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# StupidDivisibilityTricks.pdf - page 12 / 15

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/ 7019

• +

6 /619

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21 /29

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24 53

Thus 27019 53 (mod 97), and we see that 97 does not divide 27019.

# Trim from the Left Trick: Let d be given, let h ≡ 100 (mod d) and

write a = [a

a0]. Let a0 be the number that results from computing a h,

and adding that to [a

1

a0] so that the ones digit of a h lines up with

a 2. Then a a0 is divisible by d.

(mod d); in particular, a is divisible by d if and only if a0

7

2

34

2

53

6

13

4

35

5

95

5

14

2

48

4

96

4

19

5

49

2

97

3

21

5

51

2

98

2

32

4

52

4

102

2

d

100

(mod d)

d

100 (mod d)

d

100 (mod d)

33

1

Why it Works: Let d be given, write a = [a If 100 h (mod d), then

a0], and assume k 2.

a = [a

a0] = a 10 + [a

1

a0]

a h10

2

• +

[a

1

a0]

(mod

d)

# The

e

ect

of

a

h10

2

to

[a

1

a0] is the same as adding a h to

[a a0] so that the ones digit of a h lines up with a 2. Running total: We now have divisibility tests for 2, 3, 4, 5, 7, 8, 9, 10, 11, 13, 14, 16, 17, 19, 20, 21, 23, 25, 27, 29, 31, 32, 33, 34, 35, 37, 39, 40, 41, 1

12

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