There are 900 four-digit multiples of both 2 and 5.
State divisibility rule of 2?
"A number is divisible by 2 if the last digit of the number is 0, 2, 4, 6, or 8."
State divisibility rule of 5?
"A number is divisible by 5 if the last digit of the number is 0 or 5."
We have
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 undancy 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.
Hence, There are 900 four-digit multiples of both 2 and 5.
Learn more about divisibility rule of 2 and 5 here
https://brainly.com/question/8226236
#SPJ2
