4 ao = 7aj + 3az +9a3 + Tas+3as +9a6 +7a7+3as (mod 10)identification number 81504216 would then have check digitn7.8+3-1+9.5+7.0+3.4+9.2+7-1+3.6=9 (mod 10)hat 815042169 would be printed on the check.This weighting scheme for assigning check digits detects any single-digit errorthe identification number. For suppose that the digit a, is replaced by a different" By the manner in which the check digit is calculated, the difference between thecorrect ag and the new a isa9ag = k(a; - a)(mod 10)where k is 7,3, or 9 depending on the position of a. Because k(a; - a)# 0 (mod 10),it follows that a9 # ag and the error is apparent. Thus, if the valid number 81504216were incorrectly entered as 81504316 into a computer programmed to calculatecheck digits, an 8 would come up rather than the expected 9.The modulo 10 approach is not entirely effective, for it does not always detectthe common error of transposing distinct adjacent entries a and b within the stringof digits. To illustrate: the identification numbers 81504216 and 81504261 havethe same check digit 9 when our example weights are used. (The problem occursb| = 5.) More sophisticated methods are available, with larger moduliwhen Ja -and different weights, that would prevent this possible error.PROBLEMS 4.31. Use the binary exponentiation algorithm to compute both 1953 (mod 503) and 14147(mod 1537).2. Prove the following statements:(a) For any integer a, the units digit of a? is 0, 1, 4, 5, 6, or 9.(b) Any one of the integers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 can occur as the units digit of a.(c) For any integer a, the units digit of a is 0, 1, 5, or 6.(d) The units digit of a triangular number is 0, 1, 3, 5, 6, or 8.3. Find the last two digits of the number 9.[Hint: 9° = 9 (mod 10); hence, 99° = 9°+10k, notice that 9° = 89 (mod100).]4. Without performing the divisions, determine whether the integers 176521221 and149235678 are divisible by 9 or 11.3. (a) Obtain the following generalization of Theorem 4.6: If the integer N is representedin the base b byN = amb" +...+azb? + a,b+ ao0 s ak

Name: 4 ao = 7aj + 3az +9a3 + Tas+3as +9a6 +7a7+3as (mod 10)identification n
Uploaded: 2024-03-20T07:06:41.000Z
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Description: 4 ao = 7aj + 3az +9a3 + Tas+3as +9a6 +7a7+3as (mod 10)identification number 81504216 would then have check digitn7.8+3-1+9.5+7.0+3.4+9.2+7-1+3.6=9 (mod 10)hat 815042169 would be printed on the check.This weighting scheme for assigning check digits de