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Define f(n) = a+7b+3c− 2d, where n is a four-digit number abcd in its usual decimal representation. Prove that f(n) is divisible by 23 if and only if n is divisible by 23. Solution n = 1000a + 100b+10c+d⇒n=11a+8b+10c + d (mod 23), So n = 0 (mod 23) ← 11a +86 +10c+d=0 (mod 23) 22a+166+20c + 2d = 0 (mod 23) - ↔ −a – 7b – 3c + 2d = 0 (mod 23) f(n) = 0 (mod 23). по change?

Define f(n) = a+7b+3c− 2d, where n is a four-digit number abcd in its usual decimal representation. Prove that f(n) is divisible by 23 if and only if n is divisible by 23. Solution n = 1000a + 100b+10c+d⇒n=11a+8b+10c + d (mod 23), So n = 0 (mod 23) ← 11a +86 +10c+d=0 (mod 23) 22a+166+20c + 2d = 0 (mod 23) - ↔ −a – 7b – 3c + 2d = 0 (mod 23) f(n) = 0 (mod 23). по change?

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Algebra & Trigonometry with Analytic Geometry

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN: 9781133382119

Author: Swokowski

Publisher: Cengage

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Transcribed Image Text:Define f(n) = a+7b+3c− 2d, where n is a four-digit number abcd in its usual decimal representation. Prove that f(n) is divisible by 23 if and only if n is divisible by 23. Solution n = 1000a + 100b+10c+d⇒n=11a+8b+10c + d (mod 23), So n = 0 (mod 23) ← 11a +86 +10c+d=0 (mod 23) 22a+166+20c + 2d = 0 (mod 23) - ↔ −a – 7b – 3c + 2d = 0 (mod 23) f(n) = 0 (mod 23). по change?

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