the hundreds digit is half the tens digit the sum of the hundreds and tens digit is 12 the ones digit is an odd square number not equal to the thousands digit
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if the digits are xhdu,
h = d/2
h+t = 12
u = 1 or 9
possibilities:
x481 where x = 2,3,4,5,6,7,8,9
x489 where x = 1,2,3,4,5,6,7,8
you have given no other information about the 1,000 digit
Let's break down the given information step by step:
Step 1: The hundreds digit is half the tens digit.
Let's assume the tens digit is "x". According to the information given, the hundreds digit will be half of "x", so it will be "x/2".
Step 2: The sum of the hundreds and tens digit is 12.
Now, we can form an equation using the information given: "x/2 + x = 12". To solve this equation, we can multiply both sides by 2: "x + 2x = 24". Combining like terms, we have "3x = 24". Dividing both sides by 3, we get "x = 8". Therefore, the tens digit is 8.
Step 3: The ones digit is an odd square number not equal to the thousands digit.
Odd square numbers are 1, 9, 25, 49, and so on. Since the ones digit is not equal to the thousands digit, we can exclude the digits 1 and 9.
Step 4: Determining the thousands digit.
Since the information provided does not give any specific detail about the thousands digit, we will leave it as an unknown, denoted by "y".
Hence, the possible solutions for the number could be "8y5", "8y25", "8y45", "8y65", "8y85", where "y" can be any digit except 1 and 9.
To find the value of each digit in the number described, we can use a systematic approach.
Let's assign variables to each of the digits:
- Thousands digit = x
- Hundreds digit = y
- Tens digit = z
- Ones digit = w
Now, let's break down the given conditions:
1. "The hundreds digit is half the tens digit": y = (1/2)z
2. "The sum of the hundreds and tens digit is 12": y + z = 12
3. "The ones digit is an odd square number not equal to the thousands digit": w is an odd square number ≠ x
Based on these conditions, we can try different possibilities for the digits:
Since y = (1/2)z, and y + z = 12:
1. If y = 1 and z = 2, then the sum is not 12, so this possibility can be eliminated.
2. If y = 2 and z = 4, then y + z = 6, so this possibility can be eliminated.
3. If y = 3 and z = 6, then y + z = 9, so this possibility can be eliminated.
4. If y = 4 and z = 8, then y + z = 12, which satisfies the condition. We can proceed with this possibility.
Now, let's find the value for w:
- Since w is an odd square number, the possible values are 1 and 9.
- However, w cannot be equal to x, so w ≠ x.
Now, let's determine the value of x:
- Since w ≠ x, and the possible values for w are 1 and 9, x can be any digit other than 1 and 9.
- Let's assign x = 7.
Now, we have the values for each digit:
- Thousands digit (x) = 7
- Hundreds digit (y) = 4
- Tens digit (z) = 8
- Ones digit (w) can be either 1 or 9
Therefore, the possible numbers satisfying all the given conditions are 7481 and 7489.