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If a number is divisible by 10 it cannot give any remainder other than 0. That is what "divisible by" means!

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Q: What is a 10 digit number when is divisible by 10 gives remainder 9 and when it is divisible by 9 it gives remainder 8 and so on what is that no.?
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Is 30030003 divisible by 8 or 9?

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.


How do you find the greatest common factor using repeated division?

An example should illustrate the process. Find the common factor of 38 and 14. Dividing 38 by 14 gives a remainder of 10. Repeat with the smaller two numbers. 14 divide by 10 gives a remainder of 4. 10 divided by 4 gives a remainder of 2. 4 divided by 2 gives a remainder of 0. The last non-zero number is the greatest common factor. In this case, 2.


Is 24 a factor of 96?

It is a factor (96/6=16). There is a trick that can help you remember if 6 is a factor. It has to be divisible by two (ie if it ends in a even number) and be divisible by 3. The trick for finding if it's divisible by three is to add up all of its digits. In this case, you would do 9+6, which gives you 15. If you do that again (1+5), that gives you 6, which is a known multiple of three.


What number is divisible by 3 6 and 9?

yes * * * * * Absolutely not! If it is divisible by 9 it must be divisible by 3. Here, if you want it, is a proof: x is divisible by 9 implies that x = 9*y for some integer y. Now 9 = 3*3, so writing 3*3 instead of 9 gives x = (3*3)*y so, by the associative property of multiplication, x = 3*(3*y) and then, by the closure of multiplication of integers, 3*y is also an integer. Say, z, for example. That is to say, x = 3*z which is equivalent to saying that x is divisible by 3.


How many leap years will there be from 2000 to 3000 inclusive?

Leap years are any years evenly divisible by four, with two exceptions. So years like 2004 and 2008 are leap years.The two exceptions:Any "century" year evenly divisible by 100 is NOT a leap year, so 1900 was not a leap year and 2100 will not be a leap year, except:Any year number divisible by 400 IS a leap year, so 2000 was and 2400 will be leap years.Of the 1000 years between 2001 and 3000, there are 250 "divisible by 4" years, minus 10 "century" years plus 2 "divisible by 400" years (2400 and 2800) gives 242 leap years between 2001 and 3000. Add 2000, and there are 243 leap years between 2000 and 3000, inclusive.

Related questions

Is 60 divisible by 8?

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).


Is 104 divisible 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


What largest number of the five digits is divisible by 99?

Largest no of 5 digits is 99999 99999/99 gives 1010 quotient 9 as remainder 99999-9=99990 is the largest 5 digit no divisible by 99.


Is 30030003 divisible by 8 or 9?

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.


Is 382169 is it divisible by 8 and explain why?

No. All multiples of 8 are even (end in 2, 4, 6, 8 or 0) but 382169 is odd (ends in 1, 3, 5, 7 or 9) so it cannot be a multiple of 8 and thus cannot be divisible by 8. To test for any [even] number being divisible by 8 add the units digit to twice the tens digit to four times the hundreds digit; if this sum is divisible by 8, then so is the original number. If the test is repeated until a single digit remains, only if this digit is 8 is the original number divisible by 8, otherwise if it is 1-7 it gives the remainder when the original number is divided by 8 (if it is 9, the remainder is 1). For 382169 this gives 9 + 2x6 + 4x1 = 9 + 12 + 4 = 25. 25 is not divisible by 8, so 382169 is not divisible by 8. Using the test on 25 gives 5 + 2x2 + 4x0 = 9, again not divisible by 8 (remainder is 1).


What number is divisible by 3 or 4 that gives you a remainder 20?

The cannot be such a number. The biggest possible remainder is 2.


What is a 5 digit number that is divisible by 9?

50,004There is a method called "casting out nines" that can be used to quickly test if any positive number is divisible by 9. To perform this test add the digits of the number, if the result has more than one digit repeat the addition of the digits, when the result is a single digit if that digit is 9 then the number is divisible by nine.Examples:99999, gives 45, gives 9, is divisible by 912345, gives 15, gives 6, is not divisible by 910008, gives 9, is divisible by 950004, gives 9, is divisible by 930303, gives 9, is divisible by 933633, gives 18, gives 9, is divisible by 9etc.


What is the smallest digit that can be placed in the blank so that 8-6935 can be divisible by 3?

2. To test a number to be divisible by 3, add up the digits and if the sum is divisible by 3, so is the original number. If not, the excess over a multiple of 3 gives the remainder when the original number is divided by 3. Subtract this remainder from 3 and this gives the smallest digit to replace the blank; unless the remainder is 0, in which case the blank can be replaced by a 0: 8 + 6 + 9 + 3 + 5 = 31 → 31 ÷ 3 = 10 remainder 1 Replace blank by 3 - 1 = 2: 826935 is divisible by 3


Is 796 divisible by 8 show the working?

Add 4 times the hundreds digit to twice the tens digit to the units digit. If this sum is divisible by 8, then the original number is divisible by 8. If this summing is repeated until a single digit remains, if this digit is 8, then the original number is divisible by 8 otherwise it gives the remainder of the original number divided by 8 (unless it is 9, in which case the remainder is 9 - 8 = 1). 7 x 4 + 9 x 2 + 6 = 28 + 18 + 6 = 52 5 x 2 + 2 = 12 1 x 2 + 2 = 4 4 is not 8, so 796 is not divisible by 8 (the remainder is 4).


Is 95 is divisible by 9?

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).


Can 345 be divisible by 9?

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.)


Is 702 divisible by 9?

702 ÷ 9 = 78 Therefore, the answer is , Yes. ---------------------------------------------------------- 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.) 702 → 7 + 0 + 2 = 9 9 is 9, so 702 is divisible by 9.