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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?

enter image description here

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

  • This question does not appear to be about electronics design within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

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    \$\begingroup\$ 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 \$? \$\endgroup\$ – Transistor Feb 10 at 9:12
  • \$\begingroup\$ 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 \$\endgroup\$ – noorav Feb 10 at 9:14
  • \$\begingroup\$ 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. \$\endgroup\$ – jonk Feb 10 at 9:19
  • \$\begingroup\$ Good work, @jonk. It looks like the ≥ is an error. \$\endgroup\$ – Transistor Feb 10 at 9:54
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    \$\begingroup\$ I'm voting to close this question as off-topic because its about maths not electronics try [Mathematics Stack Exchange](math.stackexchange.com) \$\endgroup\$ – Warren Hill Feb 11 at 12:12
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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.

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