be of the form 6k+1 or 6k-1. 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

64

61

61

62

62

33

61