Solve the Factorials Sum Puzzle: Find the Last Two Digits

[MadScientist 1000]

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.

I got 73, but the answer key says 13. Can someone please help me? thx

 

[courtrigrad]

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!

 

[HallsofIvy]

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!.

 

[Edgardo]

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

 

[X=7]

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

 

[chaoseverlasting]

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

 

[CRGreathouse]

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.

 

[MadScientist 1000]

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

 

[dimitri151]

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.

 

[gowtham_012]

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

 

[dimitri151]

31035053229546199656252032972759319953190362094566672920420940313

 

[Mentallic]

lol ripped

 

[CRGreathouse]

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!

 

[Mentallic]

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

 

[dimitri151]

necroposted!

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?

 

[Mentallic]

What grade did you guys say you were in?

We didn't

 

[piyushon2411]

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

 

[chiro]

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.

 

[Mensanator]

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.

I got 73, but the answer key says 13. Can someone please help me? thx

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

 

[piyushon2411]

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

 

[piyushon2411]

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