Question

Find the remainder when 1!+2!+3!...100! Is divided by 15​

Answer 1

Answer: 33

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Explanation:

The exclamation mark means factorial.

Factorial's are operations in which we start with the given number, then count our way down to 1, multiplying along the way.

Examples:

• 7! = 7*6*5*4*3*2*1
• 4! = 4*3*2*1

The original number must be a positive whole number (side note: other inputs are possible but we have to involve more complicated math which I'll ignore for this solution).

Let's expand out the first few terms. I'll leave them in factored form and not multiply out the values

1! = 1

2! = 2*1

3! = 3*2*1

4! = 4*3*2*1

5! = 5*4*3*2*1 ... note the 3 and 5 in bold

6! = 6*5*4*3*2*1

7! = 7*6*5*4*3*2*1

This keeps going until we reach 100!

As you can see in the factors above, we have 3 and 5 show up as soon as we reach 5! which means we'll have 3*5 = 15 as a factor of the following items: {5!, 6!, 7!, 8!, ..., 99!, 100!}

Each higher factorial just builds in the last factorial so to speak.

So what this means is that we can write this

15q = 5!+6!+7!+...+99!+100!

where q is some integer. We don't need to worry about what q is

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Therefore,

1!+2!+3!+4!+5!+6!+7!+...+99!+100!

1!+2!+3!+4!+( 5!+6!+7!+...+99!+100! )

1!+2!+3!+4! + 15q

1+2+6+24 + 15q

33+15q

15q+33

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Recall that if we divide integers x and y, then

x/y = q remainder r

x/y = q + r/y

x = yq+r

Where q and r are the quotient and remainder.

Since 1!+2!+3!...100! = 15q+33, we're in the form yq+r showing that r = 33 is the remainder.