In the land of Binary, the unit of currency is called Ben and currency notes are available in denominations $$1, 2, 2^{2}, 2^{3}...$$ Bens. The rules of the Government of Binary stipulate that one can not use more than two notes of any one denomination in any transaction. For example, one can give a change for $$2$$ Bens in two ways: $$2$$ one Ben notes or $$1$$ two Ben note. For $$5$$ Ben one can give $$1$$ one Ben note and $$1$$ four Ben note or $$1$$ one Ben note and $$2$$ two Ben notes. Using $$5$$ one Ben notes or $$3$$ one Ben notes and $$1$$ two Ben notes for a $$5$$ Ben transaction is prohibited. Find the number of ways in which one can give change for $$100$$ Bens, following the rules of the Government.

Let the number of notes of each denomination be

$$\begin{aligned} a &\;:\;64 \text{-Ben notes}\\[-2pt] b &\;:\;32 \text{-Ben notes}\\[-2pt] c &\;:\;16 \text{-Ben notes}\\[-2pt] d &\;:\; 8 \text{-Ben notes}\\[-2pt] e &\;:\; 4 \text{-Ben notes}\\[-2pt] f &\;:\; 2 \text{-Ben notes}\\[-2pt] g &\;:\; 1 \text{-Ben notes} \end{aligned}$$

Government rule: each of $$a,b,c,d,e,f,g$$ can take the values $$0,1,2$$, except $$a$$ which can be only $$0$$ or $$1$$ (because $$2\times64=128\gt100$$).

All possible combinations satisfy the Diophantine equation

$$64a + 32b + 16c + 8d + 4e + 2f + g = 100 \qquad -(1)$$

We count the integral solutions of (1) under the above limits by systematic case-work.

Equation (1) becomes

$$32b + 16c + 8d + 4e + 2f + g = 36 \qquad -(2)$$

Here $$b\in\{0,1\}$$ (two 32-Ben notes would overshoot the remaining 36).

Then (2) reduces to

$$16c + 8d + 4e + 2f + g = 4 \qquad -(3)$$

The left side now involves only denominations $$\le 4$$, whose total value is at most $$2\times4 + 2\times2 + 2\times1 = 14$$, so every variable can still be $$0,1,2$$.

Because $$4$$ itself is a denomination, list the possibilities for (3):

• one 4-Ben note : $$(c,d,e,f,g)=(0,0,1,0,0)$$
• two 2-Ben notes : $$(0,0,0,2,0)$$
• one 2-Ben and two 1-Ben notes : $$(0,0,0,1,2)$$

Hence, Case 1.1 contributes $$3$$ valid combinations.

Equation (2) now reads

$$16c + 8d + 4e + 2f + g = 36 \qquad -(4)$$

Choose $$c \in \{0,1,2\}$$.

Sub-case 1.2.1: $$c = 2$$  ⇒ $$8d + 4e + 2f + g = 4$$ Exactly the same equation as (3), therefore contributes $$3$$ more solutions.

Sub-case 1.2.2: $$c = 1$$  ⇒ $$8d + 4e + 2f + g = 20$$ Take $$d \in \{0,1,2\}$$.

• $$d = 2$$ ⇒ $$4e + 2f + g = 4$$ ⇒ again yields the same 3 solutions.
• $$d = 1$$ ⇒ $$4e + 2f + g = 12$$.
  Pick $$e = 2$$ ⇒ $$2f + g = 4$$ gives     - $$(f,g)=(2,0)$$, $$(1,2)$$  ⇒ 2 solutions.
• $$d = 0$$ is impossible because the maximum from lower denominations is 14.

So Sub-case 1.2.2 adds $$3 + 2 = 5$$ solutions.

Sub-case 1.2.3: $$c = 0$$ The maximum attainable value with the remaining denominations is $$14 \lt 36$$, so no solution.

Adding the contributions of Case 1:

$$3\;(\text{Case 1.1}) + 3 + 5\;(\text{Case 1.2}) = 11$$ solutions.

Equation (1) becomes

$$32b + 16c + 8d + 4e + 2f + g = 100 \qquad -(5)$$

Now $$b \in \{0,1,2\}$$.

Case 2.1: $$b = 2$$  ⇒ remaining amount $$= 36$$.

Exactly the same equation (4) analysed above, and we saw it has $$8$$ solutions (3 when $$c=2$$ and 5 when $$c=1$$). Hence Case 2.1 contributes $$8$$ solutions.

Case 2.2: $$b = 1$$  ⇒ remaining amount $$= 68$$. The maximum possible with $$16,8,4,2,1$$ is $$62 \lt 68$$, so no solution.

Case 2.3: $$b = 0$$  ⇒ need full 100 with denominations $$\le16$$. Again impossible because their maximum total is $$62$$.

Therefore Case 2 contributes only $$8$$ solutions.

Adding both main cases:

$$\boxed{11 + 8 = 19}$$

Hence, the change for 100 Bens can be given in exactly 19 different ways following the Binary Government’s rules.