Question

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 !

Answer 1

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
_______________

$\rm\red{(A)}\blue{\underline{\bf\green{ \quad XYZ = 145 \quad }}}$

$\mid \underline{\underline{\LARGE\bf\green{Brainliest \: Answer}}}\mid$

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Answer 2

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  5

1! = 1  

5!  = 120

1 Y 5     or  1 5 Z

Check 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!  = 4   ( 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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