/ 7019

+

6 /619

+

21 /29

+

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

a_{0}]. Let a^{0 }be the number that results from computing a h,

and adding that to [a

1

a_{0}] so that the ones digit of a h lines up with

a _{2}. Then a ≡ a^{0 }is divisible by d.

(mod d); in particular, a is divisible by d if and only if a^{0 }

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

a_{0}], and assume k ≥ 2.

a = [a

a_{0}] = a 10 + [a

1

a_{0}]

≡ a h10

2

+

[a

1

a_{0}]

(mod

d)

# The

e

ect

of

adding

a

h10

2

to

[a

1

a_{0}] is the same as adding a h to

[a a_{0}] 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