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

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
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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**Text Transcription for Educational Website** --- where \(k\) is 7, 3, or 9 depending on the position of \(a_i'\). Because \(k(a_i - a_i') \not\equiv 0 \pmod{10}\), it follows that \(a_9 \not\equiv a_9'\) and the error is apparent. Thus, if the valid number 81504216 were incorrectly entered as 81504316 into a computer programmed to calculate check 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 the common error of transposing distinct adjacent entries \(a\) and \(b\) within the string of digits. To illustrate: the identification numbers 81504216 and 81504261 have the same check digit 9 when our example weights are used. (The problem occurs when \(|a - b| = 5\).) More sophisticated methods are available, with larger moduli and different weights, that would prevent this possible error. **PROBLEMS 4.3** 1. Use the binary exponentiation algorithm to compute both \(19^{53} \pmod{503}\) and \(141^{47} \pmod{1537}\). 2. Prove the following statements: (a) For any integer \(a\), the units digit of \(a^2\) 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 \(a^3\). (c) For any integer \(a\), the units digit of \(a^4\) 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 \(9^9\). [Hint: \(9^9 \equiv 9 \pmod{10}\); hence, \(9^9 = 9^{9 + 10k}\); notice that \(9^9 \equ
Transcribed Image Text:**Text Transcription for Educational Website** --- where \(k\) is 7, 3, or 9 depending on the position of \(a_i'\). Because \(k(a_i - a_i') \not\equiv 0 \pmod{10}\), it follows that \(a_9 \not\equiv a_9'\) and the error is apparent. Thus, if the valid number 81504216 were incorrectly entered as 81504316 into a computer programmed to calculate check 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 the common error of transposing distinct adjacent entries \(a\) and \(b\) within the string of digits. To illustrate: the identification numbers 81504216 and 81504261 have the same check digit 9 when our example weights are used. (The problem occurs when \(|a - b| = 5\).) More sophisticated methods are available, with larger moduli and different weights, that would prevent this possible error. **PROBLEMS 4.3** 1. Use the binary exponentiation algorithm to compute both \(19^{53} \pmod{503}\) and \(141^{47} \pmod{1537}\). 2. Prove the following statements: (a) For any integer \(a\), the units digit of \(a^2\) 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 \(a^3\). (c) For any integer \(a\), the units digit of \(a^4\) 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 \(9^9\). [Hint: \(9^9 \equiv 9 \pmod{10}\); hence, \(9^9 = 9^{9 + 10k}\); notice that \(9^9 \equ
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