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# Finding prime factor with binomial theorem

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Find The Largest Prime factor of 1,005,010,010,005,001.

How would I do this presumably using the binomial theorem?

Mar 12, 2019

#1
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xxxxxxxxxxxx

Mar 12, 2019
edited by Guest  Mar 12, 2019
edited by Guest  Mar 12, 2019
edited by Guest  Mar 12, 2019
#2
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13 is the largest factor.

Mar 12, 2019
#4
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1,005,010,010,005,001

As you see the pattern 1,5,10,10,5,1 in the number, the number is possibly $$(10^n + 1)^5$$, where n is a natural number.

As the number has 16 digits and $$10^{15} = \left(10^3\right)^5$$ also contains 16 digits, the number is possibly $$(10^3 + 1)^5$$ which is 10015.

What we need to do now is to find the prime factorisation of the number, and we can find the prime factorisation of that huge number.

After some trial divisions, $$1001 = 7\times 11\times 13$$.

Therefore, it immediately follows that $$1\;005\;010\;010\;005\;001 =(7\times 11\times 13)^5 = 7^5 \times 11^5 \times 13^5$$.

As a result, the largest prime factor of this number is 13 :)

Mar 13, 2019
#5
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to make this a tad clearer

Max Wong has noted that

$$1,005,010,010,005,001 = \\ 1\times 10^{15} + 5 \times 10^{12} + 10\times 10^9 + 10\times 10^6 + 5 \times 10^3 + 1\times 10^0\\ \text{This is }\\ \sum \limits_{k=0}^5~\dbinom{5}{k} \left(10^{3}\right)^k \cdot 1^{5-k} = (10^3+1)^5 = 1001^5$$

$$\text{Now the problem becomes find the largest prime factor of }1001\\ \text{I don't know any way of doing this other than brute force which is what Max Wong used}$$

Rom  Mar 13, 2019