Study guides

Q: What three digit number divided with a single digit will result in a remainder of 4?

Write your answer...

Submit

Related questions

111

100

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

81 is.

0.0003

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).

It is divisible by 8 because 202008/8 = 25251 But divided by 9 it will leave a remainder ----------------------- To test if a number is divisible by 8 add the ones digit to twice the tens digit to 4 times the hundreds digit; if this sum is divisible by 8 then so is the original number. By repeating the test on the sum until a single digit remains, only if this single digit is 8 is the original number divisible by 8, otherwise it gives the remainder when divided by 8 (except if it is 9, in which case the remainder is 1 - the excess of 9 over 8). For 202008: ones_digit + 2 × tens_digit + 4 × hundreds_digit = 8 + 2 × 0 + 4 × 0 = 8; so 208008 is divisible by 8. To test if a number is divisible by 9, add up the digits of the number; if this sum is divisible by 9, then so is the original number. By repeating the test of the sum until a single digit remains, only if this single digit is 9 is the original number divisible by 9, otherwise it gives the remainder when the original number is divided by 9. (This single digit is also called the "digital root" of the number.) For 202008: 2 + 0 + 2 + 0 + 0 + 8 = 12; 1 + 2 = 3; so 202008 is not divisible by 9; it has a remainder of 3 when divided by 9.

No. To check if a number is divisible by 9 add the digits together and if the sum is divisible by 9 then so is the original number. The check can be used on the sum so keep summing until a single digit remains. If this digit is 9, then the number is divisible by 9, otherwise it gives the remainder when the number is divided by 9. (This single digit is known as the digital root of the number.) 104 → 1 + 0 + 4 = 5 5 is not 9, so 104 is not divisible by 9; it has a remainder of 5 when divided by 9

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 !

You won't get a remainder of 1 if you divide by 1. If you want the same for all the numbers from 2-12, you must get a common denominator of all the numbers from 2-12. Any common denominator might do, so you could just multiply all those numbers together; or (to get all solutions) get the least common denominator. Next, find a multiple of this common denominator that has 13 digits, and add one to the result. That's it!

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.

16

A digit is a number or numeral. A dividend is a number to be divided. A dividend is divided by a divisor to yield a quotient. A digit dividend is a single digit number that is being divided some other (non-specified) number. The word single is assumed to be there. It could be written like this: A (single) digit dividend is a single digit number that is being divided by another number. In mathematics, there are 1-digit dividends, 2-digit dividends, etc.

Firstly, the LCM f a single number is the number itself.The LCM of many numbers is found by dividing the numbers with the smallest prime numbers until the numbers are completely divided and the remainder is zero.Then all the prime numbers used for dividing is multiplied and the LCM is found.

No. To be divisible by 9 the sum of the digits must also be divisible by 9. As the sum must be divisible by 9, the check can also be applied to the sum; so keep summing until a single digit remains, then if, and only if, this single digit is 9 is the original number divisible by 9 (the single digit gives the remainder when the original number is divided by 9 and is known as the digital root of the number). For 95: 95 → 9 + 5 = 14 14 → 1 + 4 = 5 5 ≠ 9 → 95 is not divisible by 9. (The remainder when 95 is divided by 9 is 5).

No. To test: add the digits together and if the sum is divisible by 9, so is the original number. The test can be repeated on the sum until a single digit remains. If this single digit is 9 then the [original] number is divisible by 9, otherwise it gives the remainder when the [original] number is divided by 9. 534 → 5 + 3 + 4 = 12 12 → 1 + 2 = 3 3 ≠ 0 so 12 is not divisible by 9 so 534 is not divisible by 9; the remainder when 534 is divided by 9 is 3.

16.6

No. 9 is less than 1239 so it cannot be divisible by it. Nor is 1239 divisible by 9. To test if a number is divisible by 9 add the digits and if this sum is divisible by 9 then so is the original number. As the test can be applied to the sum, repeating the summing until a single digits remains means the original number is divisible by 9 only if this single digit (also called the digital root of the number) is 9. (if it is not 9, it gives the remainder when the original number is divided by 9). For 1239: 1 + 2 + 3 + 9 = 15 → 1 + 5 = 6 which is not 9, so 1239 is not divisible by 9 (it has a remainder of 6 when divided by 9).

A single number cannot have a variance.A single number cannot have a variance.A single number cannot have a variance.A single number cannot have a variance.

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

No, it is not divisible by 9. To check if a number is divisible by 9 add all the digits together and if this sum is divisible by 9, then so is the original number. As the check can be applied to the sum, keep adding the digits together until a single digit remains. If this digit is 9, then the original number is divisible by 9 (otherwise it gives the remainder when the original number is divided by 9). for 345: 3 + 4 +5 = 12 for 12: 1 + 2 = 3 which is not 9, so 345 is not divisible by 9. (The remainder when 345 is divided by 9 is 3.)

a single digit divisor is as single digit number ( 1 - 9) that can be evenly divided into a number. like the single digit numbers for 5670 is 1,2,3,5,6,7,9 . you know this cause all of though numbers can be divided evenly into 5670.. don't believe me ... ask your calculator? ps: i <3 jb