Calculating memory size from address and word size?

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I'm currently studying for my computer architecture exam and I found this question online

A digital computer has a memory unit with 26 bits per word. The instruction set consists of 756 different operations. All instructions have an operation code part (opcode) and an address part (allowing for only one address). Each instruction is stored in one word of memory.

1-How many bits are needed for the opcode? 10 bits

2- How many bits are left for the address part of the instruction? 16 bits

3- What is the maximum allowable size for memory? 65536 bits

4- What is the largest unsigned binary number that can be accommodated in one word of memory? 67108864

I understand why 1 and 2 are correct but I didn't understand 3 and 4

Shouldn't question 3 be (2^16) x 26 bits?

And shouldn't question 4 be (2^26)- 1?

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computer-sciencecpu-architecture
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1 Answers

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For part 4, you're calculating the largest immediate that can be part of an instruction word. But the question asked about using the whole 26-bit word for a number, as data. (Also, you did 2^15, so the largest signed 16-bit integer, but the question asked for unsigned where all n bits have positive place-value).

For part 3, yes your answer looks correct. Assuming there is no indirect addressing (which would let you use 26-bit addresses), then the usable memory size = the amount an instruction can address. So 2^16 = 65536 words. Are you sure the given answer really meant 65536 bits, not words?