Trick question all numbers are divisible by four ,but only 175 result in whole numbered answers
7777
All 4 digit numbers that are divisible by 9 are also divisible by 3. The first is 1008 and the last is 9999.
2999 of them.
To get the three number that is divisible by 4 you will take the difference 4^3 and 4^2.
The smallest three-digit number divisible by the first three prime numbers (2, 3, and 5) and the first three composite numbers (4, 6, and 8) is 120.
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.
Trick question all numbers are divisible by four ,but only 175 result in whole numbered answers
what is the largest three-digit number that is divisible by 4, 5, and 6
The first 3 digit natural number is 100: 100 ÷ 4 = 25 → first 3 digit natural number divisible by 4 is 4 × 25 The last 3 digit natural number is 999: 999 ÷ 4 = 249 r 3 → last 3 digit natural number divisible by 4 is 4 × 249 → number of 3 digit natural numbers divisible by 4 is 249 - 25 + 1 = 225.
7777
There are 300 three digit numbers that are divisible by neither 2 nor 3. There are 999 - 100 + 1 = 900 three digit numbers. 100 ÷ 2 = 50 → first three digit number divisible by 2 is 50 × 2 =100 999 ÷ 2 = 499 r 1 → last three digit number divisible by 2 is 499 × 2 = 998 → there are 499 - 50 + 1 = 450 three digit numbers divisible by 2. 100 ÷ 3 = 33 r 1 → first three digit number divisible by 3 is 34 × 3 = 102 999 ÷ 3 = 333 → last three digit number divisible by 3 is 333 × 3 = 999 → there are 333 - 34 + 1 = 300 three digit numbers divisible by 3. The lowest common multiple of 2 and 3 is 6, so these have been counted in both those divisible by 2 and those divisible by 3. 100 ÷ 6 = 16 r 4 → first three digit number divisible by 6 is 17 × 6 = 102 999 ÷ 6 = 166 r 3 → last three digit number divisible by 6 is 166 × 6 = 996 → there are 166 - 17 + 1 = 150 → there are 450 + 300 - 150 = 600 three digit numbers that are divisible by either 2 or 3 (or both). → there are 900 - 600 = 300 three digit numbers that are divisible by neither 2 nor 3.
18 numbers are there
All 4 digit numbers that are divisible by 9 are also divisible by 3. The first is 1008 and the last is 9999.
40,000, 80,000
2999 of them.
A number which which ends with a 2 digit number that is divisible by 4. For example, 1028 ends with 28 which is divisible by 4.