Theory
Problems on periodicity.
Spot patterns!
Problems
A. Decimal expansions and repeating fractions
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Verify these decimal expansions:
- 1/3 = 0.(3)
- 1/6 = 0.1(6)
- 7/30 = 0.2(3)
- 7/11 = 0.(63)
- Find the 100th digit after the decimal point in the expansion of 1/7.
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Division by strings of 9s; purely periodic fractions:
- Compute by long division: 1 ÷ 9; 1 ÷ 99; 1 ÷ 9,999.
- Prove the general rule: 1/(99…9) = 0,(00…01), where the denominator has n nines and the repeating block has length n with (n−1) zeros followed by 1.
- Show that any purely periodic proper fraction equals a fraction whose numerator is the period and whose denominator is 10r − 1 (a number of r nines), where r is the period length.
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Convert to decimal form:
- 23/99
- 1234/999,999
B. Repunits and divisibility
- In the sequence 1, 11, 111, 1111, …, how many of the first 100 terms are divisible by 13?
- If a repunit 11…11 (all digits 1) is divisible by 7, prove that it is also divisible by 11, 13, and 15,873.
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In the sequence 1, 11, 111, 1111, …:
- Prove there exist two terms whose difference is divisible by 196,673.
- Deduce that there exists a repunit divisible by 196,673.
- For any natural number a not divisible by 2 or 5, prove there exists a natural b such that the product ab is a repunit (written using only the digit 1).
- If natural numbers a and m are coprime, prove there exists n such that an − 1 is divisible by m.
C. Cyclic digit moves and special base‑10 constructions
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A k-digit multiple of 13 has its first digit moved to the end. For which k is the resulting number still a multiple of 13?
Example: 503,906 → 39,065 remains divisible by 13; 7,969 → 9,697 does not.
- Find a six‑digit decimal number that becomes 5 times smaller when its first digit is moved to the end of the number.
- The last digit of a (decimal) number is 2. If this digit is moved to the front, the number doubles. Find the smallest such number.
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A five‑digit number is divisible by 41. Prove that any cyclic permutation of its digits is also divisible by 41.
Example: knowing that 93,767 is divisible by 41, conclude that 37,679 is divisible by 41.
D. Base‑16 (hexadecimal) curiosities
- Does there exist a hexadecimal number that, when multiplied by 2, 3, 4, 5, and 6, yields numbers that are permutations of its digits (in base 16)?
- Find a hexadecimal number that increases by an integer factor when its last digit is moved to the front (both operations understood in base 16).
E. Last digits and modular arithmetic
- Determine the last digit of 3377 + 7733.
- Find the last two digits of 22000.
- What is the last digit of 9,999,999,999?
- Find the last digit of 9999(99999999).