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
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.
The numbers that are divisible by 33 are infinite. The first four are: 33, 66, 99, 132 . . .