Quant Boosters  Hemant Malhotra  Set 9

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

Minimum occurs when the numbers are 6, 6, 6, 2, 2, 1.
Maximum occurs when the numbers are 7, 5, 4, 3, 2, 2.
To achieve min LCM look for max number of hcf among the 6C2 pairs.
So we need number in form of coprime.Thus, M = 414 = 2 * 3^2 * 23, M has 2 * 3 * 2 = 12 divisors
M^2 has 3 * 5 * 3 = 45 divisors.so 45  12 = 33 divisors

Q13) N = 98765432109876543210.... ( 1000 digits ).What is the least positive value of n such that N + n is divisible by 11 ?

Divisiblity rule of 11: Numerals whose alternating sum of digits is divisible by 11 represent numbers divisible by 11. Here the "alternating sum" means we alternate the signs from positive to negative to positive to negative, and so on. 10 to an odd power plus 1 is divisible by 11, and 10 to an even power minus 1 is divisible by 11. This test can also be applied recursively
Example: 12342
1  2 + 3  4 + 2 = 0 so div by 11
Another one, 2131415
2  1 + 3  1 + 4  1 + 5 = 11 so div by 11In given case, make a group of 10 numbers and add all such group, there will be 100 groups so total sum is 9876543210 * 100 = 987654321000
Now use alternate sum approach so remainder is 5 ; so u need to add 5

Q14) Exactly one of the ﬁve numbers listed below is a prime number. Which one is the prime number ?
(a) 999,991
(b) 999,973
(c) 999,983
(d) 1,000,001
(e) 7,999,973