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09 Apr 2024, 11:44
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We can use Fermat’s Theorem here which says
Rem [a^(p-1)/p] = 1 where p is a prime number
=> Rem [2^88/89] = 1
=> Rem [2^89/89] = Rem [2^88/89] * Rem [2/89] = 1*2 = 2
This is what I found online
Rem [a^(p-1)/p] = 1 where p is a prime number
=> Rem [2^88/89] = 1
=> Rem [2^89/89] = Rem [2^88/89] * Rem [2/89] = 1*2 = 2
This is what I found online
09 Apr 2024, 11:52
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Arungmat612
Nice! I don’t think This is tested on the GMAT. You don’t have to know this.
09 Apr 2024, 18:14
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bb
bb,
There are various formulas when it comes to remainder:, Fermat’s remainder theorem, Chinese remainder theorem. Luckily, GMAT doesn’t test them.
Having said that, I believe remainder questions can almost always be solved with pattern. If a question can be answered through a formula, I am sure it must be following some pattern.
Here too it should be the case:
2^1:2,
2^2: 4,
2^3: 8,
2^4: 16,
2^5: 32,
2^6: 64,
2^7: 128 or 128-89 or 39,
2^8: 39*2 or 78
2^9: 78*2-89 or 156-89 or 67
2^10: 67*2-89 or 134-89 or 45
2^11: 45*2 or 90-89 or 1
2^12: 1*2 or 2
So, we are repeating the remainders now and there are 11 different remainders in a pattern.
89=11*8+1
So, 2^89 will have same remainder as 2^1 or 2.
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10 Apr 2024, 01:57
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bb
Yes, the question is out of scope for the GMAT. You can solve it either by binomial expansion or using Fermat's Little Theorem (if p is a prime number, then for any integer a, a^p – a will be a multiple of p). Since 89 is prime, then 2^89 - 2 is a multiple of 89. Thus, 2^89 is 2 more than a multiple of 89, making the remainder equal to 2.
Having said that, you absolutely don't need this for the GMAT and you can ignore the above altogether.
