Q: What is the largest remainder possible if a number is divided by 9?

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14. The largest possible number for a remainder is 1 less than the divisor.

The largest possible remainder when dividing by any number N is N-1.

The largest possible number for a remainder is 1 less than the number of the divisor, so it is 5.

5

The largest [integer] remainder is 10. If the remainder was any more you would get one (or more) lots of 11.

14, which is 15 minus 1.

2

The remainder is the number that is left over after the initial value has been divided as much as it can. If any numbers greater than 48 were present as a remainder, then these could be divided further into 48. If 48 is present as the remainder, then this can be divided by 48 to give 1, leaving no remainder. Thus, the largest possible remainder if the divisor is 48 is 47.

9

That's not possible. The largest single-digit number by which you might divide is 9. And, by definition, the remainder is always LESS than the number by which you divide. Thus, the largest remainder you can get, when you divide by 9, is 8.

If you mean as in a whole number then it is 7

1.5

6

799

9

It is not possible, because the number 4 is divisible by 2, and it's remainder is divisible by 2 also, so whatever number works for the "4 with a remainder of 2", will never work for "2 with a remainder of 1.

It is 997.

8 is the greatest possible whole number remainder, eg seventeen divided by nine...

85. A number when divided by 238, leaves a remainder 79. What will be the remainder when the number is divided bv 17 ? (1) 8 (2) 9 (3) 10 @) 11

A number leaves reminder 6 when divided by 10. What is the remainder when the number is divided by 5? Justify your reasoning.

There is no such number. You can have arbitrarily large numbers with this property - therefore there is no largest.

2519

58

59

29

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