Quant Boosters  Number Theory

Number of Questions  30
Topic  Number Theory
Solved?  Yes
Source  Curated Content solved by Rajendra Rajput

Q1) What is the least multiple of 7, which when divided by 2, 3, 4, 5 and 6 leaves the remainders 1, 2, 3, 4 and 5 respectively?

N/2 = 1 or 1
N/3 = 2 or 1
N/4 = 3 or 1
N/5 = 4 or 1
N/6 = 5 or 1
you can see that 1 is constant
Formula to find number
N = LCM of (2,3,4,5,6) k + constant
N = 60k 1Now
60k1 / 7 should be divisible
To find K lets simplify it more and make remainder 0
4k1/7
Now obviously k should be 2 to make it perfectly divisible
4 (2)1/7
= 0 remWe Found K = 2
Put it in that number
N = 60k  1
N = 60 (2)  1
N = 119

Q2) What is the remainder when (2^100+3^100+4^100 5^100) is divided by 7?

2^100 + 3^100 + 4^100 + 5^100/7
= (2^3)^33 * 2 + (3^3)^33 * 3 + (4^3)^33 * 3 + (5^3)^33 * 3 / 7
= 2 * 8^33 + 3 * 27^33 + 4 * 64^33 + 5 * 125^33 / 7
= 2 * 1^33 + 3 * 1^33 + 4 * 1^33 + 5 * 1^33 / 7
= 2 * 1 + 3 * 1 + 4 * 1 + 5 * 1 / 7
= 2  3 + 4 5 / 7
= 2/7
= 72
= 5 remainder

Q3) If a number is divided by 14 and the remainder is 5, then if the same number is divided by 7, what is the remainder?

N/14 = 5 remainder
N could be 19, 33, 47,....
Least number is 19
So 19/7 = 5 remainderOr
N = 14k + 5
Now
N/7
14k+5/7
0*k+5/7
5/7
5 remainder

Q4) Find the remainder when 17^53 is divided by 27

Euler of 27 = 18
So Euler says
17^18/27 = 1 remainder
17^36/27 = 1
.
Similarly 17^54/27 = 1
17^53 * 17 /27 = 1
R * 17 = 27k+1
R = 27k+1/17
Let K be any number which perfectly divides the equation
R = 27 (5)+1/17
R = 8 remainder

Q5) Find the remainder when 5^100000 is divided by 196

5^100000/196
= 5^100000/ 4*49Lets divide it separately
5^100000/4
= 1^100000/4
= 1 remainder5^100000/49
Euler of 49 = 49 (1  1/7) = 49 (6/7) = 42
Euler says
5^42/49 = 1 remainder
5^84/49 = also 1Same way
5^100002/49 = 1
5^100000 * 5^2/49 = 1
R*25 = 49k+1
R = 49k+1/25
Let k be any lesser value which perfectly divides the equation
Let k be 1
R = 50/25 = 2 remainderSo our answer is that number which leaves remainder 1 when divided by 4 & remainder 2 when divided by 49
Apply Chinese remainder theorem and get your answerAnswer is 149

Q6) Find the remainder when 2017^22 is divided by 23

Fermat thereom says that
When a and p are coprime to each other
a^(p1)/p = 1 remainder
Where, p is a prime number
Here 23 is prime
So
2017^22/23
= 1 remainder

Q7) What is the unit digit of 123^4567! * 765^4321?

123^4567! * 765^4321
We have to find unit digit so we are concerned of unit digit only
3^4567! * 5^4321
(3^4)^(4567!/4) * 5^4321
(xxxx1)^(4567!/4) * 5^4321
[ Note : 1^n = 1 unit and 5^n = 5 unit]
1^(4567!/4) * 5^4321
= 1 * 5
= 5 unit digit

Q8) What is the remainder when (49^15  1) is divided by 14?

49^15  1/14
= (7^2)^15  1/14
= 7^30  1/14
[ Note : 7^n/14 = 7 remainder ]
= 71/14
= 6/14
= 6 remainder

Q9) What is the remainder when 2000^2001 is divided by 26?

2000^2001/26
24^2001/26
26 = 13*2Divide separately
24^2001/13
2^2001/13
2 * 2^2000/13
2 * (2^6)^333 * 2^2/13
8 * 64^333/13
8 * 1^333/13
(1 ^odd =  1 )
8*1/13
8/13
8 remainder24^2001/2
= 0 remainderSo our answer is that number which leaves remainder 8 when divided by 12 & remainder 0 when divided by 2
N/13 = 8
N could be 8, 21, 29,...N/ 2 = 0
N could be 0,2,4,6,8,...The very first number common in both term is 8
So final remainder is 8
Answer 8
Or
24^2001/26
2^2001/26
Eulee of 26 is 12
(2^12)^166 * 2^9/26
1 * 2^9/26
512/26
18/26
2618
8 remainder

Q10) When 3n is divided by 7, the remainder is 4. what is the remainder when 2n is divided by 7?