That would be 660.
-10
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.
144 is divisible by 2, 3, 4, 6, 9 and not divisible by 5 or 10.Divisible by 2The whole number is divisible by 2 if the number is even which is shown by the last digit being divisible by 2. The last digit of 144 is 4 and 4 is divisible by 2, thus 144 is divisible by 2.Divisible by 3The number is divisible by 3 if the sum of its digits is also divisible by 3. Sum of the digits of 144 is 1+4+4 = 9 which is divisible by 3, thus 144 is divisible by 3Divisible by 4The number is divisible by 4 is the last two digits is also divisible by 4. Last two digits of 144 are 44 which are divisible by 4, thus 144 is divisible by 4An alternative test: If the last digit plus twice the preceding digit is divisible by 4 then the whole number is divisible by 4.For 144, last digit + twice preceding digit is 4+2x4 = 12 which is divisible by 4, so 144 is divisible by 4Divisible by 5If the last digit is 0 or 5 then the number is divisible by 5 Last digit of 144 is 4 which is neither 0 nor 5, thus 144 is not divisible by 5Divisible by 6To be divisible by 6, the number must be divisible by both of 2 and 3. 144 is divisible by both 2 and 3 (see above), thus 144 is divisible by 6Divisible by 9If the sum of the digits of the number is divisible by 9, then the original number is divisible by 9. For 144, 1+4+4 = 9 which is divisible by 9, thus 144 is divisible by 9Divisible by 10To be divisible by 10, the last digit must be 0. The last digit of 144 is 4 which is not 0, thus 144 is not divisible by 10
997
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.
900
030
990
-10
990.
There is no 4-digit number that is divisible by 2356 and 10.
Any whole number between 10 and 99, with a zero tacked on to the end of it, is.
If you multiply 2*9, you get 18. Multiply that by 10 to get a three digit number, and you get 180.
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.
There are many answers to it. 100,110,120,130,140,150,160,170,180,190,200,210,220,230,240,250,260,270, 280,290
Anything from 180 to 990 counting by 90.
There is no such number. The smallest number divisible by 2356 and 10 is 11780.