Computer Number Systems

ACSL tests your ability to convert between bases and do arithmetic in non-decimal systems. You need to be fast - there’s no time for fumbling through long division on contest day. This lesson gives you the shortcuts that actually matter.

The four bases you need

Base	Name	Digits used
$2$	Binary	$0$ , $1$
$8$	Octal	$0$ - $7$
$10$	Decimal	$0$ - $9$
$16$	Hex	$0$ - $9$ , $A$ - $F$

Memorize these hex values - you’ll use them constantly:

Hex	Decimal	Binary
$A$	$10$	$1010$
$B$	$11$	$1011$
$C$	$12$	$1100$
$D$	$13$	$1101$
$E$	$14$	$1110$
$F$	$15$	$1111$

Powers you must know cold

$2^{n}$	Value	$1 6^{n}$	Value
$2^{0}$	$1$	$1 6^{0}$	$1$
$2^{1}$	$2$	$1 6^{1}$	$16$
$2^{2}$	$4$	$1 6^{2}$	$256$
$2^{3}$	$8$	$1 6^{3}$	$4096$
$2^{4}$	$16$	$1 6^{4}$	$65536$
$2^{5}$	$32$
$2^{6}$	$64$
$2^{7}$	$128$
$2^{8}$	$256$
$2^{9}$	$512$
$2^{10}$	$1024$
$2^{11}$	$2048$
$2^{12}$	$4096$

The fast shortcut: binary as the bridge

This is the single most important trick for ACSL number systems. Never convert between octal and hex through decimal. Use binary as a bridge instead.

Each octal digit = exactly $3$ binary digits
Each hex digit = exactly $4$ binary digits

Example: Convert $3AF_{16}$ to octal.

Step 1 - Expand each hex digit to $4$ bits:

3    A    F
0011 1010 1111

Step 2 - Regroup into chunks of $3$ from the right:

001 110 101 111
 1   6   5   7

Answer: $165 7_{8}$ . No decimal math needed.

Example: Convert $573 0_{8}$ to hex.

Step 1 - Expand each octal digit to $3$ bits:

5   7   3   0
101 111 011 000

Step 2 - Regroup into chunks of $4$ from the right:

1011 1101 1000
 B    D    8

Answer: $BD8_{16}$ .

Decimal to other bases

The repeated division method - divide by the target base, read remainders bottom-to-top.

Example: Convert $2024$ to octal.

2024 / 8 = 253 remainder 0
 253 / 8 =  31 remainder 5
  31 / 8 =   3 remainder 7
   3 / 8 =   0 remainder 3

Reading bottom-to-top: $375 0_{8}$ .

Try it: trace the repeated division method

Code

1Convert 2024 (decimal) to octal:232024 / 8 = 253  remainder 04 253 / 8 =  31  remainder 55  31 / 8 =   3  remainder 76   3 / 8 =   0  remainder 378Read remainders bottom-to-top: 3750

Variables

Name	Value
n	2024
base	8

Output

Divide by 8 repeatedly

Step 1 of 6

Speed tip: For hex, divide by $16$ . Remember that remainders $10$ - $15$ become $A$ - $F$ .

Other bases to decimal

Multiply each digit by its place value and sum.

Example: Convert $7BE_{16}$ to decimal.

7 \times 256 B \times 16 E \times 1 Total = 1792 = 176 (B = 11) = 14 (E = 14) = 1982

Arithmetic in other bases

The key insight: it works exactly like decimal arithmetic, but you carry/borrow at the base value instead of $10$ .

Hex addition

Example: $9876_{16} + ABC_{16}$

Work right to left:

  9876
+  ABC
------

$6 + C = 6 + 12 = 18 = 16 + 2$ , so write $2$ , carry $1$
$7 + B + 1 = 7 + 11 + 1 = 19 = 16 + 3$ , so write $3$ , carry $1$
$8 + A + 1 = 8 + 10 + 1 = 19 = 16 + 3$ , so write $3$ , carry $1$
$9 + 0 + 1 = 10 = A$ , so write $A$

Answer: $A332_{16}$

Hex subtraction

Example: $F5AD_{16} - 69EB_{16}$

Work right to left, borrowing when needed:

$D - B = 13 - 11 = 2$
$A - E$ : can’t ( $10 < 14$ ), so borrow from next column: $(16 + 10) - 14 = 12 = C$ . The $5$ becomes $4$ .
$4 - 9$ : can’t ( $4 < 9$ ), so borrow from next column: $(16 + 4) - 9 = 11 = B$ . The $F$ becomes $E$ .
$E - 6 = 14 - 6 = 8$

Answer: $8BC2_{16}$

Speed tip for subtraction: If you make an error borrowing, check by adding your answer to the smaller number. You should get the larger number back.

Octal multiplication

Example: $3 7_{8} \times 5_{8}$

Work like decimal multiplication, but carry at 8 instead of 10:

$7 \times 5 = 35 = 4 \times 8 + 3$ , so write $3$ , carry $4$
$3 \times 5 = 15$ , plus carry $4 = 19 = 2 \times 8 + 3$ , so write $3$ , carry $2$
Write the final carry: $2$

Answer: $23 3_{8}$

Verify: $3 7_{8} = 3 1_{10}$ , and $31 \times 5 = 155$ . Converting $15 5_{10}$ : $155/8 = 19$ r $3$ , $19/8 = 2$ r $3$ , $2/8 = 0$ r $2$ . That gives $23 3_{8}$ . Correct.

Note: ACSL does not test division in other bases.

Common ACSL traps

Forgetting leading zeros when regrouping binary. $101$ in binary needs to be padded to $0101$ when converting to hex.
Confusing octal $8$ and $9$ . These digits don’t exist in octal. If your answer has an $8$ or $9$ , you forgot to carry.
Going through decimal for octal-to-hex. This wastes time and invites errors. Always use the binary bridge.
Mixing up the direction of remainders. Remainders are read bottom-to-top (last remainder is the most significant digit).

Contest strategy

Binary-octal-hex conversions through binary: ~15 seconds each
Decimal conversions: ~30 seconds each
Base arithmetic: ~45 seconds each, verify with a quick sanity check
If a problem asks you to convert between octal/hex, and one answer choice has an $8$ or $9$ in octal, eliminate it immediately