Maths Olympiad Prep

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Number theory Difficulty 5.6 AIME, harder Prove it

16. (I) Let n=ck10k++c110+c0n=c_{k} \cdot 10^{k}+\cdots+c_{1} \cdot 10+c_{0}. Prove:
(i) 2n2c02|n \Longleftrightarrow 2| c_{0};
(ii) 5n5c05|n \Longleftrightarrow 5| c_{0};
(iii) 3n3(ck++c0)3|n \Longleftrightarrow 3|\left(c_{k}+\cdots+c_{0}\right);
(iv) 9n9(ck++c0)9|n \Longleftrightarrow 9|\left(c_{k}+\cdots+c_{0}\right);
(v) 11n11(ckck1++(1)kc0)11|n \Longleftrightarrow 11|\left(c_{k}-c_{k-1}+\cdots+(-1)^{k} \cdot c_{0}\right).
(II) Let n=ck(100)k++c1(100)+c0n=c_{k} \cdot(100)^{k}+\cdots+c_{1} \cdot(100)+c_{0}. Prove:
(i) 11n11(ck++c0)11|n \Longleftrightarrow 11|\left(c_{k}+\cdots+c_{0}\right);
(ii) 101nn(ckck1++(1)kc0)101|n \Longleftrightarrow n|\left(c_{k}-c_{k-1}+\cdots+(-1)^{k} c_{0}\right).
(III) Let n=ck(1000)k++c1(1000)+c0n=c_{k} \cdot(1000)^{k}+\cdots+c_{1} \cdot(1000)+c_{0}. Prove:
(i) 37n37(ck++c0)37|n \Longleftrightarrow 37|\left(c_{k}+\cdots+c_{0}\right);
(ii) 7n7(ckck1++(1)kc0)7|n \Longleftrightarrow 7|\left(c_{k}-c_{k-1}+\cdots+(-1)^{k} \cdot c_{0}\right);
(iii) 13n13(ckck1++(1)kc0)13|n \Longleftrightarrow 13|\left(c_{k}-c_{k-1}+\cdots+(-1)^{k} c_{0}\right).
(IV) Use the above results to propose corresponding divisibility tests for integers by 2,3,5,7,9,11,13,372,3,5,7,9,11,13,37, or 101. Use this method to factorize the numbers 1535625, 1158066, 82798848, and 81057226635000 into their prime factors.

Solution

 16. (IV) 1535625=3354713;1158066=232721310182798848=2835113;81057226635000=23335473112172337\begin{array}{l}\text { 16. (IV) } 1535625=3^{3} \cdot 5^{4} \cdot 7 \cdot 13 ; \quad 1158066=2 \cdot 3^{2} \cdot 7^{2} \cdot 13 \cdot 101 \text {; } \\ 82798848=2^{8} \cdot 3^{5} \cdot 11^{3} ; \\ 81057226635000=2^{3} \cdot 3^{3} \cdot 5^{4} \cdot 7^{3} \cdot 11^{2} \cdot 17 \cdot 23 \cdot 37\end{array}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.