Chee Wan's answer to Nancy's Secondary 4 A Maths Singapore question.
done
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Problem is trivial once you learn modular arithmetics
Date Posted:
3 years ago
Last sentence should be -891. Forgot the - sign
For the second line, the last term's power should be 999 rather than 1000.
Also, strictly speaking ,
999 mod 1000 = 999
9999 mod 1000 = 9999, and not -1.
Perhaps you meant:
999 ≡ -1 (mod 1000) and 9999 ≡ -1 (mod 1000)
@Nancy : Alternative style of presentation in the last two steps
Based on the multiplication property:
If a ≡ b (mod m) and c ≡ d (mod m), then ac = bd (mod m)
Then 999 ≡ 9999 ≡ 99999 ≡ ... ≡ (999 '9's) ≡ -1 (mod 1000)
Then 999 × 9999 × 99999 × ... × (999 '9's) ≡ (-1)(-1)(-1)... (mod 1000) = (-1)^997 (mod 1000) = -1 (mod 1000)
And since 891 ≡ -109 (mod 1000),
Then 9 × 99 × 999... × (999 '9's)
= 891 × 999... × (999 '9's)
≡ -109 × (-1) (mod 1000) = 109 (mod 1000)
999 mod 1000 = 999
9999 mod 1000 = 9999, and not -1.
Perhaps you meant:
999 ≡ -1 (mod 1000) and 9999 ≡ -1 (mod 1000)
@Nancy : Alternative style of presentation in the last two steps
Based on the multiplication property:
If a ≡ b (mod m) and c ≡ d (mod m), then ac = bd (mod m)
Then 999 ≡ 9999 ≡ 99999 ≡ ... ≡ (999 '9's) ≡ -1 (mod 1000)
Then 999 × 9999 × 99999 × ... × (999 '9's) ≡ (-1)(-1)(-1)... (mod 1000) = (-1)^997 (mod 1000) = -1 (mod 1000)
And since 891 ≡ -109 (mod 1000),
Then 9 × 99 × 999... × (999 '9's)
= 891 × 999... × (999 '9's)
≡ -109 × (-1) (mod 1000) = 109 (mod 1000)