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2002 AIME I Problems/Problem 7

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Problem

The Binomial Expansion is valid for exponents that are not integers. That is, for all real numbers x,y and r with |x|>|y|,

(x+y)^r=x^r+rx^{r-1}y+\dfrac{r(r-1)}{2}x^{r-2}y^2+\dfrac{r(r-1)(r-2)}{3!}x^{r-3}y^3 \cdots

What are the first three digits to the right of the decimal point in the decimal representation of (10^{2002}+1)^{\frac{10}{7}}?

Solution

1^n will always be 1, so we can ignore those terms, and using the definition (2002 / 7 = 286):

(10^{2002} + 1)^{\frac {10}7} = 10^{2860}+\dfrac{10}{7}10^{858}+\dfrac{15}{49}10^{-1144}+\cdots

Since the exponent of the 10 goes down extremely fast, it suffices to consider the first few terms. Also, the 10^{2860} term will not affect the digits after the decimal, so we need to find the first three digits after the decimal in

\dfrac{10}{7}10^{858}.

(The remainder after this term is positive by the Remainder Estimation Theorem). Since the repeating decimal of \dfrac{10}{7} repeats every 6 digits, we can cut out a lot of 6's from 858 to reduce the problem to finding the first three digits after the decimal of

\dfrac{10}{7}.

That is the same as 1+\dfrac{3}{7}, and the first three digits after \dfrac{3}{7} are 428.


An equivalent statement is to note that we are looking for 1000 \left\{\frac{10^{859}}{7}\right\}, where \{x\} = x - \lfloor x \rfloor is the fractional part of a number. By Fermat's Little Theorem, 10^6 \equiv 1 \pmod{7}, so 10^{859} \equiv 3^{6 \times 143 + 1} \equiv 3 \pmod{7}; in other words, 10^{859} leaves a residue of 3 after division by 7. Then the desired answer is the first three decimal places after \frac 37, which are \boxed{428}.

See also

2002 AIME I (ProblemsResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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