The numbers divisible by both 3 and 4 are multiples of 12, thus between 10 and 99:
12, 24, 36, 48, 60, 72, 84, 96
are the numbers divisible by both 3 and 4.
There are 9 numbers between 10 and 100 that are divisible by 11. They are: 11, 22, 33, 44, 55, 66, 77, 88, and 99
The numbers divisible by nine up to a 100 are 9,18,27,36,45,54,63,72,81,90,and,99
The numbers that are divisible by 33 are infinite. The first four are: 33, 66, 99, 132 . . .
48.The first two digit number is 10, the last two digit number is 99, so there are 99 - 10 + 1 = 90 two digit numbers→ 10 ÷ 3 = 31/3 → first two digit number divisible by 3 is 4 x 3 = 12→ 99 ÷ 3 = 33 → last two digit number divisible by 3 is 33 x 3 = 99→ 33 - 4 + 1 = 30 two digit numbers divisible by 3→ 10 ÷ 5 = 2 → first two digit number divisible by 5 is 2 x 5 = 10→ 99 ÷ 5 = 194/5 → last two digit number divisible by 5 is 19 x 5 = 95→ 19 - 2 + 1 = 18 two digit numbers divisible by 5→ 30 + 18 = 48 two digit numbers divisible by 3 or 5 OR BOTH.The numbers divisible by both are multiples of their lowest common multiple: lcm(3, 5) = 15, and have been counted twice, so need to be subtracted from the total→ 10 ÷ 15 = 010/15 → first two digit number divisible by 15 is 1 x 15 = 15→ 99 ÷ 15 = 69/15 → last two digit number divisible by 15 is 6 x 15 = 90→ 6 - 1 + 1 = 6 two digit numbers divisible by 15 (the lcm of 3 and 5)→ 48 - 6 = 42 two digit numbers divisible by 3 or 5.→ 90 - 42 = 48 two digit numbers divisible by neither 3 nor 5.
There are 18 numbers with 2 digits that are divisible by 5. First 2 digit number is 10 → 10 ÷ 5 = 2 → first 2 digit number divisible by 25 is 5 × 2 Last 2 digit number is 99 → 99 ÷ 5 = 19 4/5 → last 2 digit number divisible by 5 is 5 × 19 → There are 19 - 2 + 1 = 18 numbers with 2 digit divisible by 5.
From 10 to 99 (inclusive) there are 90 numbers.18 are evenly divisible by 5.Therefore 72 are not.
Just 1.
There are 17,999 such numbers.
There are 9 numbers evenly divisible by 10 : 10, 20, 30, 40, 50, 60, 70, 80, and 90
A number divisible by 99 is any integer that can be expressed as (99n), where (n) is an integer. For example, 198 (which is (99 \times 2)) and 297 (which is (99 \times 3)) are both divisible by 99. The key characteristic of these numbers is that they must meet the divisibility rule for both 9 and 11, since 99 is the product of these two primes.
There are a total of 33 numbers - starting with 12, and ending with 99.
No.
There are 9 numbers between 10 and 100 that are divisible by 11. They are: 11, 22, 33, 44, 55, 66, 77, 88, and 99
There are multiple numbers which meet this criteria. 71,73,74,75,77,78,79,81,82,83,85,86,87,89,91,93,94,95,97,98 & 99.
The numbers divisible by nine up to a 100 are 9,18,27,36,45,54,63,72,81,90,and,99
There are 11 divisible by 9 (which are 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99); There are 9 (or 10) divisible by 10 (which are 10, 20, 30, 40, 50, 60, 70, 80, 90; and 100 if it is 1 to 100 inclusive); There is 1 divisible by both 9 and 10 (which is 90).
99 is divisible by: 1, 3, 9, 11, 33, 99.