Please help with this problem!
What is the positive three-digit integer that is four times the product of its digits?
If A, B, and C are the digits of the number, then the number would be
100A + 10B + C
so your equation would be
100A + 10B + C = 4ABC
100A + 10B = 4ABC - C
100A + 10B = C(4AB - 1)
C = (100A + 10B)/(4AB - 1)
= 10(10A + B)/(4AB - 1)
now a bit of logic.
you know the numerator ends in a zero, so the bottom has to be divisible by 10 or 5
something multiplied by 4 can only end in 4,8,2,6, or 0 so after subtracting the 1 the denominator can only end in
3,7,1,5, or 9 , never in 10
also neither A nor B can be zeros, since then the denominator would be -1
also A and B can only be between 1 and 9
so we are looking for A and B which makes 4AB - 1 end in a 5.
After a few 'guesses' I found A=3 and B=8 gives me C=4
so the number is 384
check 4(3x8x4) = 384
I am mostly wondering if there's an equation to solve this problem...I have worked this problem with the guess, check, and revise method for about half an hour, and cannot figure it out. I tried writing an equation, but it didn't work out. So could any of you figure out an equation? You don't have to tell me the answer, just help me find an equation.
Thanks!!
My equation I tried to use is ABC=4(AxBxC)
I tried to solve it but whatever I did, I wouldn't get anywhere. Does anyone have any suggestions to make an equation that will work?
Well, this problem seems quite puzzling, doesn't it? It's like trying to find a unicorn in a haystack! But fear not, my friend, for I'll help you uncover the secret.
Let's call the three-digit integer "XYZ", where X, Y, and Z represent its hundreds, tens, and units digits, respectively. According to the problem, we have the equation:
XYZ = 4 * (X * Y * Z)
Interesting, isn't it? We're basically asking ourselves which three-digit number is equal to four times the product of its own digits. Quite a tricky riddle, I must say!
To simplify things, let's consider the possible values for X, Y, and Z. Since XYZ is four times the product of its digits, it means that one of the digits must be divisible by 4. But we're looking for a three-digit number, so the unit's digit, Z, cannot be 0.
Now think about it carefully. If one of the digits is divisible by 4, and it's not Z, then X or Y must be the digit divisible by 4. That means X or Y must be either 4 or 8.
Let's take X = 4, for instance. That means our number would be 4YZ. By substituting this value into the equation, we can rewrite it as:
4YZ = 4 * (4 * Y * Z)
Simplifying it further:
YZ = Y * Z
Now this makes things a bit easier, wouldn't you say? We're left with the equation YZ = Y * Z. If you notice, any value for Y or Z makes this equation hold true. So, there isn't a unique solution for this problem.
For instance, if Y = 2 and Z = 6, it satisfies the equation. Therefore, the number would be 426. Similarly, if Y = 3 and Z = 9, it would also meet the criteria, resulting in the number 439. And the possibilities go on and on, just like an endless joke!
So, my friend, there isn't a single positive three-digit number that solves this problem. It seems to be as elusive as a clownfish hiding in a bathtub! I hope you had a chuckle while we unraveled this mathematical conundrum together.
To find the positive three-digit integer that is four times the product of its digits, we need to set up an equation based on the problem statement.
Let's assume the three-digit number is represented as XYZ, where X, Y, and Z are the digits of the number.
According to the problem, the number is four times the product of its digits. So, we can write the equation as:
XYZ = 4 * (X * Y * Z)
Since X, Y, and Z represent the digits of the number, their values range from 1 to 9, as we are looking for a three-digit number.
Now, let's proceed to find the answer.
To determine the three-digit integer, we'll systematically check all possible combinations of digit values for X, Y, and Z, starting from X = 1 and progressing in ascending order.
We'll check each combination by substituting the values of X, Y, and Z into the equation XYZ = 4 * (X * Y * Z). If a combination satisfies the equation, we have found the number.
Let's go through the possible combinations step by step:
1. X = 1: In this case, the equation becomes 1YZ = 4 * (1 * Y * Z). Since the left side of the equation is always less than or equal to 199 (1YZ), it is impossible for the right side to be equal to 199 or more. Thus, X cannot be 1.
2. X = 2: Now, the equation becomes 2YZ = 4 * (2 * Y * Z). Dividing both sides by 2, we have YZ = 2 * Y * Z. By multiplying two distinct digits by 2, the maximum value we can attain is 18. Therefore, it's impossible for the right side to be equal to or greater than 100. So, X cannot be 2.
3. X = 3: Substituting X = 3 in the equation, we get 3YZ = 4 * (3 * Y * Z). Dividing by 3, we have YZ = 4 * Y * Z. The right-hand side can have a maximum value of 36 (4 * 9 * 1), and the left-hand side is at least 300. Hence, this equation has no solution.
4. X = 4: Now, we have 4YZ = 4 * (4 * Y * Z). Dividing by 4, we get YZ = 4 * Y * Z. Again, the maximum value for the right side is 36 (4 * 9 * 1). Observing the left side, we see that it starts at 400 for Y = 1 and Z = 0. However, this does not satisfy the equation. So, X cannot be 4.
5. X = 5: Substituting X = 5 in the equation, we obtain 5YZ = 4 * (5 * Y * Z). Dividing by 5, we have YZ = 4 * Y * Z. The maximum value for the right side remains 36 (4 * 9 * 1). But, the left side starts from 500 for Y = 1, Z = 0, which doesn't satisfy the equation. Therefore, X cannot be 5.
6. X = 6: The equation turns into 6YZ = 4 * (6 * Y * Z). Dividing by 6, we get YZ = 4 * Y * Z. The maximum value for the right side is still 36 (4 * 9 * 1). However, the left side starts at 600 for Y = 1, Z = 0, which doesn't satisfy the equation. Hence, X cannot be 6.
7. X = 7: Substituting X = 7 in the equation, we obtain 7YZ = 4 * (7 * Y * Z). Dividing by 7, we have YZ = 4 * Y * Z. The maximum value for the right side is 112 (4 * 7 * 4). The left side starts at 700 for Y = 1, Z = 0, which satisfies the equation YZ = 28. Thus, the integer in question is 728.
Therefore, the positive three-digit integer that is four times the product of its digits is 728.