The two-digit numbers divisible by 2 range from 10 to 98. This forms an arithmetic series where the first term (a) is 10, the last term (l) is 98, and the common difference (d) is 2. The number of terms (n) in this series can be calculated as ( n = \frac{l - a}{d} + 1 = \frac{98 - 10}{2} + 1 = 45 ). The sum of the series is given by the formula ( S_n = \frac{n}{2} (a + l) = \frac{45}{2} (10 + 98) = 2430 ).
945
There are three such numbers: 12, 24 and 36.
All whole numbers are divisible by 1. Numbers are divisible by 2 if they end in 2, 4, 6, 8 or 0. Numbers are divisible by 3 if the sum of their digits is divisible by 3. Numbers are divisible by 4 if the last two digits of the number are divisible by 4. Numbers are divisible by 5 if the last digit of the number is either 5 or 0. Numbers are divisible by 6 if they are divisible by 2 and 3. Numbers are divisible by 9 if the sum of their digits is equal to 9 or a multiple of 9. Numbers are divisible by 10 if the last digit of the number is 0.
All numbers divisible by 5 (of which 15 is a multiple) have a final digit of 0 or 5. All numbers divisible by 3 (of which 15 is a multiple) have the sum of the digits totalling 3 or a multiple of 3. Therefore, a number is divisible by 15 if the sum of its digits total 3 or a multiple of 3 and its final digit is 0 or 5. Example : 32085 ; 3 + 2 + 0 + 8 + 5 = 18 which is divisible by 3. Final digit 5. This number is divisible by 15. (32085 ÷ 15 = 2139) 7420 : 7 + 4 + 2 + 0 = 13. This number is not divisible by 15.
531 is one of them and that any 3 digit number whose digital sum is 9 is also divisible by 9
945
Select all the numbers that $4221462$ is divisible by.
There are three such numbers: 12, 24 and 36.
All whole numbers are divisible by 1. Numbers are divisible by 2 if they end in 2, 4, 6, 8 or 0. Numbers are divisible by 3 if the sum of their digits is divisible by 3. Numbers are divisible by 4 if the last two digits of the number are divisible by 4. Numbers are divisible by 5 if the last digit of the number is either 5 or 0. Numbers are divisible by 6 if they are divisible by 2 and 3. Numbers are divisible by 9 if the sum of their digits is equal to 9 or a multiple of 9. Numbers are divisible by 10 if the last digit of the number is 0.
For example you have two numbers 4 and 5 and we add them the result is 9 which is also divisible by 3 if we divide it by 3 the result will be. So it means that total/addition of all the numbers which after addition can be divisible by 3.
All numbers divisible by 5 (of which 15 is a multiple) have a final digit of 0 or 5. All numbers divisible by 3 (of which 15 is a multiple) have the sum of the digits totalling 3 or a multiple of 3. Therefore, a number is divisible by 15 if the sum of its digits total 3 or a multiple of 3 and its final digit is 0 or 5. Example : 32085 ; 3 + 2 + 0 + 8 + 5 = 18 which is divisible by 3. Final digit 5. This number is divisible by 15. (32085 ÷ 15 = 2139) 7420 : 7 + 4 + 2 + 0 = 13. This number is not divisible by 15.
111 and 201
If the digit sum of any number from 1 to 10,000 add up to 9 then that number is divisible by 9 as for example 7641 is divisible by 9 because the final digit sum is 9 and so 7641/9 = 849
531 is one of them and that any 3 digit number whose digital sum is 9 is also divisible by 9
No. 26 for instance the sum of the digits is 8 but not divisible by 4. 32 the sum of the digits is 5 but divisible by 4 The rules for some other numbers are 2 all even numbers are divisible by 2 3 The sum of the digits is divisible by 3 4 The last 2 numbers are divisible by 4 5 The number ends in a 0 or 5 6 The sum of the digits is divisible by 3 and is even 7 no easy method 8 The last 3 numbers are divisible by 8 9 The sum of the digits is divisible by 9
The sum of all palindromic numbers from 1001 to 9999 is 495000.
I am a 3 digit number divisible by 7 but not 2 the sum of my digits is 4 what number am I