# Solve the Factorials Sum Puzzle: Find the Last Two Digits

In summary: The answer is C) 493. To understand this, we need to consider the fact that any set of consecutive numbers can be expressed as the average of the first and last number, multiplied by the number of elements in the set. For example, the set [1,2,3] can be expressed as (1+3)/3 * 3 = 2 * 3 = 6. This means that for any set of consecutive numbers, we can find the sum by multiplying the average of the first and last number by the number of elements in the set. Now, for the set N, the smallest possible sum is 3 (1+2). The largest possible sum is 499 (249+250). This
Here is a sum from MATHCOUNTS:

What are the last two digits in the sum of the factorials of the first 100 positive integers?

From 1! to 4! you can add the units digits, since 5! to ... have 0 in their units place.

From that I get 13, and I carry over the 1 over to the next column and add the tens digits of 1! to 9! since 10! to ... have 0 in their tens and units place.

So $$\sum_{n=1}^{100} n!$$$$0!: 01$$

$$1! : 01$$

$$1!+2!: 03$$

$$1! + 2! + 3!: 09$$

$$1!+2!+3!+4!: 33$$

$$1!+2!+3!+4!+5!: 53$$

Can you see a pattern?

Go up to $$9!$$ because $$\sum_{n=1}^{10} n!$$ has the last two digits $$00$$. Therefore $$\sum_{n=1}^{\19} n!$$ has last two digits $$13$$ as does $$\sum_{n=1}^{100} n!$$

Last edited:
So $$\sum_{n=1}^{100} n!$$

$$0!: 01$$

$$1! : 01$$

$$1!+2!: 03$$

$$1! + 2! + 3!: 09$$

$$1!+2!+3!+4!: 33$$

$$1!+2!+3!+4!+5!: 53$$

Can you see a pattern?

Go up to $$9!$$ because $$\sum_{n=1}^{10} n!$$ has the last two digits $$00$$. Therefore $$\sum_{n=1}^{\19} n!$$ has last two digits $$13$$ as does $$\sum_{n=1}^{100} n!$$

I presume you meant to say that 10! has last two digits 00, not $$\sum_{n=1}^{100} n!$$.

1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
7! = 5040
8! = ...20 (the dots stand for some digits, but I didn't calculate them, since I am only interested in the last two digits)
9! = ...80
10! = ...800

Which numbers in the sum of factorials contribute to the last digit? It's
1! = 1
2! = 2
3! = 6
4! = 24
the other numbers have 0 as their last digit.
Thus, the last digit of our factorial sum is 3 because 1+2+6+24=33

Which numbers in the sum of factorials contribute to the "10" digit (the digit left to the last digit)?
It's

5! = 120
6! = 720
7! = 5040
8! = ...20
9! = ...80

AND the 33 (= sum from 1! to 4!)

I wrote down the relevant numbers again and behind the numbers their "10" digit in brackets:

5! = 120 (2)
6! = 720 (2)
7! = 5040 (4)
8! = ...20 (2)
9! = ...80 (8)
33 (3)

Let us add the numbers in the brackets:
2+2+4+2+8+3 = 21

Thus, for you sum of factorials the "10" digit is:
1

HallsofIvy said:
I presume you meant to say that 10! has last two digits 00

Hi HallsofIvy. I think you've misread courtrigrad's post. It (correctly) says that the sum to 100 has last two digits 13.

Best wishes

X = 7

I know this sounds dumb, but I am sorry, I don't see the pattern...

chaoseverlasting said:
I know this sounds dumb, but I am sorry, I don't see the pattern...

Did you calculate them out, or are you just looking at the above? I suggest you calculate the (last two digits of) the factorials from 1 to 10, then find the sums of those. The pattern should be obvious.

Sorry guys... I reworked it and found out I made a mistake... The answer was 71 (taking the last 2 digits of the sum).

It's natural that the last few digits of of the sum of the factorials 1!+2!+3!+..should be the same. The succeeding terms of the series have all zeros (the number of zeroes depending on how many times 2 and 5 appear in the prime factorization of n!) and so do not affect the first digits of the sums representation.

1!+2!+3!...50!=
wats the answer guys help me out...

31035053229546199656252032972759319953190362094566672920420940313

lol ripped

dimitri151 said:
It's natural that the last few digits of of the sum of the factorials 1!+2!+3!+..should be the same. The succeeding terms of the series have all zeros (the number of zeroes depending on how many times 2 and 5 appear in the prime factorization of n!) and so do not affect the first digits of the sums representation.

necroposted!

At least he can use the search button! So they kind of cancel each other out in a way.

CRGreathouse said:
necroposted!
Mentallic said:
At least he can use the search button! So they kind of cancel each other out in a way.
What grade did you guys say you were in?

dimitri151 said:
What grade did you guys say you were in?

We didn't

How to find out number of digits in 1!+2!+3!...+100!?

piyushon2411 said:
How to find out number of digits in 1!+2!+3!...+100!?

Hey piyushon2411 and welcome to the forums.

For this problem in base b you need to calculate log_b(z) = ln(z)/ln(b) where b is the number of possibilities in each digit. Round up if you have a fractional part in your answer.

The z is your expression 1!+ 2! + 3! + ... blah

What you should realize is that for this problem you only need to evaluate the highest term which is 100!.

Using properties of logs you can use the property log(ab) = log(a) + log(b) which means the answer for this problem is:

log(1) + log(2) + ... log(100) from 1 up to 100 like the sum suggests.

Here is a sum from MATHCOUNTS:

What are the last two digits in the sum of the factorials of the first 100 positive integers?

From 1! to 4! you can add the units digits, since 5! to ... have 0 in their units place.

From that I get 13, and I carry over the 1 over to the next column and add the tens digits of 1! to 9! since 10! to ... have 0 in their tens and units place.

The answer IS 13. I think you miscounted the 10s. I get 8+2+4+2+2+2

Hey chiro, thanks for the solution,
i doubt that i got it fully ,can u please explain wt u did after we consider just 100! to get the number of digits,thanks in advance

For the above problm to get the last 2 digits,
going by the conventnl method, last 2 digits from 1 to 9 factorial goes like
01+02+06+24+20+20+40+20+80= 13 in the last 2 digits

Also ,please can someone explain how to do these kind of questions ,forex-

N is a set of all natural numbers less than 500 which can be written as the sum of 2 or more consecutive natural numbers,find the max number of elements possible in N?
A.)250
B.)492
C.)493
D.)None of these

## 1. What is the "Factorials Sum Puzzle"?

The Factorials Sum Puzzle is a mathematical puzzle that involves finding the last two digits of a sum of factorials.

## 2. How do I solve the Factorials Sum Puzzle?

To solve the puzzle, you need to first find the individual factorials of each number in the given sum. Then, add all the factorials together and find the last two digits of the resulting number.

## 3. What is the significance of finding the last two digits?

In mathematics, finding the last two digits of a number can sometimes reveal patterns or help in simplifying calculations. In the context of the Factorials Sum Puzzle, it is the ultimate goal of the puzzle and requires critical thinking and problem-solving skills.

## 4. What are some tips for solving the Factorials Sum Puzzle?

Some tips for solving the puzzle include breaking down the given sum into smaller parts, recognizing patterns in the factorials, and using divisibility rules to simplify calculations.

## 5. Are there any real-world applications for the Factorials Sum Puzzle?

The Factorials Sum Puzzle is a purely mathematical puzzle and does not have any direct real-world applications. However, it can help in developing critical thinking skills and problem-solving abilities, which are valuable in various fields of science and technology.

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