Give a simple rule to multiply a given number by 5 in the base-5 system that we created.

Question

Homi · Accepted Answer

Multiplying a number by its base in any number system shifts its digits one place to the left and adds a zero.

Step 1 — Understanding Place Values in Base 5

In our base-5 number system, we use five digits: 0, 1, 2, 3, 4. Each position in a number represents a power of 5. The rightmost position is for $5^0$ , which is 1. The next position to the left is for $5^1$ , which is 5. The next position is for $5^2$ , which is 25. And so on, for higher powers of 5.

Step 2 — Identifying Landmark Symbols

The problem uses special "landmark symbols" for these place values. We are told that 5 is represented by Square. So, we can match the symbols to their place values: The $5^0$ place (value 1) is represented by Triangle. The $5^1$ place (value 5) is represented by Square. The $5^2$ place (value 25) is represented by Hexagon. The $5^3$ place (value 125) is represented by Circle. These symbols represent the "landmark numbers" in our system.

Step 3 — Property of Landmark Numbers

Let us see what happens when we multiply these landmark numbers by 5 (Square). Multiplying Triangle (1) by Square (5) gives Square (5). $1 \times 5 = 5$ Multiplying Square (5) by Square (5) gives Hexagon (25). $5 \times 5 = 25$ Multiplying Hexagon (25) by Square (5) gives Circle (125). $25 \times 5 = 125$ This shows that the product of a landmark number with another landmark number (Square) results in a higher landmark number.

$\boxed{\text{The Product of a landmark number with another landmark number gives a landmark number.}}$

Step 4 — Deriving the Multiplication Rule

Consider any number in our base-5 system. Let it be $d_2 \text{ Hexagon } d_1 \text{ Square } d_0 \text{ Triangle}$ . This number means $d_2 \times 25 + d_1 \times 5 + d_0 \times 1$ . Now, let us multiply this entire number by 5 (which is Square). $(d_2 \times 25 + d_1 \times 5 + d_0 \times 1) \times 5$ We distribute the multiplication: $d_2 \times (25 \times 5) + d_1 \times (5 \times 5) + d_0 \times (1 \times 5)$ Using the property from Step 3: $d_2 \times 125 + d_1 \times 25 + d_0 \times 5$ In terms of our landmark symbols, this becomes: $d_2 \text{ Circle } + d_1 \text{ Hexagon } + d_0 \text{ Square } + 0 \text{ Triangle}$ Notice how each digit has moved. The digit $d_0$ (originally with Triangle) is now with Square. The digit $d_1$ (originally with Square) is now with Hexagon. The digit $d_2$ (originally with Hexagon) is now with Circle. The Triangle place (units place) now has a zero. This means each digit's associated symbol has shifted to the next higher landmark symbol.

Answer

Rule: The Product of a landmark number with another landmark number gives a landmark number. To multiply by 5 (which is represented by Square), we shift each symbol to the next higher landmark symbol (e.g., Triangle becomes Square, Square becomes Hexagon, Hexagon becomes Circle, etc.).

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