Proposition 6.1.21. A number is divisible by 11 if and only if the alternat- ing sums of the digits is divisible by 11. (Note: alternating sums is where the signs of the number alternate when summing.) PROOF. Given an integer with digits do . dn where the number is writeen as dndn-1...d1do we can write n = dm · 10" + dm-1 · 10m-1 +...+ do · 10º it follows that: mod (n, 11) = mod (dm · 10m + dm-1 · 10m-1 + ..+ do · 10º, 11) = mod (dm · (-1)™ + dm-1 · (-1)"m-1 + ... = mod ((-1)"(dm – dm-1+ ·· [substitution] + do · (–1)º, 11) [mod(10, 11) = –1] + do · 1), 11) [factor out (-1)"] Therefore, mod (n, 11)=0 if and only if the alternating sums of the digits of
Proposition 6.1.21. A number is divisible by 11 if and only if the alternat- ing sums of the digits is divisible by 11. (Note: alternating sums is where the signs of the number alternate when summing.) PROOF. Given an integer with digits do . dn where the number is writeen as dndn-1...d1do we can write n = dm · 10" + dm-1 · 10m-1 +...+ do · 10º it follows that: mod (n, 11) = mod (dm · 10m + dm-1 · 10m-1 + ..+ do · 10º, 11) = mod (dm · (-1)™ + dm-1 · (-1)"m-1 + ... = mod ((-1)"(dm – dm-1+ ·· [substitution] + do · (–1)º, 11) [mod(10, 11) = –1] + do · 1), 11) [factor out (-1)"] Therefore, mod (n, 11)=0 if and only if the alternating sums of the digits of
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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