Number Theory · step 10 of 13

Base-b representations & digit arguments

Every positive integer has a unique representation in base b as Σdᵢbⁱ with digits 0 ≤ dᵢ < b. Conversion is repeated division by b, reading remainders bottom-up. The digit-sum divisibility rules follow immediately from congruences: 10 ≡ 1 (mod 9) makes every power of 10 congruent to 1, so a number is congruent to its digit sum mod 9.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

Convert by repeated divisionDivide by b, record the remainder, repeat on the quotient. Read the remainders in reverse.
Convert back by positional expansionMultiply each digit by its power of b and add, or use Horner’s method for speed.
Derive divisibility rules from b mod nSince 10 ≡ 1 (mod 9), Σdᵢ10ⁱ ≡ Σdᵢ (mod 9). Since 10 ≡ −1 (mod 11), the ALTERNATING digit sum governs divisibility by 11.
Count digits with logarithmsA number n has ⌊log_b n⌋ + 1 digits in base b.

Worked example

Convert 45 to binary, and use digit sums to test whether 12345 is divisible by 9.

45 ÷ 2 = 22 r 1; 22 ÷ 2 = 11 r 0; 11 ÷ 2 = 5 r 1; 5 ÷ 2 = 2 r 1; 2 ÷ 2 = 1 r 0; 1 ÷ 2 = 0 r 1.
Reading remainders bottom-up: 101101₂. Check: 32 + 8 + 4 + 1 = 45 ✓.
Digit sum of 12345: 1+2+3+4+5 = 15.
15 is not divisible by 9 (though it is by 3).

Answer. 45 = 101101₂; and 12345 is divisible by 3 but not by 9, since its digit sum is 15.

Where marks get dropped

These are the specific errors that cost credit on base-b representations & digit arguments questions — QED's rubric penalises each of them separately.

Reading the remainders top-down. The FIRST remainder is the least significant digit, so the list must be reversed.
Using a digit ≥ b. In base 8 the digits run 0–7 only.
Applying the digit-sum rule to divisibility by 7 or 11. Nine and three work because 10 ≡ 1 (mod 9); 11 needs the alternating sum.

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Base-b representations & digit arguments — frequently asked questions

Why does the digit sum test divisibility by 9?

Because 10 ≡ 1 (mod 9), so 10ⁱ ≡ 1 for all i, making n ≡ Σdᵢ (mod 9). The same argument gives the rule for 3.

What is the rule for 11?

Alternate the signs of the digits from the right, since 10 ≡ −1 (mod 11). If the alternating sum is divisible by 11, so is the number.

How do I convert between binary and hexadecimal?

Group binary digits in fours from the right; each group is one hex digit. No arithmetic is needed because 16 = 2⁴.

The rest of Number Theory

Divisibility, primes, gcd, modular arithmetic. Each subtopic below has its own method, worked example and mark-losing traps.

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