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

2 Answers

6 votes
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.
User Peter Ivan
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4.4k points
9 votes

Answer:

3,600

Step-by-step 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:

User DavidWhitney
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4.1k points