Find that Factorial!

Note that 40585 = 4! + 0! + 5! + 8! + 5! (i.e. the number can be expressed as the sum of the factorial of each of its digits).

There is only one three digit number that can be expressed this way. Find the sum of the digits of that number.

(P.S: I would really appreciate it if somebody could share their complete solutions to this problem.)


The answer is 10.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

4 solutions

Mudra Desai
Apr 5, 2014

Step 1) First of all please note that the required three digit number will never have the number >=6 in it as 6! = 720 you will see by trial and error that for this case you will get an exceeding number. Step 2) Now left are the numbers 5,4,3,2 and 1 . You will see in this case by placing 1! , 2! and 3! you will never get a 3 digit number. Step 3) Now simply manipulating the digit 4! and 5! in a set and with them joining 1!, 2!,3! in a set you will get 145 i.e 1!+4!+5!=145 whose sum of digits is 1+4+5=10

This is by trail and error method : 6! equals 720 but 7! is not 3 digit number so digits must be less than 6. below 6 only 5! is 3 digit number so required nuber contains 5 if all digits 5 the num is 360 only so 5 must be in 10's pos (or) 1's pos let it in 1's pos the num must contain either of these pairs 0,4(or) 1,4 from those 1,4 pair gives the riquire sollution so the num is 145 and sum is 10

Swati Shukla
Feb 22, 2014

Step 1) First of all please note that the required three digit number will never have the number >=6 in it as 6! = 720 you will see by trial and error that for this case you will get an exceeding number. Step 2) Now left are the numbers 5,4,3,2 and 1 . You will see in this case by placing 1! , 2! and 3! you will never get a 3 digit number. Step 3) Now simply manipulating the digit 4! and 5! in a set and with them joining 1!, 2!,3! in a set you will get 145 i.e 1!+4!+5!=145 whose sum of digits is 1+4+5=10

Also you have a magical trick too see 5+4+3+2+1=10 .. the sum of number <=6 will always be the answer (to use the short cut)

Swati Shukla - 7 years, 3 months ago
Himadri Shee
Feb 21, 2014

1! + 4! + 5! = 1 + 24 + 120 = 145

So, 145 is the required 3-digit number. Now, 1+4+5= 10 is the answer

was it a hit and trial method that u guessed 145 or there is any specific method????

Anmoldeep Singh - 7 years, 3 months ago

Log in to reply

Yeah, I also want to know if there is a particular mathematical way of doing this.

Mark Mottian - 7 years, 3 months ago

145 is the answer. For doing this u should know that only 5 is the number with a 3 digit factorial.This helps to eliminate more than half the 3 digit nos then use numerical approximation. FOLLOW PLEASE. CHEERS

pratyush ranjan Tiwari - 7 years, 3 months ago

How did you realise that 145 was the correct number? Did you just exhaust all possibilities until you reached 145?

Mark Mottian - 7 years, 3 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...