Question 1 :
Find the least positive value of x such that
(i) 71 ≡ x (mod 8)
Solution :
71 - x = 8n
71 - x is the multiple of 8
if x = 1, 71 - x = 70 is not the multiple of 8,
The number which is nearest to 71 is 64 (multiple of 8).In order to make this 71 as 64, we have to subtract 7.
So, the value of x is 7.
Hence 7 is the least number.
(ii) 78 + x ≡ 3 (mod 5)
Solution :
78 + x = 3 (mod 5)
78 + x - 3 = 5n
75 + x = 5n
If x = 5, then 75 + 5 = 80(which is the multiple of 5)
Hence 5 is the least number.
(iii) 89 ≡ (x + 3) (mod 4)
Solution :
89 = (x + 3) (mod 4)
89 - x - 3 = 4n
86 - x = 4n
If x = 2, then 86 - 2 = 84(which is the multiple of 4)
Hence 2 is the least number.
(iv) 96 ≡ (x/7) (mod 5)
Solution :
96 ≡ (x/7) (mod 5)
By dividing 96 by 5, we get 1 as remainder
1 = x/7
x = 7
Hence the least value of x is 7.
(v) 5x ≡ 4 (mod 6)
Solution :
5x ≡ 4 (mod 6)
5x - 4 = 6n
If x = 2, then 10 - 4 = 6 (which is the multiple of 6)
Hence 2 is the least number.
Question 2 :
If x is congruent to 13 modulo 17 then 7x -3 is congruent to which number modulo 17?
Solution :
Question 3 :
Solve 5x ≡ 4 (mod 6)
Solution :
5x - 4 = 6n
n = (5x - 4)/6
n = [(6 - 1)x - (6 - 2)]/6
n = 6(-1x + 2)/6
n = (-1x + 2)
If x = 2, then n = -2 + 2 = 0
If x = 8, then n = -8 + 2 = -6
If x = 14, then n = -14 + 2 = -12
Hence the values of x are 2, 8, 14,.......
Question 4 :
Solve 3x −2 ≡ 0 (mod 11)
Solution :
3x - 2 = 11n
n = (3x - 2)/11
If x = 8, then n = (24 - 2)/11 = 2
(8 + 11) = 19
If x = 19, then n = (57 - 2)/11 = 5
Hence the values of x are 8, 19, 30.......
Question 5 :
What is the time 100 hours after 7 a.m.?
Solution :
On day consists of 24 hours, by dividing 100 by 24, we get 96 as quotient and 4 as remainder.
7 + 4 = 11 a.m
Question 6 :
What is the time 15 hours before 11 p.m.?
Solution :
11 - 15 = -4
Which is not the divisor of 12.
Find the next integer after 4 which is exactly divisible by 12. That is 12. Now -4 can be written as -4 = -12 + 8
Hence the answer is 8 am.
Question 7 :
Today is Tuesday. My uncle will come after 45 days. In which day my uncle will be coming?
Solution :
By dividing 45 by 7, we get 6 as quotient and 3 as remainder.
"0" corresponds to "Sunday"
"1" corresponds to "Monday"
"2" corresponds to "Tuesday"
"3" corresponds to "Wednesday"
"4" corresponds to "Thursday"
"5" corresponds to "Friday"
"6" corresponds to "Saturday"
2(Tuesday) + 45 = 47
By dividing 47 by 7, we get 5 as remainder.
5 - Friday.
Hence the answer is Friday.
Question 8 :
Prove that 2^{n} + 6×9^{n} is always divisible by 7 for any positive integer n.
Solution :
If n = 1
= 2^{1} + 6×9^{1}
= 2 + 54
= 56
56 is divisible by 7.
For n = 1, it is true. Let n = k
Assume that for n = k the given statement is true.
= 2^{k} + 6×9^{k }is divisible by 7.
2^{k} + 6×9^{k}^{ }= 7m ----(1)
To prove for n = k + 1
= 2^{k+1} + 6×9^{k+1}
= 2^{k} ⋅ 2 + 6 × (9^{k }⋅ 9)
= 2^{k} ⋅ 2 + 6 × 9^{k }(9)
By applying the value of 6×9^{k }from (1)
= 2^{k} ⋅ 2 + (7m - 2^{k}) (9)
= 2^{k} ⋅ 2 + (63m - 9 ⋅ 2^{k})
= 2^{k} (2 - 9) + 63m
= 2^{k} ( -7) + 63m
= 63m - 7 ⋅ 2^{k}
= 7(9 - 2^{k})
Hence it is divisible by 7.
Question 9 :
Find the remainder when 2^{81} is divided by 17.
Solution :
2⁸¹
= 2⁸⁰ ⋅ 2
= 2⁴ˣ²° ⋅ 2
= (2⁴)²⁰ ⋅ 2
= (16)²⁰ ⋅ 2
2⁸¹/17 Reminder
= (-1)²⁰ ⋅ 2
= 1 ⋅ 2
= 2
Question 10 :
The duration of flight travel from Chennai to London through British Airlines is approximately 11 hours. The airplane begins its journey on Sunday at 23:30 hours. If the time at Chennai is four and half hours ahead to that of London’s time, then find the time at London, when will the flight lands at London Airport.
Solution :
As journey time is 11 hrs, flight will reach London airport at 23:30 + 11 hrs = 10:30 Chennai time.
But, we know that Chennai time is 4:30 hrs ahead of London time, so at the time of landing time will be 10:30 - 4:30 hrs = 6:30 London time.
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