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Oh, isn't that a happy little question! To find the two-digit numbers divisible by 3, we start by finding the first two-digit number divisible by 3, which is 12. Then, we find the last two-digit number divisible by 3, which is 99. Now, we can count how many numbers there are between 12 and 99 that are divisible by 3.
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How many two digit numbers are divisible by either 3 or 5?
How many two digit number are divisible by 5
How many 2 digit numbers are divisible by 3?
Starting at 12 and ending at 99, there are 30 two-digit numbers divisible by three.
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 or 3-digit numbers can be formed using the digits 134568 and 9 which are diisible by 4?
48. Assuming no digit can be used more than once, the two digit numbers divisible by 4 are: 16, 36, 48, 56, 64, 68, 84, 96 8 of them. For any number to be divisible by 4, only the last two digits need be divisible by 4; so for three digit numbers, each of the two digit numbers above can be preceded by any of the remaining 5 digits and still be divisible by 4. → 5 x 8 = 40 three digit numbers are divisible by 4 → 40 + 8 = 48 two or three digit numbers made up of the digits {1, 3, 4, 5, 6, 8, 9} are divisible by 4. If repeats are allowed, there are an extra 2 two digit numbers (44 and 88) and each of the two digit numbers can be preceded by any of the 7 digits, making a total of 7 x 10 + 10 = 80 two and three digits numbers divisible by 4 make up of digits from the given set.
What two digit numbers greater than 74 are divisible by 2 and 3?
78,84,90,96
How many two digit numbers are divisible by either 3 or 5?
How many two digit number are divisible by 5
How many 2 digit numbers are divisible by 3?
Starting at 12 and ending at 99, there are 30 two-digit numbers divisible by three.
How many two digit numbers are divisible by 7?
13
What is the probability that a four digit number is divisible by 2?
Even numbers are divisible by two, and half of all numbers are even, so there is a 50 percent chance that a four-digit number is divisible by two.
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 and 6?
If n is divisible by both 5 and 6, then it should be divisible by 30 (5 * 6). Considering you are asking for only two-digit numbers, the answer(s) would be 30, 60, and 90. So, three numbers.
How many 2-digit or 3-digit numbers can be formed using the digits 134568 and 9 which are diisible by 4?
48. Assuming no digit can be used more than once, the two digit numbers divisible by 4 are: 16, 36, 48, 56, 64, 68, 84, 96 8 of them. For any number to be divisible by 4, only the last two digits need be divisible by 4; so for three digit numbers, each of the two digit numbers above can be preceded by any of the remaining 5 digits and still be divisible by 4. → 5 x 8 = 40 three digit numbers are divisible by 4 → 40 + 8 = 48 two or three digit numbers made up of the digits {1, 3, 4, 5, 6, 8, 9} are divisible by 4. If repeats are allowed, there are an extra 2 two digit numbers (44 and 88) and each of the two digit numbers can be preceded by any of the 7 digits, making a total of 7 x 10 + 10 = 80 two and three digits numbers divisible by 4 make up of digits from the given set.
How many two-digit numbers are divisible by both 9 and 2?
There are five numbers: 18, 36, 54, 72, 90.
What is the sum of all two digit numbers which are divisible by 5?
945
How many four digit numbers which are divisible by 7?
Best way to work this out: find the highest number below 10,000 that is divisible by 7 (9,996) and divide that by 7 (1,428). 1,428 is the amount of one-, two-, three- and four-digit numbers divisible by 7. Now find the number of one-, two- and three-digit numbers divisible by 7 (which is 994/7 = 142) and subtract this number from 1,428, giving us 1,286.
What two digit numbers greater than 74 are divisible by 2 and 3?
78,84,90,96
What two digit numbers are greater than 80 and divisible by 2 and 3?
Three of them.
