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Monday, February 10, 2003 - 03:37 PM
How can you quickly know if one number will divide evenly into another number, leaving no remainder? For example, will 3 divide evenly into 2,169,252? Well, I wouldn't have brought the subject up if I didn't know some curious shortcuts...Division by 2
No surprise here. Any number that ends in 0,2,4,6 or 8 is evenly divisible by 2. Division by 3 Add the number's digits. If the sum is evenly divisible by 3, then so is the number. So, will 3 divide evenly into 2,169,252? Yes it will, because the sum of the digits is 27, and 27 is divisble by 3. If you want, you can keep adding numbers until one digit remains. For example, keep going with 27. 2 + 7 = 9, which is also evenly divisible by 3. Division by 4 If the number's last 2 digits are 00 or if they form a 2-digit number evenly divisible by 4, then number itself is divisible by 4. How about 56,789,000,000? Last 2 digits are 00, so it's divisible by 4. Try 786,565,544. Last 2 digits, 44, are divisible by 4 so, yes, the whole number is divisible by 4. Division by 5 Any number that ends in a 0 or 5 is evenly divisible by 5. Easy enough. Division by 6 The number has to be even. If it's not, forget it. Otherwise, add up the digits and see if the sum is evenly divisible by 3. It it is, the number is evenly divisible by 6. Try 108,273,288. The digits sum to 39 which divides evenly into 13 by 3, so the number is evenly divisible by 6. If you want, you can keep adding numbers until only one digit remains and do the same thing. So in this case, 3 + 9 = 12 and 1 + 2 = 3, and 3 is evenly divisible by 3! Division by 7 Multiply the last digit by 2. Subtract this answer from the remaining digits. Is this number evenly divisible by 7? If it is, then your original number is evenly divisible by 7. Try 364. 4, the last digit, multiplied by 2 = 8. 36, the remaining digits, minus 8 = 28. The last time I checked, 28 is evenly divisble by 7, and thus, so is 364! Division by 8 If the number's last 3 digits are 000 or if they form a 3-digit number evenly divisible by 8, then the number itself is divisible by 8. How about 56,789,000,000? Last 3 digits are 000, so it's divisible by 8. Try 786,565,120. The last 3 digits, 120, divide by 8 into 15, so yes, the whole number is divisible by 8. Division by 9 Sum the number's digits. If it divides by 9, you're in luck. As with the tests for 3 and 6, you can keep adding numbers until you're left with only one digit. Division by 10 Any number that ends in 0 is evenly divisible by 10. Division by 11 Here are four ways for different types of numbers:
If the number can be evenly divided by 3 and 4, the same can also be said for 12. Use the methods for Division by 3 and Division by 4 above. If they both work, your number is also evenly divisible by 12. Division by 15 If the number can be evenly divided by 3 and 5, the same can also be said for 15. Use the methods for Division by 3 and Division by 5 above. If they both work, your number is also evenly divisible by 15. Division by 24 If the number can be evenly divided by 3 and 8, the same can also be said for 24. Use the methods for Division by 3 and Division by 8 above. If they both work, your number is also evenly divisible by 24. Division by 33 If the number can be evenly divided by 3 and 11, the same can also be said for 33. Use the methods for Division by 3 and Division by 11 above. If they both work, your number is also evenly divisible by 33. Division by 36 If the number can be evenly divided by 4 and 9, the same can also be said for 36. Use the methods for Division by 4 and Division by 9 above. If they both work, your number is also evenly divisible by 36.
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Comments
As far as I know the rest work so thanks for the good info! :-)
-Ren
If you take a three digit number, abc, then the number is made of of 100 lots of a, 10 of b and a single c.
We split up the 100 lots of a into 99 lots of a and one single a, we split the 10 lots of b into 9 lots of b and a single b.
We can split the 99 lots of a into 3 groups (33a's each) and we can split the 9 lots of b into 3 groups (3bs each).
So, if the sum of the single a, b and c values can be split into 3 groups as well then so can the entire number.
You can convince yourself that this works for larger numbers by using the same thinking and that the same argument holds for the divisibilty by 9 rule.
to be divisible by 6 the number needs to be even, and 1521893673 isn't even.
By 13: Multiply the last digit by 4. Add this answer from the remaining digits. Is this number evenly divisible by 13? If it is, then your original number is evenly divisible by 13. Try 221, the last digit, multiplied by 4 = 4. And , the remaining digits 22, plus 4 = 26. The last time I checked, 26 is evenly divisible by 13, and thus, so is 221!
By 19: Multiply the last digit by 2. Add this answer from the remaining digits. Is this number evenly divisible by 19? If it is, then your original number is evenly divisible by 19. Try 304, the last digit, multiplied by 2 = 8. And , the remaining digits 30, plus 8 = 38. The last time I checked, 38 is evenly divisible by 19, and thus, so is 304! If this is not clear, we can do 38 --> 3+8*2 = 19
Greetings from Spain
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