If a number is divisible by 10 it cannot give any remainder other than 0. That is what "divisible by" means!
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.
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.
Well, honey, let me break it down for you. A number is a factor of another number if it divides into it evenly, no remainder drama. So, if you do the math, you'll see that 8 is not a factor of 4788 because 4788 divided by 8 gives you a decimal, and we don't do decimals in the world of factors.
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.
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.
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
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).
The largest 5-digit number divisible exactly by 99 is 99990.
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. 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).
The cannot be such a number. The biggest possible remainder is 2.
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
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).
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, 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.)
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.
A 5-digit number that is divisible by 9 must have its digits sum to a multiple of 9. The smallest 5-digit number that fits this criterion is 10,008 (1 + 0 + 0 + 0 + 8 = 9). Other examples include 10,017, 10,026, and so on, up to the largest 5-digit number that is divisible by 9, which is 99,999.