There will be more than one answer.
To have a remainder of 8, the divisor must be greater than 8. The only single digit greater than 8 is 9, so 9 must be the divisor. Since 8 is 1 less than 9, the dividend needs to be one less than a multiple of 9. Candidates are values of 3 digits, as the problem states. The lowest 3-digit number satisfying the condition is 107 (which is 9x12 - 1); the highest is 998 (which is 9x111 - 1). Some answers: 107 / 9 = 11 R8 125 / 9 = 13 R8 998 / 9 = 110 R8
An outcome is the result of a single trial.examples: one toss of a coin
The probability is 0%. The result will be heads or it will be tails but it cannot be heads and tails.
It is 100%. The coin will result in heads or tails since there are no other possible outcomes.
Your single number is your only information of the median value of the population, so the median value is the same as your single number. It is also the mode and mean of your sample.
The probability of getting an odd number in a single throw of a fair die (not dice!) is 1/2.The probability of getting an odd number in a single throw of a fair die (not dice!) is 1/2.The probability of getting an odd number in a single throw of a fair die (not dice!) is 1/2.The probability of getting an odd number in a single throw of a fair die (not dice!) is 1/2.
100
111
Probably many answers, but for example 100 / 8 = 96 remainder 4
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.
Regardless of the dividend (the number being divided), no divisor can produce a remainder equal to, or greater than, itself..... dividing by 4 cannot result in a remainder of 5, for example, Therefore the only single-digit number which can return a remainder of 8 is 9. 35 ÷ 9 = 3 and remainder 8
27.2222
506
0.0003
81 is.
2 x 6 + 0 = 12 2 x 1 + 2 = 4 4 is not [divisible by] 8, so 60 is not divisible by 8. (The remainder when 60 is divided by 8 is 4). To test divisibility by 8: Add together the hundreds digit multiplied by 4, the tens digit multiplied by 2 and the units (ones) digit. If this sum is divisible by 8 so is the original number. (Otherwise the remainder of this sum divided by 8 is the remainder when the original number is divided by 8.) If you repeat this sum on the sum until a single digit remains, then if that digit is 8, the original number is divisible by 8 otherwise it gives the remainder when the original number is divided by 8 (except if the single digit is 9, in which case the remainder is 9 - 8 = 1).
Not possible ! No single number will match allof your criteria ! Additionally - you've used the number 6 twice in your question - bothconditions can't be true !
6 + 4 + 6 = 16 1 + 6 = 7 → No; 646 is not divisible by 9 (there is a remainder of 7). ----------------------------------------- Only if the sum of the digits is divisible by 9 is the original number divisible by 9. Repeat the test on the sum until a single digit remains; only if this single digit is 9 is the original number divisible by 9, otherwise this single digit is the remainder when the original number is divided by 9.