3. Find the last two digits of the number 99°. [Hint: 9° = 9 (mod 10); hence, 99 = = 99+10k. notice that 9° = 89 (mod100).]

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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#3

(a For any integer a, the of a? is 0, 1, 4, 5, 6, or 9. deCLoanson
a9 – ag = k(a; -a)
6p
(mod 10)
herekis 7,3, or 9 depending on the position of a.Because k(a; – a') # 0 (mod 10),
it1oincorrectly entered as 81504316 into a computer programmed to calculate
heck digits, an 8 would come up rather than the expected 9.
The modulo 10 approach is not entirely effective, for it does not always detect
he common error of transposing distinct adjacent entries a and b within the string
he vcits To illustrate: the identification numbers 81504216 and 81504261 have
O ame check digit 9 when our example weights are used. (The problem occurs
uben la - b| = 5.) More sophisticated methods are available, with larger moduli
and different weights, that would prevent this possible error.
%3D
PROBLEMS 4.3
Use the binary exponentiation algorithm to compute both 1953 (mod 503) and 14147
Ymod 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 99°
[Hint: 9° = 9 (mod 10); hence, 99° = 99+10k, notice that 9° = 89 (mod100).]
Without performing the divisions, determine whether the integers 176521221 and
149235678 are divisible by 9 or 11.
5. (a) Obtain the following generalization of Theorem 4.6: If the integer N is represented
in the base b by
%3D
Op + qlD+,q7p +.+ 9"D = N
then b-1| N if and only if b- 1|(am +.+ az + a1 + ao).
Transcribed Image Text:(a For any integer a, the of a? is 0, 1, 4, 5, 6, or 9. deCLoanson a9 – ag = k(a; -a) 6p (mod 10) herekis 7,3, or 9 depending on the position of a.Because k(a; – a') # 0 (mod 10), it1oincorrectly entered as 81504316 into a computer programmed to calculate heck digits, an 8 would come up rather than the expected 9. The modulo 10 approach is not entirely effective, for it does not always detect he common error of transposing distinct adjacent entries a and b within the string he vcits To illustrate: the identification numbers 81504216 and 81504261 have O ame check digit 9 when our example weights are used. (The problem occurs uben la - b| = 5.) More sophisticated methods are available, with larger moduli and different weights, that would prevent this possible error. %3D PROBLEMS 4.3 Use the binary exponentiation algorithm to compute both 1953 (mod 503) and 14147 Ymod 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 99° [Hint: 9° = 9 (mod 10); hence, 99° = 99+10k, notice that 9° = 89 (mod100).] Without performing the divisions, determine whether the integers 176521221 and 149235678 are divisible by 9 or 11. 5. (a) Obtain the following generalization of Theorem 4.6: If the integer N is represented in the base b by %3D Op + qlD+,q7p +.+ 9"D = N then b-1| N if and only if b- 1|(am +.+ az + a1 + ao).
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