-- List the digits that are used to write the number.
-- Add up the list.
1002 = 10000 which is the smallest possible 5 digit number. So we have to go 1 smaller which is 992 = 9801
No. To be divisible by 6 the number must be divisible by both 2 and 3. To be divisible by 2, the last digit of the number must be even (ie one of {0, 2, 4, 6, 8}). The last digit of 35 is 5 which is not even and so 35 is not divisible by 2. To be divisible by 3, sum the digits of the number and if this sum is divisible by 3 then so is the original number. As the test can be applied to the sum, repeatedly summing the digits of the sums until a single digit remains, then the original number is divisible by 3 only if this single digit is one of {3, 6, 9}. For 35 3 + 5 = 8 which is not divisible by 3 (nor is is one of {3, 6, 9}, thus 35 is not divisible by 3. As 35 is not divisible by both 2 and 3 (in fact it is divisible by neither 2 nor 3) it is not divisible by 6.
100
no, numbers go up to infinity and down to negative infinity.
If the last digit is a 5-9 you round up. Ex: 585 go up to 590. If the last digit is 1-4 you round down ex: 584 go down to 580
Well I believe that when you divide the number 1st you have to see if the number will go in to the three digit number and if don't then see if another can go into that one
The formula for finding the sum of interiors angles is... S(sum) = 180(n(number of sides) -2) S = 180(n-2) so... S = 180(35 - 2) S = 180(33) S = 5940 And there you go!
1002 = 10000 which is the smallest possible 5 digit number. So we have to go 1 smaller which is 992 = 9801
No, it does not. It is finding a number that they all go into evenly.
No. To be divisible by 6 the number must be divisible by both 2 and 3. To be divisible by 2, the last digit of the number must be even (ie one of {0, 2, 4, 6, 8}). The last digit of 35 is 5 which is not even and so 35 is not divisible by 2. To be divisible by 3, sum the digits of the number and if this sum is divisible by 3 then so is the original number. As the test can be applied to the sum, repeatedly summing the digits of the sums until a single digit remains, then the original number is divisible by 3 only if this single digit is one of {3, 6, 9}. For 35 3 + 5 = 8 which is not divisible by 3 (nor is is one of {3, 6, 9}, thus 35 is not divisible by 3. As 35 is not divisible by both 2 and 3 (in fact it is divisible by neither 2 nor 3) it is not divisible by 6.
100
Yes. It goes in 188 times. --------------------------------------- To be divisible by 3, sum the digits of the number and if this sum is divisible by 3, then the original number is divisible by 3. As the test can be repeated on the sum, repeat the summing until a single digit remains; only if this number is one of {3, 6, 9} is the original number divisible by 3. For 564 this gives: 564 → 5 + 6 + 4 = 15 15 → 1 + 5 = 6 6 is one of {3, 6,9} so 564 is divisible by 3.
no, numbers go up to infinity and down to negative infinity.
If the last digit is a 5-9 you round up. Ex: 585 go up to 590. If the last digit is 1-4 you round down ex: 584 go down to 580
For this kind of problems, go from left to right, and use the smallest possible digit for each position.
I would go with -999'999'999.However, if you are looking for positive number, it would be 100'000'000.
The smallest number would be 10,000,000 because the one is the lowest number that could possibly go in the ten millions place. After that, you can put all zeros.