0
6! which equals 6x4x3x2x1=24x6=144
721 that's 6! +1
6
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Is 121 divisible by 3?
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)
What four digit number is divisible by 123456789 with a remainder of one?
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.
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 is special about the number 2520?
it's the smallest number divisible without remainder by the numbers 1 to 10.
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 121 divisible by 3?
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)
What whole number when divided by 2 has the remainder of 1 and divided by 4 has the remainder of 2?
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.
Is the number 33121 evenly divisible by the number 3?
No because it will have a remainder of 1
Is 15345 divisible by 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.
Is 234 divisible by 9?
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
Why do prime numbers have a remainder of 1 or 5 when divided by 6?
If the number has a remainder of 2 or 4 when divided by 6, it is an even number, so it is not prime number. 6n + 2 = 2 (3n + 1), so it is divisible by 2. 6n + 4 = 2 (3n + 2), so it is divisible by 2. If the number has a remainder of 3 when divided by 6, it is divisible by 3, so it is not a prime number. 6n + 3 = 3 (2n + 1), so it is divisible by 3. However, there are two prime numbers that do not have a remainder of 1 or 5 when divided by 6 : 2 and 3
What four digit number is divisible by 123456789 with a remainder of one?
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.
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).
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 is special about the number 2520?
it's the smallest number divisible without remainder by the numbers 1 to 10.
What number when divided by 5 gives remainder of 3 ang divided by 7 gives remainder of 1?
8
753248 is divisible by 11?
Of course it is. Any number can be divided by another number (the result in this case is not an integer (whole number)) 753248 / 11 = 68477,0909... or 68477 1/11 (There is a remainder of 1) -------------------------------------------------------------------------------------------------------------- Given the nature of the question the answer is no - assuming that the definition of divisible is that no remainder is left.
