# Question Bank - Number Theory - Hemant Malhotra

• a + b + c = 51
max product = 16 * 17 * 18 = 4896
minimum= 48, 2, 1 so 96
As a,b,c are distinct
OA = B

• @hemant_malhotra ans 2100

• Q71) For prime numbers a and b , a + b = 102 and a > b. What is the least possible value of a - b

• Minimum will be when a and b are as close as possible so
a = 51 + m and b = 51 - m
where 51 + m and 51 - m both are primes
at m = 8, 59 and 43 both are prime so minimum difference will be 16

• Q72) a, b and c are real numbers such that a > 6 , b < 0 and c > -2 . how many of the following statements are definitely true ?
i) a+ b + c > 0
ii) ab + bc + ca =0
iii) a + b^2 + c^4 < 0

• Q73) The first four terms of an arithmetic sequence are a, 9, 3a - b, and 3a + b . find remainder when 2010th term of this sequence is divided by 1000

• Q74) What is the maximum number of natural numbers that one can select from the first 99 natural numbers and be sure that no three of the selected numbers add up to 99?

• Q75) A is a natural number which has 4 factors. If A < = 70 then find how many values are possible for A

• A has 4 factors so Form of A will be a^3 or a*b form so that total number of factors will be 4
where a and b are prime numbers
Case1- when A is in form of a^3 here a=2 and 3 possible because for a=5 , A=5^3=125 but we want A < = 70
Case2- when A is in form of a * b here a=2 then b will vary from 3 to 31 so all prime from 3 to 31 are 3,5,7,11,13,17,19,23,29,31=10
When a=3 then b will vary from 5 to 23 so 5,7,11,13,17,19,23 so 7 numbers
when a=5 then b will vary from 7 to 13 so 7,11,13 so 3 numbers
when a=7 then b=11 not possible
so total 22 numbers are possible

• Q76) Find the remainder when 17^49 divided by 101.

• Method1- (289)^24 * 17mod101
(-14)^24 * 17
(196)^12 * 17mod101
((-6)^4)^3 * 17mod101
(-1
612/101-7)^3 * 17 mod101
84 * 84 * 84 * 17 mod101
-14 * -17 * 17 mod101
238 * 17mod 101
36 * 17 mod 101=6

Method 2 - Reverse Euler --- 17^100 mod 101 = 1
so 17^50 mod 101 = 1 or -1;
so it might be possible that +1 or -1 will be remainder in case of 17^50
but 17^5 mod 101=-1
so 17^50 mod 101 = 1;
17^49 * 17 mod 101 = 1;
r * 17 mod 101 = 1; so r = 6

• Q77) Martian bus tickets have six-digit numbers, so that all tickets are numbered from 000001 to 999999. Martians think that the ticket is lucky if the sum of the ﬁrst three digits is equal to the sum of the last three digits. Find the remainder when sum of all 6-digit numbers which appear on the lucky tickets is divided by 13

• Q78) How many prime numbers are are there which when divided by another prime number gives a quotient which is same as remainder.

• Q79) Find sum of digits of N where N = 10^2015 - 2015

• 10^2-2=98
10^3-3=997
10^4-4=9996
So we get the equation from the relation which is
9*(N-n) + each digit of ( 10^n -N)
Wheras,
N: the number which 10 is raised to and also the number that is minused to the number
n: the number of the digits of N
So substituting to the equation
N=2015
n=4
So we get
9*(2015-4) + sum of the digits of (10^4 - 2015)
= 18099 + (7+9+8+5) = 18128

• Q80) How many three digit numbers increased by 495 when their digits are reversed?
a) 4
b) 20
c) 40
d) 100

145

104

89

151

106

200

103

144