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If a number is divisible by six, then it must also be divisible by both two and three. To check it then, you simply apply both rules for each of those:
Is it an even number?
Is the sum of it's digits divisible by three?
If both answers are yes, then the number is divisible by six.
To find out if a number is divisible by seven take it's last digit, double it, and subtract it from the remaining digits. If the result is divisible by seven, then the original number is as well.
For example is three 343 divisible by 7? Let's find out:
3 * 2 = 6, and 34 - 6 = 28. Is 28 divisible by seven? Yes, but if you're not sure, you can repeat the process. 8 * 2 = 16, 2 - 16 = -14. 14 is of course divisible by 7.
If the last three digits of a number are divisible by 8, then the entire number is. For example, I know that 10923485710234985723908471859256 is divisible by eight, because I know that the last three digits, 256, form a number that's divisible by eight. Not sure about those last three digits? Simply divide them by two, three times in a row. If the result is a whole number, then it's divisible by eight. 256 / 2 = 128, 128 / 2 = 64, 64 / 2 = 32, so 256 is divisible by eight, and therefore 10923485710234985723908471859256 is also.
If the sum of the digits is divisible by 9, then the number itself is. For example, is 8936523 divisible by 9? Well, 8 + 9 + 3 + 6 + 5 + 2 + 3 = 36. Is 36 divisible by 9? 3 + 6 = 9. 9 is obviously divisible by 9, so yes, 8936523 is also.
If the last digit is a zero, then the number is divisible by 10. For example, 12340 is divisible by ten, but 12345 is not.
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What is the divisibility rule for 21?
For any practical purpose, it is easier to simply divide, instead of looking for fancy divisibility rules. However, you can apply the divisibility rules for 3 and for 7. This works because (a) their product is 21, and (b) these numbers are relatively prime.
What is 3 over 10 divided by 7 over 6?
(3/10) / (7/6) = (3/10) *(6/7) = 18/70 = 9/35(3/10) / (7/6) = (3/10) *(6/7) = 18/70 = 9/35(3/10) / (7/6) = (3/10) *(6/7) = 18/70 = 9/35(3/10) / (7/6) = (3/10) *(6/7) = 18/70 = 9/35
What are the divisibility rules for 23?
Take the last digit and add 7 times to the rest of the number. Ex: 5888 588+ (7x8)=644, 644/23= 28
How do you get the factor pairs of 35?
By the rules of divisibility, you know that 35 is divisible by 1 and 5 and not divisible by 2, 3, 4 and 6. Dividing those two numbers into 35 gives you 7 and 35. That makes the factors of 35 (1, 5, 7, 35) Now pair them up. (35,1)(7,5)
What are the divisibility rules of 42?
A number is a multiple of 42 if it's a multiple of 2 and 21 at the same time 126 is a multiple of 42 because it ends in 6, which means it's a multiple of 2, and 12 - 6 x 2 = 0, which means it's a multiple of 21
Write about a reallied situation that can be solved using divisibility rules solve it?
3+7=10
Who invented the divisibility rules?
The divisibility rules were not invented by a single individual, but rather developed over time by mathematicians through observation and exploration of number patterns. The rules for divisibility by 2, 3, 5, and 10 can be traced back to ancient civilizations such as the Egyptians and Greeks. The more complex rules for divisibility by numbers like 7, 11, and 13 were further refined by mathematicians in the Middle Ages and beyond. These rules are now fundamental concepts in elementary number theory.
What is the greatest common factor of 91 and 49 using the divisibility rules?
It is: 7
What is the divisibility rules of 7?
If the last digit doubled subtracted from the rest is a multiple of 7, the whole number is divisible by 7.
What is the divisibility rule for 21?
For any practical purpose, it is easier to simply divide, instead of looking for fancy divisibility rules. However, you can apply the divisibility rules for 3 and for 7. This works because (a) their product is 21, and (b) these numbers are relatively prime.
How can you tell if 53 is a prime number?
The square root of 53 is between 7 and 8. If 53 was composite, half of its factors would be less than its square root. By the rules of divisibility, we know that 53 is not divisible, by 2, 3, 4, 5, 6 or 7. It has to be prime.
How can you check to be sure that you have found all the factors of 100?
You know that the square root of 100 is 10, so half the remaining factors will be greater than 10, half lesser. By the rules of divisibility, you know that 100 is not divisible by 3, 6, 7, 8 or 9. That leaves 1, 2, 4 and 5. Divide them into 100. 1, 2, 4, 5, 10, 20, 25, 50, 100
What is 3 over 10 divided by 7 over 6?
(3/10) / (7/6) = (3/10) *(6/7) = 18/70 = 9/35(3/10) / (7/6) = (3/10) *(6/7) = 18/70 = 9/35(3/10) / (7/6) = (3/10) *(6/7) = 18/70 = 9/35(3/10) / (7/6) = (3/10) *(6/7) = 18/70 = 9/35
What are the divisibility rules for 23?
Take the last digit and add 7 times to the rest of the number. Ex: 5888 588+ (7x8)=644, 644/23= 28
What is the pattern for 85491761032?
spell out the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 in English then alphabetize them.
What is the divisibility for the number 7?
7 is only divisible by 1 and itself.
How do you find the factors of 54?
The square root of 54 is between 7 and 8. That means that half of the factors will be 7 or less and half will be 8 or higher. From the rules of divisibility, we know that 2, 3, 6 and 9 are factors. Dividing those numbers into 54 gives us the rest. 1, 2, 3, 6, 9, 18, 27, 54
