It is 9996
a 3 digit number that is divisible by on is a three digit number that is a multiple of one.
There is no 4-digit number that is divisible by zero.
There is no limit to the number of digits.If, for example, a X is a k-digit number which is divisible by 4 then 10*X is divisible by 4 and 10*X will be a (k+1)-digit number.
to be divisible by 5 the last digit of the number has to be 0 or 5
yes- the number is 1260.
A number divisible by 123456789 must be 0 or bigger than 123456789. It must, therefore have 1 digit or 9 digits (or more). A remainder of 1 makes no difference to the number of digits. In any case, there can be no number of 4 digits that is divisible by 123456789.
-1
2520
-1
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.
yes 552 is divisible by 6 and the result is 92... to find out if a number is divisible by 6 then that number must be divisible by 2 and 3.. 1.If a number is divisble by three then the last digit must be divisible by 2 (so that the last digit must be 0,2,4,6,8) in this number the last digit is 2 so that it is divisible by 2 2.If a number is divisble by 3 then the sum of its digit must be divisible by 3. In this case the sum of the digits of 552 is 5+5+2= 12 and 12 is divisible by 3.... If this two rules fit in then that number is divisible by 6
Try 1008
The trick to this question is to realize that you can tell a number is divisible by nine by observing whether the sum of it's digits is divisible by nine.For example, we know that the number 25731 is divisible by nine, because 2 + 5 + 7 + 3 + 1 = 18, and 18 is divisible by nine.So in this case, what we want is a value for "A" in which 3 + A + A + 1, or 2A + 4, is divisible by 9. "A" also has to be less than ten (as we're dealing with a specific digit). That leaves us with only one possible digit, 7. 3 + 7, + 7 + 1 = 18, and 18 is indeed divisible by nine, so we know that 3771 is also divisible by nine.
no the smallest number that is divisible by four different primes would be 210
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.
This depends on what you mean by smallest. The smallest four-digit whole number is 1,000; the smallest four-digit real number is .0001 (1/10,000); the smallest four-digit integer is -9,999.