a,b,c are 3 digits?

Question

(abc)base10 is a three digit number , with non-zero digits a,b and c, such that the following conditions are satisfied

(i)a
(ii)a!+b!+c!=(abc)base10

Find the value of (axbxc)^b


how to solve it ?

-----------------------------------------------------------
i want to find the value a,b,c values without trial qand error

-------------------------------------------
Dr . can you provide simpler solution :-(

Answer 1

There really is no way that does not involve some sort of trial and error.

Let b = a + x
c = a + y
where 0 < x < y < 5

So a! + b! + c! =
a! * [1 + (a+x)*(a+x-1)*...*(a+1) + (a+y)*(a+y-1)*...*(a+1) ]
= 111*a + 10x + y

From here you just have to try values until you find the answer.

**EDIT**
Since the sum of the factorials is a 3 digit number, and 4! = 24, c = 5. That way c! = 3 digit no.

So the possibilities are:
1,2,5 ; ∑ !'s = 123
1,3,5 ; ∑ !'s = 127
1,4,5 ; ∑ !'s = 145
2,3,5 ; ∑ !'s = 128
2,4,5 ; ∑ !'s = 146
3,4,5 ; ∑ !'s = 150

So 1,4,5 works

Now (1*4*5)^4 = 20^4 = 160,000

Answer 2

Maybe you mean solving it by deduction. In that case, let's make the problem a little more challenging:

(1) a, b, and c are non-zero digits from 0 to 9, while a, being the first digit of abc, cannot be zero.

(2) Repeating digits are allowed, as are the relationships between the numbers. For example, a > b, a < b, or a = b are all permitted.

(3) a! + b! + c! = (abc) base 10


OK, to solve this, first note that 7!, 8!, and 9! are all at least four digit numbers. That rules out 7, 8, and 9 as candidates, since abc must be a three-digit number.

Also, 6 can be ruled out because 6! is 720, meaning the three digit number has a digit 7 or higher in it, which we just ruled out. So, 6 can't be a digit.

So far we know that all three digits must be 5 or less. Now suppose all of the digits are less than 5. In that case, the sum of the three factorials is at most 72 -- not a three-digit number. From that we can conclude that one of the digits MUST be a 5.

Next, all three digits can't be 5. We can test that one directly: 5! + 5! + 5! = 360, not 555. Can two of the digits be 5s? In that case, the third digit would have to be 2, since 5! + 5! + x!, where x < 5, is more than 240 but less than 300. However, 2! + 5! + 5! = 242, not 255. So, two 5s is not a possibility.

We now know that EXACTLY one digit is a 5, the other two less than 5. That means abc is more than 120 and at most 120 + 24 + 24 = 168. So, the first digit of abc has to be a 1. We know a = 1.

We're down to abc = 1x5 or 15x, where x < 5. However, the largest number we can get now is 1 + 24 + 120 = 145, which rules out 15x. So, now we know that c = 5.

The only remaining digit is the middle one. It has to be at least 2, since 5! makes abc at least 120. So, we are down to three possible numbers: 125, 135, and 145. From here we might as well test them individually, and we find that abc = 145 is the only solution.

Answer 3

By trial & error. There are so few numbers to check < 6 that abc is quickly found to be 145, so that (1*4*5)^4 = 160,000