What number has a remainder of 3 when divided by 8?
NUMBERS
55 47 19 24 30 68
Answers
Answer:
the question is wrong none of them will not
Answer:
Basic Divisibility Rules
1) How do I find the remainder when 12345678910…99100 is divided by 16?
Ans: The divisibility test of 2^n is that you need to check the last ‘n’ digits of the number.
= To find out the remainder from 16, you need to check the last 4 digits
= Rem [12345….99100 / 16] = Rem [9100/16] = 12
2) Quantitative Aptitude: What is the remainder when (111…) + (222…) + (333…) + … + (777…) is divided by 37?
Ans: aaa = a*111 = a*3*37
= aaa is divisible by 37
= aaaa….. repeated 3n number of times is divisible by 37
= (1111…..1) 108 times is divisible by 37
Now, 1111…111 (110 times) = 111…..1100 (108 1s and 2 0s)+ 11
= Rem [1111…111 (110 times) /37] = 11
= Rem [2222…222 (110 times) /37] = 22
.
.
= Rem [7777…777 (110 times) /37] = 77
So, we can say that
Remainder when (111…) + (222…) + (333…) + … + (777…) is divided by 37
= Rem [11 + 22 + 33 + 44 + 55 + 66 + 77 / 37]
= Rem [308/37]
= 12
3) What is the remainder when 123456789101112131415161718192021222324252
627282930313233343536373839404142434481 is divided by 45?
Ans: We need to find out Rem [1234…..434481/45]
This looks a little difficult if you do not any theorems for finding out remainders. Let us take a simpler approach and break down the problem into smaller parts.
These type of questions become really simple if you understand the concept of negative remainders. Always try and reduce the dividend to 1 or -1.
45 = 9*5
Let us find out the remainders separately and combine them later
Rem [1234…..434481/9]
The divisibility test of 9 is to divide the sum of the digits by 9.
The sum in this case is 1 + 2 + 3 … 43 + 44 + 81 = 44*45 + 81
We can see that this is divisible by 9
= The number is divisible by 9
= Rem [1234…..434481/9] = 0
Rem [1234…..434481/5] = 1 (It only depends on the last digit)
So, our answer is a number which leaves a remainder of 1 when divided by 5 and is divisible by 9.
Consider multiple of 9,
9, does not leave a remainder of 1 from 5. Invalid.
18, does not leave a remainder of 1 from 5. Invalid.
27, does not leave a remainder of 1 from 5. Invalid.
36, leaves a remainder of 1 from 5. Valid. (This is our answer)
4) What is the remainder when 123456………….4647484950 is divided by 16?
Ans: To find out the remainder from 2^n, we just need to look at the last ‘n’ digits.
= Rem [123…484950 / 16]
= Rem [4950/16]
= 6
5) What is the remainder when 1! + 2! + 3! … 100! is divided by 18?
Ans: We have to find out Remainder of when divided by 18.
= Rem [(1! + 2! + 3! … 100!)/18]
6! is divisible by 18
7! is divisible by 18
.
.
100! is divisible by 18
= We have to find out Rem[(1! + 2! + 3! + 4! + 5!)/18]
= Rem [ (1 + 2 + 6 + 24 + 120)/18]
= Rem [153/18] = 9
6) What is the remainder when the infinite sum (1!)² + (2!)² + (3!)² + ··· is divided by 1152?
Ans: We have to find out the remainder when (1!)² + (2!)² + (3!)² + ··· is divided by 1152
1152 = 2^7 * 3^2
= (6!)^2 is divisible by 1152
= All (n!)^2 are divisible by 1152 as long as n > 5
So, our problem is now reduced to
Rem [((1!)² + (2!)² + (3!)² + (4!)² + (5!)²)/1152]
= Rem[(1 + 4 + 36 +576 + 14400) / 1152]
= Rem [15017/1152]
= 41
7) What is the remainder when 3^21 + 9^21 + 27^21 + 81^21 is divided by (3^20+1)?
Ans: 3^21 + 9^21 + 27^21 + 81^21
= 3^21 + (3^21)^2 + (3^21)^3 + (3^21)^4
= x + x^2 + x^3 + x^4
where x = 3^21
Now, 3^21 = 3(3^20) = 3(3^20 + 1) – 3
= Rem [3^21 / (3^20+1)] = -3
= Rem [x / (3^20 + 1)] = -3
We can use this to find out out answer
Rem [(x + x^2 + x^3 + x^4) / (3^20 + 1) ] = (-3) + (-3)^2 + (-3)^3 + (-3)^4
= -3 + 9 – 27 + 81 = 60
8) What is the remainder when 2222…..300 times is divided by 999?
Ans: To check divisibility by 999, check the sum of the digits taken 3 at a time
Sum of the digits of 222…. 300 times (taken 3 at a time)
= 222 + 222 + 222…. 100 times
= 22200
Rem [22000/999] = 222