Number system :Remainder | |

What is the highest possible value of ‘n’ for which (3^1024) – 1 is divisible by 2^n? (a) 13 (b) 10 (c) 11 (d) 12 |

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ques 2 . How many 10-digit natural numbers in binary are there in which at least two 1’s come together? kch hi samaj nhi aaya |

Re: Number system :Remainder | |

Here, (3^1024)-1 = ((2+1)^1024)-1 = (1024C0 * 2^1024+....+1)-1. i.e (1024C0 *2^1024+...+1024C1023 * 2)/2^n and 1024C1023 * 2 =2^11. Hence highest value of n will be 11. |

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ans is 12 ? |

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Hi Rakshit This question is exactly similar what I posted about three years back in an article on "Fibonacci Recursion". Slight difference is that now you are forming numbers by using the digits 0 and 1. So answer will nt be exactly same. Read the article and you'll be able to understand the funda. If you start making cases to solve this problem, then it's going to be a good lengthier question. I am pasting a picture of solved question from the article here. Kamal Lohia |

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Hi Rakshit There is no problem with Euler. phi(2 So 3 Or 3 Kamal Lohia |

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sir, one more Find the remainder when (10^3+9^3)^752 divided by 12^12 ? |

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yes sir, got it ..thanx a lot you reminded me a special property about 1729..12^3 + 1^3 = 10^3 + 9^3 |

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For second question,i am getting 7 as the answer. is it correct? |

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no..it is not ryt..i dont have ans..but it a two digit no. |

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Hi Rakshit 2. It should be 16. Place the points symmetrically in L shape. |

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12 |

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3^1024 - 1 = (2+1)^1024 - 1 = 2^1024 + .... + C(1024, 1022)*2^2 + (1024)2 + 1 - 1 = 2^11(k) Approaching in dis way, why different answer is coming? |

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I think it should be 36 First of all making all the possible rectangles and then 4 rt angles in each triangle |

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Hi navin, how you concluded that the max power of 2 is 12 from the thing that 3+1 is divisible by 4. |