Hi guys! I have a question for you all. If u r a genius, then answer this!
a + b + c = abc. Find the values of a, b, and c.
Correct answer will be marked as brainliest, and wrong answer or spam will be reported.
Answers
Answer:
My strategy to find the answer to this problem is by using elimination of impossible combination.
First, let's list all the possible digits:
0! = 1
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
7! = 5040
8! = 40320
9! = 362880
Digit 7, 8, and 9 can't be used since the value is more that three digits even before they are added.
Digit 6 also can't be used since the value contains a digit that is higher than 6 which will make the condition A! + B! + C! = ABC impossible.
So, we only have below list of possible digits now:
0! = 1
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
Can we further eliminate digit 5? Let's see..if 5 is eliminated, then the biggest combination will not add to a three-digit number:
4! + 3! + 2! = 24 + 6 + 2 = 32
So, now we know that digit 5 should be used somewhere.
Let's list the sum range of the possible answer. We'll start with the biggest possible sum:
5! + 4! + 3! = 150
and the smallest possible sum:
5! + 1! + 0! = 122
from the range of possible sum (122 to 150) we know that A is 1. That also means, 5 is either at the B or C digit.
If B is 5, then C must be 0 to satisfy the condition. This is not possible since 1! + 5! + 0! ≠ 150
So now we are sure that 5 is at the C digit. We only need to find the value of B digit then to solve this problem.
From the possible sum range, we know that the value of B should be 2 ≤ B ≤ 4 so there are only 3 possible digits to try: 2, 3, and 4. We may do trial and error here if we want to. But let's deduce it mathematically.
Let's take a look again at the problem:
A! + B! + C! = ABC
1! + B! + 5! = 1B5
1 + B! + 120 = 1B5
B! + 1 + 120 = 1B5
B! + 121 = 1B5
We want to make the last digit of the sum to be 5. so we need B! value that ends in 4. What are the possible B digit again?
2! = 2. It doesn't end in 4.
3! = 6. It doesn't end in 4.
4! = 24. It ends in 4.
Let's see if B = 4 fits:
1! + 4! + 5! = 1 + 24 + 120 = 145
1! + 4! + 5! = 145 -> it fits! Solved.
With A = 1, B = 4, C = 5
A! = 1, B! = 24, C! = 120
ABC = 145
and A!+B!+C! = 1 + 24 + 120 = 145
Sum of A, B, and C = 1+4+5 = 10