Q: What 3 digit number is 4 5 11 and 3 divisible by?

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Converse:If a number is divisible by 3, then every number of a digit is divisible by three. Inverse: If every digit of a number is not divisible by 3 then the number is not divisible by 3? Contrapositive:If a number is not divisible by 3, then every number of a digit is not divisible by three.

132

a 3 digit number that is divisible by on is a three digit number that is a multiple of one.

the number 3 is divisible by 3

The only 3-digit number that is divisible by 624 is 624.

no

660

100. The highest 3 digit number divisible by 9 is 999 (111 times). The highest 2 digit number divisible by 9 is 99 (11 times). The difference is 100.

1002 is the smallest 4 digit number divisible by 3 and 2.

-1

Yes.

There are 180 3-digit numbers divisible by five.

36360 is a 5 digit number that is divisible by both 3 and 9.

It is: 924

No. (Assuming a three digit number is in the range 100-999 and excludes leading zeros, that is 080 does not count as it is really 80 which is a two digit number) To be divisible by 11, the difference in the sums of the alternate digits of the number must be divisible by 11 (or 0). For a three digit number, this means that the sum of the first and last digits less the second digit must be a multiple of 11 (or 0). For a three digit number with all the digits the same, this calculation results in the value of one of the digits (eg 333 → 3 + 3 - 3 = 3) which will not be 0, and cannot be a multiple of 11 as a single digit is less than or equal to 9 which is less than 11 and thus not a multiple of 11.

The number 100,002 is divisible by 3. The largest 5 digit number is 99996 (6 less).

108 is the smallest 3-digit number divisible by 2,3,4 & 6

261, 264, or 267

111

To be divisible by 2 the last digit must be even, ie one of {0, 2, 4, 6, 8}; The last digit is 0, which is one of these so 330 is divisible by 2. To be divisible by 3, sum the digits of the number and if this sum is divisible by 3, then the original number is divisible by 3. As the test can be repeated on the sum, repeat the summing until a single digit remains; only if this number is one of {3, 6, 9} is the original number divisible by 3. 330→ 3 + 3 + 0 = 6 6 is one of {3, 6,9} so 330 is divisible by 3. To be divisible by 5, the last digit must be one of {0, 5}. The last digit is 0 which is one of {0, 5} so 330 is divisible by 5. There is no real check for 7 which is not much slower than just dividing by 7 to see if there is no remainder. One check: Write the digits in blocks of 3 starting from the right hand end (like you would for reading the number): in each block of 3 add twice the first digit to three times the second digit to the third digit. Alternately subtract and add the blocks starting from the right hand end of the number. If the result is divisible by 7, then so is the original number. 330 → 2×3 + 3×3 + 0 = 15 15 is not divisible by 7, so 330 is not divisible by 7. To be divisible by 11, alternately subtract and add the digits of the number from the right hand end; only if this sum is divisible by 11 (or is 0) is the original number divisible by 11. 330 → 0 - 3 + 3 = 0 0 is 0, so 330 is divisible by 11. Therefore 330 is divisible by 2, 3, 5, 11 But not divisible by 7.

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

That would be 660.

No. Consider 13 or 23.

There is no 4-digit number that is divisible by zero.

1000 is the smallest 4 digit number but of the two factors, it is only divisible by 2. As 999 is obviously divisible by 3 then 999 + 3 = 1002 is also divisible by 3 and, as an even number, is also divisible by 2. The answer is 1002