If a ! = a ( a − 1 ) ( a − 2 ) ( a − 3 ) . . .1 , then find X + Y + Z , given that X Y Z is a three digit number such that X Y Z = X ! + Y ! + Z !
Answers
Given : a ! = a ( a − 1 ) ( a − 2 ) ( a − 3 ) . . .1 . X Y Z is a three digit number such that X Y Z = X ! + Y ! + Z !
To find : X + Y + Z
Solution:
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
7! = 5040 ( Larger than 3rd digit )
XYZ is a 3 digit number
6! = 720 also not possible as 7 > 6
Hence XYZ can have digits 1 to 5
Three digit number hence > 100
largest 5 ! = 120
Hence maximum possible sum = 120 + 120 + 120 = 360
but for that 555 ( not possible)
120 + 120 = 240 => 255 = 2! + 5! + 5! = 242 ( hence not possible )
So we left with 100th digit as 1 and one digit as 51! = 1 5! = 120
1 Y 5 or 1 5 Z
heck 15Z 100 + 50 + Z = 1! + 5! + Z! => 150 + Z = 121 + Z!
=> Z! = 31 + Z
but maximum 4 ! = 24
Hence not possible
Check 1 Y 5
100 + 10Y + 5 = 1! + Y! + 5!
=> 105 + 10Y = 121 + Y !
=> Y! = 10Y - 16
Y can not be 1 as then 10Y - 16 will be -ve
Y = 2=> 2! = 20-16 ( not possible )
Y = 3=> 3! = 14 ( not possible )
Y = 4=> 4! = 40 - 16 = 24
Hence 145 is the number
X = 1
Y = 4
Z = 5
X + Y + Z = 1 + 4 + 5 = 10
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Step-by-step explanation:
If a ! = a ( a − 1 ) ( a − 2 ) ( a − 3 ) . . .1 , then find X + Y + Z , given that X Y Z is a three digit number such that X Y Z = X ! + Y ! + Z !
X = 1
Y = 4
Z = 5
_______________
145 = XYZ
= X! + Y! + Z!
= 1! + 4! + 5!
= 1 + 24 + 120
= 145
_______________
XYZ = 145