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ReneNumbers that are divisible by both 3 and 4 are also divisible by their least common multiple, which is 12. To find all the numbers between 10 and 99 that are divisible by 12, we can divide the range by 12 and find the multiples within that range. The multiples of 12 between 10 and 99 are 12, 24, 36, 48, 60, 72, 84, and 96.
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How many numbers between 10 and 100 are divisible by 11?
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
What numbers 1-100 are divisible by 9?
The numbers divisible by nine up to a 100 are 9,18,27,36,45,54,63,72,81,90,and,99
What numbers are divisible by 33?
The numbers that are divisible by 33 are infinite. The first four are: 33, 66, 99, 132 . . .
How many two digit numbers divisible by neither 3 nor 5?
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.
How many 2 digit numbers are divisible by 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.
