27

Number Theory Level pending

The a b c ‾ \overline{abc} is a three-digit positive integer, so that both of a ‾ \overline{a} and b ‾ \overline{b} are bigger than 0 0 . We know that 27 27 is a divisor of a b c ‾ \overline{abc} . What is the probality that the b c a ‾ \overline{bca} number is also divisible by 27 27 ?


The answer is 1.

Áron Bán-Szabó by Áron Bán-Szabó , 17, Hungary

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2 solutions

H K
Jun 7, 2017

The number abc satisfies 0 ≡ 10 0 ≡ 10 ( 100 a + 10b + c ) ≡ 100b + 10c + a ≡ bca (mod 27) such that bca is also divisible by 27. The simplification in step 4 holds true because 1000 = 999 + 1 ≡ 1 (mod 27) since 999 = 27 37.

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b c a ‾ = 100 b + 10 c + a \overline{bca}=100b+10c+a

10 ∗ a b c ‾ − 999 ∗ a = 1000 a + 100 b + 10 c − 999 a = a + 100 b + 10 c = 100 b + 10 c + a 10*\overline{abc}-999*a=1000a+100b+10c-999a=a+100b+10c=100b+10c+a

If a b c ‾ = 27 ∗ k \overline{abc}=27*k , then

b c a ‾ = 10 ∗ a b c ‾ − 999 a = 27 ∗ ( 10 k − 37 a ) \overline{bca}=10*\overline{abc}-999a=27*(10k-37a) ,

so b c a ‾ \overline{bca} is always divisible by 27 27 and the anserw is 1 \boxed{1} .

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