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Show that 19^93-13^99 is a positive integer divisible by 162. Can you answer it without using modular arithmetics?

Show that 19^93-13^99 is a positive integer divisible by 162. Can you answer it without using modular arithmetics?

Grade:

1 Answers

SHAIK AASIF AHAMED
askIITians Faculty 74 Points
7 years ago
Hello student,
Please find the answer to your question below
Both 19^93 and 13^99 are odd, so odd – odd is even and is divisible by 2,
so it’s enough we need to show that 19^93 and 13^99 is divisible by (162 / 2)
which is 81.
Now, 19^93 (modulo 81) = (18 + 1)^93 (modulo 81) = 93(18) + 1 = 55 (modulo 81) Similarly,13^99 (modulo 81) = (12 + 1)^99 (modulo 81)
= 99(12) + 1 = 1189 (modulo 81)
= 55 (modulo 81)
Hence, 19^93 – 13^99 = 0 (modulo 81)
Therefore, 19^93 and 13^99 is divisible by 162
2)Another method is to show that
Only problem is to show that [19^{93} - 13^{99}] is divisible by [3^4=81]
because[19^{93} - 13^{99}] is odd-odd=even which is divisible by 2 so 162/2=81.
[19^{93} - 13^{99} = (16+3)^{93} - (16-3)^{99}]
Expanding using binomial, taking modulo [3^4] , almost everything vanishes, only lefts : [4779] which is indeed divisible by [3^4]

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