No.
(Assuming a three digit number is in the range 100-999 and excludes leading zeros, that is 080 does not count as it is really 80 which is a two digit number)
To be divisible by 11, the difference in the sums of the alternate digits of the number must be divisible by 11 (or 0).
For a three digit number, this means that the sum of the first and last digits less the second digit must be a multiple of 11 (or 0).
For a three digit number with all the digits the same, this calculation results in the value of one of the digits (eg 333 → 3 + 3 - 3 = 3) which will not be 0, and cannot be a multiple of 11 as a single digit is less than or equal to 9 which is less than 11 and thus not a multiple of 11.
102.
120
The first 3 digit number divisible by 19 is 114 (= 19 x 6) The last 3 digit number divisible by 19 is 988 (= 19 x 52) That means that there are 52 - 6 + 1 = 47 three digit numbers that are divisible by 19.
Every three-digit number that ends with a zero or a 5 is divisible by 5.It doesn't matter what the first 2 digits are.
The greatest three digit number that is divisible by 7 is 994.
A three digit number cannot be divisible by a 5 digit number - in any base.
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.
102.
120
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.
The first 3 digit number divisible by 19 is 114 (= 19 x 6) The last 3 digit number divisible by 19 is 988 (= 19 x 52) That means that there are 52 - 6 + 1 = 47 three digit numbers that are divisible by 19.
Converse:If a number is divisible by 3, then every number of a digit is divisible by three. Inverse: If every digit of a number is not divisible by 3 then the number is not divisible by 3? Contrapositive:If a number is not divisible by 3, then every number of a digit is not divisible by three.
Every three-digit number that ends with a zero or a 5 is divisible by 5.It doesn't matter what the first 2 digits are.
The greatest three digit number that is divisible by 7 is 994.
a 3 digit number that is divisible by on is a three digit number that is a multiple of one.
100. The highest 3 digit number divisible by 9 is 999 (111 times). The highest 2 digit number divisible by 9 is 99 (11 times). The difference is 100.
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.