Number patterns?

Question

My teacher always talks about how to easily find out what numbers divide into other numbers, like adding the sum of the even and odds and finding out what that number is, which divides into the original number ... I don't know if that is correct, but is there any way to figure out how to do this with large numbers? I am looking for relations in perfect squares, radicals, etc. Thanks :)

Answer 1

The divisibility rules for numbers from 2 to 15 are the following:

2: A number is divisible by 2 if it ends in 0,2,4,6, or 8.

3: A number is divisible by 3 if the sum of its digits can be divided by 3.
(36 is divisible by 3 since (3+6)=9 is divisible by 3)

4: A number is divisible by 4 if the last two digits form a number which is divisible by 4
(112 is divisible by 4 since 12 is divisible by 4)
0 is considered as the number, which is divisible by all numbers. Numbers consisting from several zeros are equal to 0.
(100 is divisible by 4 since 00=0 is divisible by 4).

5: A number is divisible by 5 if its last digit is a 0 or 5.

6: A number is divisible by 6 if it divisible both on 2 and 3.

7: Remove the last digit of the number, double it, and subtract it from the rest of the number. If the result is divisible by 7, the number is also divisible by 7.
(203 is divisible by 7: remove 3, subtract 3*2 from 20, which is 20-6=14 divisible by 7)

8: A number is divisible by 8 if the last three digits form a number (including 0) which is divisible by 8.
(7640 is divisible by 8 since 640 is divisible by 8)

9: A number is divisible by 9 if the sum of the digits is divisible by 9.
(4581 is divisible by 9 since (4+5+8+1)=18 is divisible by 9)

10: A number is divisible by 10 if its last digit is a 0.

11: A number is divisible by 11 if the difference of the sums of alternating digits is divisible by 11.
(946 is divisible by 11 since (9+6)-4=11 is divisible by 11;
3267 is divisible by 11 since (3+6)-(2+9)=0 is divisible by 11)

12: A number is divisible by 12 if it divisible both on 3 and 4.


13: Delete the last digit of the number. Add the deleted digit fourfold. If the result is divisible by 13, then the original number is divisible by 13.
(91 is divisible by 13: deleting last digit gives 9, adding fourfold deleted digit gives 9+4=13, which is divisible by 13)

14: A number is divisible by 14 if it divisible both on 2 and 7.

15: A number is divisible by 15 if it divisible both on 3 and 5.

Answer 2

a million) it fairly is Fibonacci's series. with a view to get a sort you upload the two previous numbers. as an occasion: 3 + 5 = 8 5 + 8 = 13 8 + 13 = 21 13 + 21 = 34 21 + 34 = fifty 5 34 + fifty 5 = 89 fifty 5 + 89 = one hundred forty four etc. 2) you will locate that the numbers get extra advantageous via 2. as an occasion 2x2 = 4 4x2 = 8 8x2=sixteen 16x2=32 32x2=sixty 4 64x2=128 the series hence is two^n the subsequent 3 words could be: 32, sixty 4 and 128 3) you will locate that each and each type is extra advantageous via 3. So as an occasion: 4x3=12 12x3=36 36x3=108 108x3=324 324x3=972 972x3=2916 so the subsequent 3 words are 324, 972 and 2916 The series is 3^n + 3^(n-a million) 4) you will locate that 7 is extra each and each time. So as an occasion: a million + 7 = 8 8 + 7 = 15 15 + 7 = 22 22 + 7 = 29 29 + 7 = 36 36 + 7 = 40 3 So your next 3 words are 29, 36 and 40 3 The series is given via 7n - 6 wish this is sensible