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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.

It is 19.

111 and 201

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.

4,905

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.

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

Almost all of the two digit composite numbers have two one digit factors that have a sum.

I am a 3 digit number divisible by 7 but not 2 the sum of my digits is 4 what number am I

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

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

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