#Is this correct?

45 messages · Page 1 of 1 (latest)

sand wyvern
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For any integer n and any natural number a, (n x 10^a) mod 10 = 0.

near elkBOT
small bough
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Yes, by definition of mod

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(Anything * m) mod m = 0

drifting heron
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or 0

sand wyvern
small bough
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Would hold for n with denominator of lower order of magnitude than 10^a in general, so if the denominator is > 1 then it won't hold for a = 1

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In most reduced form

drifting heron
small bough
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Otherwise any single digit denominator rational will work for a>2

drifting heron
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modulo would get distributed and become congruent to

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wait if denominator of n is 10^a then things might change

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they would cancel

small bough
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If it's of the same order of magnitude in general things change
10/3 = 3.3 for example

drifting heron
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yes

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yes

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if denom(n) is of the form 10^k where k is a natural number >=a then things will change mod can become undefined or numer(n)mod 10

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eg check this

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,w (3/10^3)*(10^2) mod 10

drifting heron
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what

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wait it did not recognise it wait

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,w (3/10) mod 10

small bough
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Is 0.3 it's correct

drifting heron
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let me explain if n is a rational number of the form a/b then a/b mod some number k can be written as ( (a mod k) (b mod inverse k) ) whole mod k if gcd(a,b) = 1 then this exists

sand wyvern
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hmm then 0.3is correct?

drifting heron
drifting heron
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(3/10) mod 10 as gcd being 1 of (3,10) can be written as 3 mod 10 10 mod inverse 10 whole mod 10 but as 10 mod inverse 10 is not defined this modulo is not defined

small bough
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You're thinking of x (mod m)

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Which is congruence

drifting heron
small bough
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Where

drifting heron
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For any integer n and any natural number a, (n x 10^a) mod 10 = 0.

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see the bracket

small bough
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Right, that's mod operation

drifting heron
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oh

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shit

small bough
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(x) mod 10 or x % 10

drifting heron
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oh oh sorry

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i see now

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i felt it was 0 mod 10

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sorry