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Q: I am a 3- digit odd number greater than 800. My Ten Digit is 2 less than my ones digit. I am divisible by 3 but not 5. The sum of digits is 12. What number am I?

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

-8

There is no number, no matter the number of digits, that is only divisible by 2.

4 and all of its multiples are divisible by 4. They end in even digits. No number that ends in an odd digit is divisible by 4.

When its digital root is 9 (divisible by 9) and it is even (divisible by 2). The digital root of a number is calculated by adding up all the digits in a number. If the answer is greater than 9, then repeat until you get an answer that is a single digit.

Any 5-digit number is greater than a 4-digit one.

A number divisible by 123456789 must be 0 or bigger than 123456789. It must, therefore have 1 digit or 9 digits (or more). A remainder of 1 makes no difference to the number of digits. In any case, there can be no number of 4 digits that is divisible by 123456789.

The digital root (sum of digit) must be divisible by 9, and the number formed by the last 4 digits must be divisible by 16. The second requirement ensures that the number is divisible by 16.

A number is divisible by 9 if the sum of its digits is divisible by 9. Thus, the number 99999 is divisible by 9 (the sum of its digits is 45, and it is clearly 11111 * 9), and because 99999 is the largest 5 digit number, it must be the largest 5 digit multiple of 9. 99999 is the greatest no divisible by 9.if u divide it u`ll get 11111.

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.

The 2-digit multiples of 11 all have identical digits.

987652431

If the number formed by the last three digits is divisible by 8. This requires that: if the digit in the hundreds place is even, the last two digits must form a number divisible by 8 and if the digit in the hundreds place is odd, the last two digits must form a number divisible by 4 but not by 8.

2,4,6,8,and 0 are divisible by 2 in the ones digit. Zero is only divisible in a number with 2 digits or more. 0 itself is not divisible by 2.

A delectable number has nine digits, using the numbers 1-9 once in each digit. The first digit of a delectable number must be divisible by one. The first and second digits must be divisible by two, the first through third must be divisible by three, etc. There has only been one delectable number discovered: 381654729.

the answer is 95

12 or 24

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.

301

Yes. If a number is divisible by 9 it is also divisible by 3 so "is divisible by 3 and 9" can just say "is divisible by 9". The only numbers that this can be applied to is 9, 90, 900, 9000 and so on. The difference of digits can only be 9 if one of them is 9 and all of the rest of the digits are 0, since there is no digit greater than 9 (in base 10) and 9 minus anything greater than 0 is less than 9.

995 is divisible by 5 evenly and is the largest 3 digit number to satisfy the condition

The largest 5-digit number divisible exactly by 99 is 99990.

If the last two digits are divisible by 4 then the number is divisible by 4. Thus, if the tens digit is even and the units digit is 0 or 4 or 8 OR if the tens digit is odd and the units digit is 2 or 6 then the number is divisible by 4.

56

Not necessarily. Consider 444. The digits are not different. The first and second digits are not multiples of 3 The first digit is not greater than the second digit. In spite of all that, 444 is a 3-digit number