-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.
To determine if a number is divisible by another number, you need to check if the first number can be divided by the second number without leaving a remainder. For 501105: Divisibility by 2: A number is divisible by 2 if its last digit is even. The last digit of 501105 is 5, which is odd, so 501105 is not divisible by 2. Divisibility by 3: A number is divisible by 3 if the sum of its digits is divisible by 3. The sum of the digits of 501105 is 5+0+1+1+0+5=12, which is divisible by 3, so 501105 is divisible by 3. Divisibility by 5: A number is divisible by 5 if its last digit is 0 or 5. The last digit of 501105 is 5, so 501105 is divisible by 5. Divisibility by 9: A number is divisible by 9 if the sum of its digits is divisible by 9. The sum of the digits of 501105 is 12, which is not divisible by 9, so 501105 is not divisible by 9. Divisibility by 10: A number is divisible by 10 if its last digit is 0. The last digit of 501105 is 5, so 501105 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.