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2 years ago

How many 4-digit numbers are neither multiples of 2 nor multiples of 5?

Mathematics

2 answers:

slamgirl [31]2 years ago

4 0

Answer:

3,600

Explanation:

There's nine thousand numbers here, 4,500 of which are even, you will need to subtract mutiples of five from this number, making it even smaller.

subtract the non-even five's (exactly half of the 1800 hundred fives that fit in our little original number, (900!) ) 4,500-900= 3,600. Which should be your answer. :D

Step-by-step explanation:

anygoal [31]2 years ago

3 0

There are 10,000 total four-digit numbers (1000 through 9999).

Multiples of 2 end in 0, 2, 4, 6, and 8. There are 9*10*10*5 = 4500 four-digit multiples of 2.

Multiples of 5 end in 0 or 5. There are 9*10*10*2 = 1800 four-digit multiples of 5.

There is redundancy between the two sets of numbers, namely those that end in 0, which are both multiples of 2 and 5. There are 9*10*10*1 = 900 four-digit multiples of both 2 and 5.

Then there are 4500 + 1800 - 900 = 5400 total four-digit numbers that are either multiples of 2 or 5, which means the remaining 4600 numbers are neither multiples of 2 nor 5.

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ANTONII [103]

Hello there!

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Solve for n

Let's start by subtracting 5n from both sides

n - 5 - 5n ≤ 5n - 5n - 1

n - 5 - 5n ≤ -1

We need to transfer -5 on the other side, we can do that by adding 5 on both sides

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Do you know why I changed the sign?

A lot of students failed to remember that rules, which is:
When you are solving an inequality, if you divide both sides by a negative sign, you MUST change the symbol as well. If it was this <, it will change to this >. Got it? Cool!

I hope the steps are clear to understand. If you have questions, feel free to let me know...

As always, I am here to help!

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