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.
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.
Any numerical value ending in zero (0) is divisioble by 10. 10,20,30,40,50,60,70,80,90,100, 1000, 1,000,000, 1,000,000,000 are all divisible by '10'. NB THere millions more numbers divisible by '10' , provided they and with a zero(0).
5715 is not divisible by 2: the last digit is not an even number or 05715 is divisible by 3: as the last two digits when added together is a multiple of 35715 is divisible by 5: as the last digit is a 5 or 05715 is divisible by 9: as if u added all the digits of the number it will be divisible by 9, in this case 5 +7+1+5=18 and 18 is a multiple of 95715 is not divisible by 10: as the last digit is not a 0
The greatest 3-digit number that is not divisible by 2, 3, 5, or 10 is 997.
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.
990
There is no 4-digit number that is divisible by 2356 and 10.
There is no such number. The smallest number divisible by 2356 and 10 is 11780.
A whole number is evenly divisible by 10 only if its last digit is a zero.
20.
No. A whole number is (evenly) divisible by 10 only if its last digit is ' 0 '.
No. In order for a number to be divisible by 10, the ones digit must be zero (0).
Yes +++ To amplify a little, if the final digit of any multi-digit number is 0, then the number is divisible by 10. (Just remove the 0.)
An even number, by definition is divisible by 2. For the number to be divisible by 10, the last digit must be 0, which ensures it is an even number. Any number, with zero as the last digit will satisfy the requirements. So 111110 will do.
180