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Chapter 5: Problem 453

The reminder when number \(1 !+2 !+3 !+4 !+\ldots \ldots+100 !\) is divided by 240 is (a) 153 (b) 33 (c) 73 (d) 187

Short Answer

Expert verified
The remainder when the sum of factorials from 1! to 100! is divided by 240 is 33.

Step by step solution

01

Find the sum of factorials from 1! to 100!

Let's denote the given sum as S. Then, \[S = 1! + 2! + 3! + 4! + \cdots + 100!\]
02

Observe a pattern and deduce properties

Notice that: - \(1! = 1\) - \(2! = 2\) - \(3! = 6\) - \(4! = 24\) When you have factorials of numbers greater than or equal to 5, they will have a 5 as a factor (for example: \(5! = 5 \times 4 \times 3 \times 2 \times 1\), \(6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1\) and so on). This means that all factorials of numbers greater than or equal to 5 will be divisible by 5, and when these factorials are divided by 240, they will leave a remainder of 0. So, to find the remainder of the given sum when divided by 240, we only need to consider the first 4 factorials, i.e., \[S = 1! + 2! + 3! + 4!\]
03

Evaluate S

We already have the values of individual factorials, so now let's find the value of S. \[S = 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33\]
04

Find the remainder

Now, let's find the remainder when S is divided by 240. Since S is already lesser than 240, the remainder when S is divided by 240 is simply the value of S. Therefore, the remainder is 33. The correct answer is (b) 33.

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Most popular questions from this chapter

If \(500 !=2^{\mathrm{m}} \times\) an integer, then (a) \(\mathrm{m}=494\) (b) \(m=496\) (c) It is equivalent to number of \(n\) is \(400 !\) is \(=2^{n} \times\) an integer (d) \(\mathrm{m}={ }^{500} \mathrm{C}_{2}\)

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