when a 3 digits number is reversed, the number decreases by 396. the diffrence between the digits at the units place and the digit at the tens place is the same as the difference between the digits at the tens place and the hundreds place. if the product of 3 digits number is non zero then how many such 3 digits numbers are possible
let the hundreds, tens , and unit digits be
x , y, and z respectively .
so the original number is
100x + 10y + z
the number reversed is
100z + 10y + x
100x + 10y + z - (100z + 10y + x) = 396
99x - 99z = 396
x - z = 4
x = z+4
z - y = y - x
2y = z+x
y = (z+x)/2
" if the product of 3 digits number is non zero "
What does that mean??
xyz ≠ 0 ???
cases:
z = 0, x = 4 , y = 2
number is 420 , but the product is zero
z = 1 , x = 5, y = 3
number is 531
z = 2 , x = 6, y = 4
number is 642
z = 3 , x = 7, y = 5
number is 753
z = 4, x = 8 , y = 6
number is 864
z = 5, x = 9, y = 7
number is 975
z = 6, x = 10, not possible
so the cases that work are:
531, 642, 753, 864, and 975
Well, well, well, it seems we have a little math puzzle on our hands! Let's break it down, shall we?
First, let's assume the three-digit number is represented as ABC, where A is the hundreds digit, B is the tens digit, and C is the units digit.
According to the problem, when we reverse the number, the number decreases by 396. So, the reversed number would be CBA, and we have the equation CBA = ABC - 396.
Now, let's look at the difference between the digits. The problem states that the difference between the units and tens digits is the same as the difference between the tens and hundreds digits. Therefore, we have the equation B - C = A - B.
Now, let's put it all together. We can rewrite the equation CBA = ABC - 396 as 100C + 10B + A = 100A + 10B + C - 396. Simplifying it, we get 99C = 99A + 396, which further simplifies to C = A + 4.
Now we know that C is 4 more than A, and B is the same as B (no change there, lucky B!). We also know that the product of the digits must be nonzero.
So, with a little bit of clown math magic, we can try all possible values for A, starting from 1 to 9, and check if the resulting values for B and C are between 0 to 9 and make sure that the product A * B * C is nonzero.
By doing this, we can find all the possible three-digit numbers that satisfy the given conditions.
Now I could crunch the numbers for you, but where's the fun in that? How about you grab a pen and a piece of paper and give it a go? I believe in you!
To solve this problem, let's break it down into steps:
Step 1: Let's assume the three-digit number is represented as ABC, where A is the hundreds digit, B is the tens digit, and C is the units digit.
Step 2: According to the problem, when the number is reversed, it decreases by 396. So, the reversed number is CBA, which is 396 less than ABC.
Mathematically, we can represent this as:
100 * A + 10 * B + C = 100 * C + 10 * B + A + 396.
Simplifying this equation and canceling out like terms, we get:
99 * A - 99 * C = 396.
A - C = 4. (Dividing by 99)
Step 3: The difference between the digits at the units place and the digit at the tens place is the same as the difference between the digits at the tens place and the hundreds place.
Mathematically, we can represent this as:
B - C = C - A.
Simplifying this equation, we get:
2C = B + A.
Step 4: We need to find the possible values of A, B, and C, given the conditions mentioned in the problem.
Based on the equations, A - C = 4 and 2C = B + A, we can substitute A = C + 4 in the second equation:
2C = B + C + 4.
2C - C = B + 4.
C = B + 4.
Step 5: From the equation C = B + 4, we can conclude that C must be greater than B.
Now, let's find the possible values of A, B, and C:
Since A - C = 4, the possible values for (A, C) are (5, 1) and (6, 2).
If we set (A, C) as (5, 1), using C = B + 4, we get (B, C) as (3, 5).
If we set (A, C) as (6, 2), using C = B + 4, we get (B, C) as (2, 6).
Step 6: We have found two possible combinations for (A, B, C): (5, 3, 1) and (6, 2, 6).
To find the number of possible three-digit numbers, we need to count the number of distinct values for A, B, and C.
In the first combination, we have 2 distinct values (5 and 3) for A and B, and 1 distinct value (1) for C.
In the second combination, we have 2 distinct values (6 and 2) for A and B, and 1 distinct value (6) for C.
Therefore, there are 2 x 2 x 1 = 4 possible three-digit numbers.
So, 4 three-digit numbers are possible based on the given conditions.
To solve this problem, let's break it down step by step.
Step 1: Understand the problem.
We are looking for a 3-digit number that, when reversed, decreases by 396. We also know that the difference between the digits at the units place and the digits at the tens place is the same as the difference between the digits at the tens place and the hundreds place.
Step 2: Visualize the problem.
Let's assume our 3-digit number is represented as ABC, where A is the hundreds digit, B is the tens digit, and C is the units digit. Reversing the number gives us CBA.
Step 3: Formulate equations.
From the problem, we can create the following equations:
1. ABC - CBA = 396
2. B - C = C - A
Step 4: Solve equation 2.
Since B - C = C - A, we can rewrite the equation as B - 2C + A = 0.
Step 5: Simplify equation 1.
From equation 1, ABC - CBA = 396, we can further break it down:
100A + 10B + C - (100C + 10B + A) = 396
99A - 99C = 396
A - C = 4
Step 6: Substitute the value of A - C into equation 5.
With A - C = 4, equation 5 becomes B - 2C + 4 = 0. Simplifying it, we get B = 2C - 4.
Step 7: Analyze the possible values for C.
Since B and C must be digits between 0 and 9, and B = 2C - 4, we can analyze the possible values for C:
When C = 2, B = 0, which is not a valid digit.
When C = 3, B = 2×3 - 4 = 2, which is a valid digit.
Step 8: Determine the possible values for A.
With C = 3 and B = 2, we can solve equation 5:
A - C = 4
A - 3 = 4
A = 7
Step 9: Formulate the final answer.
We have found one possible value for the 3-digit number: 723. We can reverse it: 327. The difference between the reversed number and the original number is indeed 396.
Therefore, there is only one 3-digit number that satisfies all the conditions: 723.
In conclusion, there is only one possible 3-digit number.
100u+10t+h = 100h+10t+u - 396
u-t = t-h
solve that and you get
t = h-2
u = h-4
h = t+2 = u+4
since no digit is zero, we can have
531
642
753
864
975