# Number systems and binary [closed]

I am new to number systems and binary.

• Can someone tell me why is X>=11 and Y>=6 in this problem?
• Can someone also explain how the values X=12 and Y=16 are obtained from the conditions shown in the answer?

## closed as off-topic by Andy aka, Warren Hill, Sparky256, Finbarr, Sean HoulihaneFeb 14 at 11:36

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• Welcome to EE.SE but (1) Why are you asking a maths problem here? (2) What is that number notation system where $(49)_{(x-1)} = 4(x-1)+9$? – Transistor Feb 10 at 9:12
• I'm sorry, this comes under my electrical and electronics syllabus so I thought of asking it here. To answer your second part, 49 base x-1 is 4(x-1) +9 when being converted to decimal. So X would be greater than 11 so that X-1 is 10. But why must y>=6 – noorav Feb 10 at 9:14
• I think the subscript means the "base" of the number. So, $49_{x-1}=4\times\left(x-1\right)^1+9\times\left(x-1\right)^0$ and also $35_y=3\times y^1+5\times y^0$. It follows then that $4\times\left(x-1\right)+9=3\times y+5$. It follows then that $4\times\left(x-1\right)+9=3\times y+5$. So $4 x=3 y$. I'm not sure, though, where the $\ge$ comes from. – jonk Feb 10 at 9:19
• Good work, @jonk. It looks like the ≥ is an error. – Transistor Feb 10 at 9:54
• I'm voting to close this question as off-topic because its about maths not electronics try [Mathematics Stack Exchange](math.stackexchange.com) – Warren Hill Feb 11 at 12:12

This feels kinda mathy for EE, but here goes.

I hope the reductions make sense. They rewrite a two digit number like '35 base y' as $$\ 3y^1+5y^0 \$$. This generalizes as $$\ digitn \times y^n+... \$$.

There's an error in the last line of math for the solution when they convert the relation from 4x = 3y to x>=3y/4. Those statements aren't equivalent.

We see a digit "9" in the left-hand number which means it's at least in base 10, and given we know it's in base x-1, then x >= 11.

By the same logic: We see a digit "5" in the right-hand number which means it's at least in base 6, and given we know it's in base y, then y >= 6.

We also know that x and y are both integers, so x is divisible by 3 so y is an integer, and y is divisible by 4 so x is an integer.

We can then check y = 8 and solve for x finding that if y = 8 then x = 6 which isn't high enough, so instead we pick the smallest value for x that is greater than 11 and divisible by 3. Then we solve for y.