If ABC is a three-digit integer and ABC = A! + B! + C!, what is the va

Question

If ABC is a three-digit integer and ABC = A! + B! + C!, what is the value of A + B + C ?

A. 8
B. 9
C. 10
D. 11
E. 12

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Babineaux · Answer

If ABC = A! + B! + C! ;

Then C must be 5

And others, A = 1 and B = 4

Thus 1! + 4! + 5! = 145

Hence, 1 + 4 + 5 = 10

IMO ans is C.

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Kinshook · Answer

Asked: If ABC is a three-digit integer and ABC = A! + B! + C!, what is the value of A + B + C ?
 
100A + 10B + C = A! + B! + C!

1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720;
7! = 5040; Digits {7,8,9} are NOT FEASIBLE
Hundredth digit of A! + B! + C! = 1 ; A= 1; One of the other 2 digits = 5
A = 1; B or C = 5

Case 1: 
A=1;B=5
150 + C = 1 + 120 + C!
C! - C = 29; Not feasible

Case 2: 
A=1; C=5 
105 + 10B = 120 + 1 + B!
B! - 10B + 16 = 0 
24 - 40 + 16 = 0 
B = 4

ABC = 145 = 1 + 24 + 120 = A! + B! + C!
A + B + C = 10

IMO C

nick1816 · Answer

Good one!

Max (A,B,C) < 7, as 7! is a 4 digit number.

Max(A,B,C) can't be equal to 6 either, as A must be greater than or equal to 7 (6! = 720).

Now at least one among A,B and C must be 5, else A!+B!+C! can not be a 3 digit number. (4!+4!+4!=72 <100)

A can't be equal to 5, as (5!+5!+5!=360<500)

120+1+1 < A!+B!+C! < 120+120+24

Hence, A can be 1 or 2

Case 1-

i)  A=1 and B=5

120+1+C! = 150+C

C! =29+C
Since C is a single digit number from 0 to 5, C! must lies between 29 and 34.

Not possible

ii)  A=1 and C=5

120+1+B! = 100+10B+5

B! = 10B -16

Unit digit of 10B -16 is 4, hence, unit digit of B! = 4 or B=4

120+1+4! = 100+40+5

145 =145 (BINGO)

We don't need to check further.

A=1, B=4 and C=5

A+B+C=10

dimri10 · Answer

hi,

can i ask how did you conclude that A is 1.? why is the assumption that A can not be 6! or 5! and B and C are other numbers.

i just tried different variations with the numbers, but knowing that A is 1 makes this question a lot easier, but i cant get why 
i get why 7! and up is out.

kind regards dear friends

GMATWhizTeam · Answer

Solution 
 • ABC = A! + B! + C!
• Since ABC is a three-digit number,
 o So, the greatest digit among A, B, and C, must be greater than or equal to 5 and must be less than 7.
  As 5! = 120 and 7! = 5040  
• Also, greatest digit among A, B, and C, cannot be 6 as 6! = 720, 
 o and in that case the greatest digit will be 7 not 6. 
• So, we have the greatest digit among A, B, and C as 5, so other two digits can be:
 o Case 1: 1 and 2
  In that case A! + B! + C! = 1! + 2! + 5! = 1+ 2 + 120 = 123
 And, ABC = 125
 So, A! + B! + C! ≠ ABC 
o Case 2: 1 and 3
  In that case A! + B! + C! = 1! + 3! + 5! = 1 + 6 + 120= 127
 And, ABC = 135.
 So, A! + B! + C! ≠ ABC 
o Case 3: 1 and 4
  In that case A! + B! + C! = 1! + 4! + 5! = 1 + 24+120 = 145
 And, ABC = 145.
 So, A! + B! + C! = ABC
 Hence, A + B+ C = 1 + 4 + 5 = 10   
Thus, the correct answer is  Option C.

Lipun · Answer

Given A! + B! + C! = ABC

7! > 1000, so we need to consider A,B,C < 7.
Again, if either of them is 6, then the hundreds place will be >= 7 meaning the overall number will not remain a 3-digit number. (since 7 cannot be one of the digits)

=> A,B,C = {0,1,2,3,4,5}. A!+B!+C! will always be less than 360. => A = {1,2,3}.

A will be 1 always because,
a) if A = 2, then only possibility will be 255 which doesn't satisfy the given condition.
b) if A = 3, then A! + B! + C! < 300.

A = 1 => B = 5 or C = 5 for ABC to be a 3-digit number. So, ABC = 15C or 1B5.

By trial and error we can determine B = 4 satisfies the given condition.

Ans: 1+4+5 = 10

Lipun · Answer

Hi  dimri10 ,

We know that A,B,C will be less than 7. Possible set = {0,1,2,3,4,5,6}

If any of the digits is 6, then the resultant number will be > 700, meaning A >= 7. From, the above statement we know that A < 7. So possible set for A,B,C = {0,1,2,3,4,5}

Let's consider the maximum value of A! + B! +C!. It will be maximum when A=B=C=5 and the maximum value will be 360. => A = {1,2,3}

Say, A = 3 => A! = 6. But max value of B! + C! = 240. We can't reach a sum of 300 with A=3. So, A=3 is discarded.
Say, A = 2 => A! = 2. To reach a sum of 200 we need B = C = 5. So, the number becomes 255. But A! + B! +C! = 242. So, A=2 is also discarded.

Thus, A can take the value of 1 only.

Hope it helps!

KarishmaB · Answer

This was my thought process behind it: 
The sum of all factorials of digits is a 3 digit number. This means that no digit can be more than 6 because 6! is 720 and 7! is a four digit number. But the sum cannot be 700+ because no digit can be 7 or 8 or 9 (because their factorials are 4 digit numbers). Hence no digit is 6 either. 
So the digits must be only 0 to 5. Hence the sum must be 100 something because 5! = 120. 
So A must be 1. One of the digits must be 5 too. Hence we have already got a sum of 120 + 1 = 121.
If one of the digits is 4, we add 24 to it to get 145 and that works.

Value of A + B + C = 1 + 4 + 5 = 10

Answer (C)

If ABC is a three-digit integer and ABC = A! + B! + C!, what is the va

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