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Nullbyte
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Oh actually it can’t. My bad
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in this case it can’t be

but in general if the value comes out as 0 then its neither +ve nor -ve
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focusedknight
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can anybody confirm the answer for this? is it B?
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Alex55
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Yes its B

Explain plz. I am not sure about the methos
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Alex55
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Can you explain the method

Plz

focusedknight
Thanks1
Can you show the method
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Bhagwan212
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explanation: (35^12+63^17)/14 = 1/2 *(5*35^11+9*63^16) = 1/2 (-----4). so last digit inside the bracket is even and if it is divided by 2 (remaining factor of 14), then remainder will be 0. so A = 0 which is less than 3. So B is the answer.
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This may not be the best explanation, but I hope it helps. If you have doubt please ask.
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i did the same way thnks

i had doubt that remainder will be 0 or not thnks for the hlp
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A quick solution is: Both 35^{12} and 63^{17} are multiples of 7, so their sum will also be multiple of 7. Next both are odd to some power, so each will be odd and the sum will be odd+odd or even. Thus sum of two terms is even and multiple of 7, that is multiple of 2 and 7 or 14, meaning remainder is 0. A=0 and B=3, so B is greater
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8^43 =(7+1)^43 When you expand it all terms except one contain 7 and the term is 1^43. Thus remainder is 1^43 or 1
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Find the remainder when 2^89 is divided by 89?

Can anyone solve this?
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Ans is correct

but how to approach this
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2^1=2
2^2=4
2^3=8
2^4=16
So for 2 after 4 steps it will repeat
4*22=88
Remaining 1 so the unit digit will be 2 now divide it by 89

I think this is the way

Anyone better suggestion
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Do you know the patter of 2 in case of power
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i understood the pattern of unit digits repeating every 4 steps
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There is 89 in power is it divisible by 4 not what is divisible 88 after that 22 repeat cycle 1st term comes again

The first term is 2 in the cycle now divide it by 89 the remainder will be 2

Did you get it now
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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
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