Only 1's and 0's counts

Number Theory Level 4

Find the sum of all 16-bit numbers $> 1$ , whose representation in the octal and hexadecimal systems contain only 1's and 0's.

Example: [ $2018$ has no 1's or 0's in its representations in both octal and hexadecimal.] $\text {in base} \ \ \ 8 , \ \ 2018_{10} = 3742_8 \\ \text {in base} \ \ 16 , \ \ 2018_{10} = 7E2_{16}$

The answer is 8193.

1 solution

Arjen Vreugdenhil
Jan 1, 2018

If a number contains only 0s and 1s in the octal system, all bits must be zero except possibly bits 0, 3, 6, 9, etc.

If a number contains only 0s and 1s in the hexadecimal system, all bits must be zero except possibly bits 0, 4, 8, 12 etc.

If both are the case, the only non-zero bits may be bits 0, 12, 24, ...

Since we are talking 16-bit numbers here, we must have all bits zero except possibly bits 0 and 12. This allows for four integers: $0000\,0000\,0000\,0000_2 = 0000_{16} = 00000_8 = 0_{10}; \\ 0000\,0000\,0000\,0001_2 = 0001_{16} = 00001_8 = 1_{10}; \\ 0001\,0000\,0000\,0000_2 = 1000_{16} = 10000_8 = 4096_{10}; \\ 0001\,0000\,0000\,0001_2 = 1001_{16} = 10001_8 = 4097_{10}.$

The question only asks for integers greater than one, we add $4096 + 4097 = \boxed{8193}$ .

Only 1's and 0's counts

The answer is 8193.

1 solution

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