# Find the four digit numbers #abcd# that satisfy, #2(abcd)+1000=dcba#?

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To find the four-digit numbers abcd that satisfy the equation 2(abcd) + 1000 = dcba, we can break down the numbers into their decimal place values and solve for each digit.

Given the equation, we can represent the digits as follows:

- a as the thousands place digit,
- b as the hundreds place digit,
- c as the tens place digit, and
- d as the units place digit.

Substitute these values into the equation: 2(1000a + 100b + 10c + d) + 1000 = 1000d + 100c + 10b + a.

Expand and simplify the equation: 2000a + 200b + 20c + 2d + 1000 = 1000d + 100c + 10b + a.

Group like terms: 1999a + 190b + 80c - 998d = 1000.

Now, we need to find values for a, b, c, and d such that this equation holds true.

Since a, b, c, and d are digits, they must satisfy the following conditions:

- 0 ≤ a, b, c, d ≤ 9.

We can iterate through all possible values of a, b, c, and d to find the solutions that satisfy the equation 1999a + 190b + 80c - 998d = 1000.

The solutions for abcd that satisfy the given equation are:

- a = 4, b = 1, c = 7, d = 8, resulting in the number 4178.

So, the four-digit number abcd that satisfies the equation 2(abcd) + 1000 = dcba is 4178.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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