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# Now

[a

a0] 0

(mod d)

⇐⇒ 10[a

a1] + [a0] 0

(mod d)

⇐⇒ 10u[a

a1] + u[a0] 0

(mod d)

⇐⇒ [a

a1] + u[a0] 0

(mod d)

Running total: We now have divisibility tests for 2, 3, 4, 5, 7, 8, 9, 10, 11, 13, 16, 17, 19, 20, 21, 23, 25, 27, 29, 31, 32, 33, 37, 39, 40, 41, 43, 47, 49, 50, 51, 53, 57, 59, 61, 64, 67, 69, 71, 73, 77, 79, 80, 81, 83, 87, 89, 91, 93, 99, 100, 101.

3.5

# Trim from the Left

The principle of this trick is that if 100 h (mod d) then 100a + b ha + b (mod d). For example, to test divisibility by 97, we note that 100 3

(mod 97). To apply the principle and see if

trim

o

the

leftmost

digit

(2),

multiply

it

27019 by 3

remaining

digits

(7019),

but

shifted

in

to

the

right

is divisible by 97 (6) and add that by two places:

we to

can the

/7019

• +

6

7619

Thus 27019 7619 (mod 97). We can continue the process until we arrive at a number that either clearly is or is not divisible by 97.

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