6! which equals 6x4x3x2x1=24x6=144
No. Add the digits of the dividend and if that is divisible by 3 then the original number is divisible by 3; if not, its remainder when divided by 3 gives the remainder when the original number is divided by 3: 1 + 2 + 1 = 4 which gives a remainder of 1 when divided by 3, so 121 divided by 3 gives a remainder of 1. (121 = 40 x 3 + 1)
A number divisible by 123456789 must be 0 or bigger than 123456789. It must, therefore have 1 digit or 9 digits (or more). A remainder of 1 makes no difference to the number of digits. In any case, there can be no number of 4 digits that is divisible by 123456789.
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
it's the smallest number divisible without remainder by the numbers 1 to 10.
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).
No. Add the digits of the dividend and if that is divisible by 3 then the original number is divisible by 3; if not, its remainder when divided by 3 gives the remainder when the original number is divided by 3: 1 + 2 + 1 = 4 which gives a remainder of 1 when divided by 3, so 121 divided by 3 gives a remainder of 1. (121 = 40 x 3 + 1)
The number would be (7x23456) + 1 = 164193
No 13 is not divisible by 3 because there would be a remainder of 1 and a number that is divisible by another should have no remainder
To determine if 65483 is divisible by a certain number, you would typically divide 65483 by that number and check if the remainder is zero. 65483 divided by 2 has a remainder, so it is not divisible by 2. 65483 divided by 3 also has a remainder, so it is not divisible by 3. However, 65483 divided by 7 gives a remainder of 0, so it is divisible by 7.
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.
No because it will have a remainder of 1
85 is not exactly divisible by 4. Dividing 85 by 4 gives 21 with a remainder of 1.
When you divide a number by 1, it remains the same. A number is considered divisible is the resulting answer is an integer with no remainder left over. Thus, if the starting number is an integer, it will be divisible by 1. 15345 is an integer, and thus is divisible by 1.
1, any number that you divide by 2 is either divisible by 2 (even) or not divisible by 2, leaving a remainder of 1 (odd).
To determine if 2421 is divisible by a certain number, we need to check if the remainder when 2421 is divided by that number is zero. For example, 2421 ÷ 3 = 807 with no remainder, so 2421 is divisible by 3. However, 2421 is not divisible by 2, as 2421 ÷ 2 = 1210 with a remainder of 1. Similarly, 2421 is not divisible by 5, as 2421 ÷ 5 = 484 with a remainder of 1.
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
Yes, 234 is divisible by 9 because on dividing 234 by 9 we get 26 as quotient and 0 as remainder. A number is divisible by another number if: 1- Quotient is a whole number 2- Remainder is zero