Question Bank - Number Theory
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Number of Questions - 30
Topic - Number Theory
Solved? - Yes
Source - Curated Content solved by Rajendra Rajput
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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?
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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
60k-1 / 7 should be divisible
To find K lets simplify it more and make remainder 0
4k-1/7
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
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Q2) What is the remainder when (2^100+3^100+4^100 5^100) is divided by 7?
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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
= 7-2
= 5 remainder
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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?
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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
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Q4) Find the remainder when 17^53 is divided by 27
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Euler of 27 = 18
So Euler says
17^18/27 = 1 remainder
17^36/27 = 1
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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
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Q5) Find the remainder when 5^100000 is divided by 196
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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
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Q6) Find the remainder when 2017^22 is divided by 23
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Fermat thereom says that
When a and p are coprime to each other
a^(p-1)/p = 1 remainder
Where, p is a prime number
Here 23 is prime
So
2017^22/23
= 1 remainder
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Q7) What is the unit digit of 123^4567! * 765^4321?
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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
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Q8) What is the remainder when (49^15 - 1) is divided by 14?
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49^15 - 1/14
= (7^2)^15 - 1/14
= 7^30 - 1/14
[ Note :- 7^n/14 = 7 remainder ]
= 7-1/14
= 6/14
= 6 remainder
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Q9) What is the remainder when 2000^2001 is divided by 26?
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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
26-18
8 remainder
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Q10) When 3n is divided by 7, the remainder is 4. what is the remainder when 2n is divided by 7?
