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Q: What is a 4 digit number that is divisible by 11 an17?

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The largest 5 digit number is 99999. If this is divided by 11 it leaves a remainder of 9. Therefore the largest 5 digit number divisible by 11 is 99999 - 9 = 99990.

-8

You're looking for a 4-digit number that is divisible by 11 and 17. Multiply them together. Put a zero on the end. 1870

1001 / 11 = 91 1001 is the smallest 4-digit number that 11 divides equally into.

10010

1496

4 is not divisible by any 3-digit number. Nor are 5, 11 or 3. The smallest positive numbers that is divisible by 4, 5, 11 and 3 is 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.

987652431

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

10296 is one.

132

660

56

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.

-11

56

It is: 924

Yes - from 11/11 = 1 to 99/11 = 9, this is the case.

No. It is divisible by 11.No. It is divisible by 11.No. It is divisible by 11.No. It is divisible by 11.

There are many possible solutions. One such is 132486970

55.

That would be 660.

The lowest common multiple of 5 and 9 is 45 and 11 times 45 is 495 which is the greatest 3 digit number divisible by 5 and 9

Let A, B, C are digits. Number ABC is divisible by eleven if and only if A+C-B=0 or A+C-B=11