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schepotkina [342]
3 years ago
9

How many integers are there that consist of four digits (from 1000 to 9999) where the first digit is a 5 or an 8 and the third d

igit is not 5 or 8
Mathematics
1 answer:
Ostrovityanka [42]3 years ago
5 0

Answer:

1600 integers

Step-by-step explanation:

Since we have a four digit number, there are four digit placements.

For the first digit, since there can either be a 5 or an 8, we have the arrangement as ²P₁ = 2 ways.

For the second digit, we have ten numbers to choose from, so we have ¹⁰P₁ = 10.

For the third digit, since it neither be a 5 or an 8, we have two less digit from the total of ten digits which is 10 - 2 = 8. So, the number of ways of arranging that is ⁸P₁ = 8.

For the last digit, we have ten numbers to choose from, so we have ¹⁰P₁ = 10.

So, the number of integers that can be formed are 2 × 10 × 8 × 10 = 20 × 80 = 1600 integers

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Step-by-step explanation: This problem states that the sum of three consecutive integers is -354 and it asks us to find the integers. Three consecutive integers can be represented as followed.

X ⇒ <em>first integer</em>

X + 1 ⇒ <em>second integer</em>

X + 2 ⇒<em> third integer</em>

<em />

Since the sum of our three consecutive integers is -354, we can set up an equation to represent this.

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Now, we can simplify on the left side of the equation.

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<em />

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