Q: What is the sum of all two digit numbers which are divisible by 5?

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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

The sum of all palindromic numbers from 1001 to 9999 is 495000.

Related questions

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.

111 and 201

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.

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

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

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

what is the least possible sum of two 4-digit numbers?what is the least possible sum of two 4-digit numbers?