0
What is the missing digit to make each number divisible by 9 .7 and 1?
Wiki User
∙ 9y agoWant this question answered?
Be notified when an answer is posted
Add your answer:
Earn +20 pts
What is the one 9 digit delectable number which can be made using the digits 1 to 9?
A delectable number has nine digits, using the numbers 1-9 once in each digit. The first digit of a delectable number must be divisible by one. The first and second digits must be divisible by two, the first through third must be divisible by three, etc. There has only been one delectable number discovered: 381654729.
Is 1110 divisble by 6?
If the sum of each digit of a number is divisible by three ((1 + 1 + 1 + 0)/3 = 1) and it gives an integer, and the number is divisible by 2 (1110/2 = 555) and also gives an integer, then the number is divisible by 6.So, yes.
What are the divisibility rules from 2-12?
14
The two leading digits of each 4-digit number are not shown which number is divisible by 4?
0
What is the least 6-digit number which is divisible by 6 with no repeated digits?
0
The number 3456 is divisible by which single digit numbers?
A very unique number; it is divisible by each single digit number except 5 and 7.
What digit can be placed 87 and 2 to make each number divisible by 3?
I don't completely understand the question, but there's a trick: A number is divisible by 3 if and only if the sum of its digits is divisible by 3. This should guide you to an answer.
A nine-digit number consists of each digit from 1 to 9 The nine digit number is therefore divisible by 9 If you remove the rightmost digit the remaining eight-digit number will be divisible by 8andson?
This is a weekly question posted by Columbus State University. See http://www.colstate.edu/mathcontest/problem.php?CategoryID=3&LinkID=CurrentPlease do not answer this question as this person is too lazy to figure it out themselves.
What is the one 9 digit delectable number which can be made using the digits 1 to 9?
A delectable number has nine digits, using the numbers 1-9 once in each digit. The first digit of a delectable number must be divisible by one. The first and second digits must be divisible by two, the first through third must be divisible by three, etc. There has only been one delectable number discovered: 381654729.
Is 1110 divisble by 6?
If the sum of each digit of a number is divisible by three ((1 + 1 + 1 + 0)/3 = 1) and it gives an integer, and the number is divisible by 2 (1110/2 = 555) and also gives an integer, then the number is divisible by 6.So, yes.
If you add the digits of each multiple of 9, you get a multiple of 9, eg:36=3 + 6 = 9783=7 + 8 + 3 = 18One digit is missing from each of the multiples of 9 below. Enter the missing digit for each number.71_2_201_1 010_000 000?
1 5 6 9
What are the divisibility rules from 2-12?
14
The two leading digits of each 4-digit number are not shown which number is divisible by 4?
0
Three digit number that is divisible by 4 and 9?
To be divisible by both 4 and 9 a number must be a multiple of the Least Common Factor of 4 and 9. The LCM is 36. The first 3 digit number in each hundred that is a multiple of 36 are as follows :- 108, 216, 324, 432, 504, 612, 720, 828 and 900.
What is the least 6-digit number which is divisible by 6 with no repeated digits?
0
What are the divisibility rules to determine which numbers 330 is divisible by from the numbers 2 3 5 7 11 and what are those numbers?
To be divisible by 2 the last digit must be even, ie one of {0, 2, 4, 6, 8}; The last digit is 0, which is one of these so 330 is divisible by 2. To be divisible by 3, sum the digits of the number and if this sum is divisible by 3, then the original number is divisible by 3. As the test can be repeated on the sum, repeat the summing until a single digit remains; only if this number is one of {3, 6, 9} is the original number divisible by 3. 330→ 3 + 3 + 0 = 6 6 is one of {3, 6,9} so 330 is divisible by 3. To be divisible by 5, the last digit must be one of {0, 5}. The last digit is 0 which is one of {0, 5} so 330 is divisible by 5. There is no real check for 7 which is not much slower than just dividing by 7 to see if there is no remainder. One check: Write the digits in blocks of 3 starting from the right hand end (like you would for reading the number): in each block of 3 add twice the first digit to three times the second digit to the third digit. Alternately subtract and add the blocks starting from the right hand end of the number. If the result is divisible by 7, then so is the original number. 330 → 2×3 + 3×3 + 0 = 15 15 is not divisible by 7, so 330 is not divisible by 7. To be divisible by 11, alternately subtract and add the digits of the number from the right hand end; only if this sum is divisible by 11 (or is 0) is the original number divisible by 11. 330 → 0 - 3 + 3 = 0 0 is 0, so 330 is divisible by 11. Therefore 330 is divisible by 2, 3, 5, 11 But not divisible by 7.
What is the divisible rule for 7?
To discover whether a number is divisible by 7, double the last digit of the number, and subtract it from the rest of the number. If the result is either 0 or divisible by 7, then the number is divisible by 7. For example:672 - last digit is 2, double 2 to make 4, subtract from 67 equals 63, 63/7 = 9, therefore, 672 is divisible by 7.Dividing by 7 (2 Tests)Take the last digit in a number.Double and subtract the last digit in your number from the rest of the digits.Repeat the process for larger numbers.Example: 357 (Double the 7 to get 14. Subtract 14 from 35 to get 21 which is divisible by 7 and we can now say that 357 is divisible by 7. NEXT TESTTake the number and multiply each digit beginning on the right hand side (ones) by 1, 3, 2, 6, 4, 5. Repeat this sequence as necessaryAdd the products.If the sum is divisible by 7 - so is your number.Example: Is 2016 divisible by 7?6(1) + 1(3) + 0(2) + 2(6) = 2121 is divisible by 7 and we can now say that 2016 is also divisible by 7.
