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Q: What is the largest 5 digit number divisible by 11?

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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.

1870 is divisible by both 17 and 11 1870 / 17 = 110 1870 / 11 = 170

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

99990 Note that if you add 11 to my answer you get a 6-digit number therefore it must be the largest 5-digit whole multiple of 11.

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

10296 is one.

132

660

56

That would be 11.

The smallest is 11 and the largest is 97.

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.

56

It is: 924

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

97 - 11 = 86

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

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