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標題:
發問:
When (1 + x)^38 is expanded in ascending powers of x, N1 of the coefficients leave a remainder of 1 when divided by 3, while N2 of the coefficients leave a remainder of 2 when divided by 3. Find N1 - N2.
最佳解答:
The coefficients are 38C0 , 38C1 , 38C2 , ... , 38C37 and 38C38.We just consider 38C0 to 38C19 since 38C38 = 38C0 , 38C37 = 38C1 etc. 38C19 = 38x37x(36)]x35x34x33]x32x31x30]x29x28x{27}]x26x25x24]x23x22x21]x20 ----------------------------------------------------------------------------------------------- 01x02x03]x04x05x06]x07x08x(09)]x10x11x12]x13x14x15]x16x17x(18)]x19 Take the first terms of the numerator and denominator respectively we have 38C1 = 38/01 = 38 = 2 (mod 3) Take the first 2 terms we have 38C2 = 38x37/01x02 = 703 = 1 (mod 3) Take the first 3 terms we have 38C3. etc. Note that 36 , 9 and 18 including a factor 32 , while 27 = 33. We can see that 38C3 = 38C4 = ... = 38C8 = 0 (mod 3) because of the number of the factor 3 of the numerator is one more than the denominator. But 38C9 , 38C10 and 38C11 = 1 or 2 (mod 3) since the number of the factor 3 of the numerator are the same as the denominator. And then 38C12 = 38C13 = ... = 38C19 = 0 (mod 3) since the number of the factor 3 of the numerator is more than the denominator. 38C9 = 38x37x36x35x34x33x32x31x30/9! = 38x37x04x35x34x11x32x31x10/1x2x4x5x2x7x8 = 38x37x34x11x31x10 = 02x01x01x02x01x01 (mod 3) = 4 (mod 3) = 1 (mod 3) 38C10 = 38x37x36x35x34x33x32x31x30x29/10! = 38x37x34x11x31x10 x 29/10 = 38x37x34x11x31x29 = 02x01x01x02x01x02 (mod 3) = 8 (mod 3) = 2 (mod 3) 38C11 = 38x37x36x35x34x33x32x31x30x29x28/11! = 38x37x34x11x31x29 x 28/11 = 38x37x34x31x29x28 = 02x01x01x01x02x01 (mod 3) = 4 (mod 3) = 1 (mod 3) We conclude that : 38C0 = 38C38 = 1 (mod 3) 38C1 = 38C37 = 2 (mod 3) 38C2 = 38C36 = 1 (mod 3) 38C9 = 38C29 = 1 (mod 3) 38C10 = 38C28 = 2 (mod 3) 38C11 = 38C27 = 1 (mod 3) Therefore N1 - N2 = 8 - 4 = 4
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Binomial Theorem發問:
When (1 + x)^38 is expanded in ascending powers of x, N1 of the coefficients leave a remainder of 1 when divided by 3, while N2 of the coefficients leave a remainder of 2 when divided by 3. Find N1 - N2.
最佳解答:
The coefficients are 38C0 , 38C1 , 38C2 , ... , 38C37 and 38C38.We just consider 38C0 to 38C19 since 38C38 = 38C0 , 38C37 = 38C1 etc. 38C19 = 38x37x(36)]x35x34x33]x32x31x30]x29x28x{27}]x26x25x24]x23x22x21]x20 ----------------------------------------------------------------------------------------------- 01x02x03]x04x05x06]x07x08x(09)]x10x11x12]x13x14x15]x16x17x(18)]x19 Take the first terms of the numerator and denominator respectively we have 38C1 = 38/01 = 38 = 2 (mod 3) Take the first 2 terms we have 38C2 = 38x37/01x02 = 703 = 1 (mod 3) Take the first 3 terms we have 38C3. etc. Note that 36 , 9 and 18 including a factor 32 , while 27 = 33. We can see that 38C3 = 38C4 = ... = 38C8 = 0 (mod 3) because of the number of the factor 3 of the numerator is one more than the denominator. But 38C9 , 38C10 and 38C11 = 1 or 2 (mod 3) since the number of the factor 3 of the numerator are the same as the denominator. And then 38C12 = 38C13 = ... = 38C19 = 0 (mod 3) since the number of the factor 3 of the numerator is more than the denominator. 38C9 = 38x37x36x35x34x33x32x31x30/9! = 38x37x04x35x34x11x32x31x10/1x2x4x5x2x7x8 = 38x37x34x11x31x10 = 02x01x01x02x01x01 (mod 3) = 4 (mod 3) = 1 (mod 3) 38C10 = 38x37x36x35x34x33x32x31x30x29/10! = 38x37x34x11x31x10 x 29/10 = 38x37x34x11x31x29 = 02x01x01x02x01x02 (mod 3) = 8 (mod 3) = 2 (mod 3) 38C11 = 38x37x36x35x34x33x32x31x30x29x28/11! = 38x37x34x11x31x29 x 28/11 = 38x37x34x31x29x28 = 02x01x01x01x02x01 (mod 3) = 4 (mod 3) = 1 (mod 3) We conclude that : 38C0 = 38C38 = 1 (mod 3) 38C1 = 38C37 = 2 (mod 3) 38C2 = 38C36 = 1 (mod 3) 38C9 = 38C29 = 1 (mod 3) 38C10 = 38C28 = 2 (mod 3) 38C11 = 38C27 = 1 (mod 3) Therefore N1 - N2 = 8 - 4 = 4
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