16. (I) Let n=ck⋅10k+⋯+c1⋅10+c0. Prove:
(i) 2∣n⟺2∣c0;
(ii) 5∣n⟺5∣c0;
(iii) 3∣n⟺3∣(ck+⋯+c0);
(iv) 9∣n⟺9∣(ck+⋯+c0);
(v) 11∣n⟺11∣(ck−ck−1+⋯+(−1)k⋅c0).
(II) Let n=ck⋅(100)k+⋯+c1⋅(100)+c0. Prove:
(i) 11∣n⟺11∣(ck+⋯+c0);
(ii) 101∣n⟺n∣(ck−ck−1+⋯+(−1)kc0).
(III) Let n=ck⋅(1000)k+⋯+c1⋅(1000)+c0. Prove:
(i) 37∣n⟺37∣(ck+⋯+c0);
(ii) 7∣n⟺7∣(ck−ck−1+⋯+(−1)k⋅c0);
(iii) 13∣n⟺13∣(ck−ck−1+⋯+(−1)kc0).
(IV) Use the above results to propose corresponding divisibility tests for integers by 2,3,5,7,9,11,13,37, or 101. Use this method to factorize the numbers 1535625, 1158066, 82798848, and 81057226635000 into their prime factors.
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