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18 Mar 2022, 05:45
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The factorial (!) of a positive integer n denotes the product of all integers from 1 to n, inclusive. If \(k = 1! + 2! + 3! + . . . + p!\), where p is a prime number greater than 10, what is the remainder when k is divided by 4?
A. 0
B. 1
C. 2
D. 3
E. 9
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B. 1
C. 2
D. 3
E. 9
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19 Mar 2022, 01:11
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Every factorial from \(4!\) to \(p!\) will have \(4\) within it and thus we need to only focus on the terms which do not have \(4\).
\(1! + 2! + 3!\)
\(= 1 + 2 + 6\)
\(= 9\)
\(\frac{9}{4} = 2\) with a remainder of \(1\).
Answer B
\(1! + 2! + 3!\)
\(= 1 + 2 + 6\)
\(= 9\)
\(\frac{9}{4} = 2\) with a remainder of \(1\).
Answer B
General Discussion
02 Sep 2022, 21:34
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Given that k = 1! + 2! + 3! + . . . + p!, where p is a prime number greater than 10. And we need to find the remainder when k is divided by 4
In these type of problems we need to find a pattern and then try to eliminate the terms based on the pattern.
Now, k = 1! + 2! + 3! + . . . + p! and we know that a higher factorial will always be a multiple of lower factorial
Ex: 4! = 1*2*3*4 = 3! * 4
So, if we can prove that the remainder of lets say n! with 4 is 0 then remainder of all higher factorials like (n+1)!, (n+2)! will be 0 too.
So, lets start finding the remainder of factorials from 1 onwards and see when we can reach a remainder of 0
Remainder of 1! by 4 = Remainder of 1 by 4 = 1
Remainder of 2! by 4 = Remainder of 1*2(=2) by 4 = 2
Remainder of 3! by 4 = Remainder of 1*2*3(=6) by 4 = 2
Remainder of 4! by 4 = Remainder of 1*2*3*4 by 4 = 0 (As it is a multiple of 4)
Now, all factorials from 5! onwards will give a remainder of 0 when divided by 4, as all of them will be a multiple of 4.
=> Remainder of k with 4 = Remainder of (1! + 2! + 3! + . . . + p!) by 4 = Remainder of 1! by 4 + Remainder of 2! by 4 and so on
(Theory: Remainder of sum of two numbers by a number is same as the sum of remainder of those two numbers (individually) by that number)
= 1 + 2 + 2 + 0 + 0....+0 = 1 + 2 + 2 = 5
But remainder of k by 4 cannot be greater than 3
(Theory: If we are dividing a number by n then we can get remainders only from 0 to n-1)
=> Remainder will be = Remainder of 5 by 4 = 1
So, Answer will be B
Hope it helps!
Watch the following video to learn the Basics of Remainders
In these type of problems we need to find a pattern and then try to eliminate the terms based on the pattern.
Now, k = 1! + 2! + 3! + . . . + p! and we know that a higher factorial will always be a multiple of lower factorial
Ex: 4! = 1*2*3*4 = 3! * 4
So, if we can prove that the remainder of lets say n! with 4 is 0 then remainder of all higher factorials like (n+1)!, (n+2)! will be 0 too.
So, lets start finding the remainder of factorials from 1 onwards and see when we can reach a remainder of 0
Remainder of 1! by 4 = Remainder of 1 by 4 = 1
Remainder of 2! by 4 = Remainder of 1*2(=2) by 4 = 2
Remainder of 3! by 4 = Remainder of 1*2*3(=6) by 4 = 2
Remainder of 4! by 4 = Remainder of 1*2*3*4 by 4 = 0 (As it is a multiple of 4)
Now, all factorials from 5! onwards will give a remainder of 0 when divided by 4, as all of them will be a multiple of 4.
=> Remainder of k with 4 = Remainder of (1! + 2! + 3! + . . . + p!) by 4 = Remainder of 1! by 4 + Remainder of 2! by 4 and so on
(Theory: Remainder of sum of two numbers by a number is same as the sum of remainder of those two numbers (individually) by that number)
= 1 + 2 + 2 + 0 + 0....+0 = 1 + 2 + 2 = 5
But remainder of k by 4 cannot be greater than 3
(Theory: If we are dividing a number by n then we can get remainders only from 0 to n-1)
=> Remainder will be = Remainder of 5 by 4 = 1
So, Answer will be B
Hope it helps!
Watch the following video to learn the Basics of Remainders
