If (1235)x = (3033)y, where x and y indicate the bases of the corresponding numbers, then :
x = 8 and y = 6
Convert both sides to decimal for each option and look for the pair that agrees — but two quick observations remove most of the work first.
Observation 1 — the digits constrain the bases. Every digit must be less than its base. The left-hand number contains a 5, so \(x\gt5\); the right-hand number contains a 3, so \(y\gt3\). That alone eliminates option 1, where x = 5 and y = 3 are both illegal.
Observation 2 — test the remaining options.
\((1235)_{x}=x^{3}+2x^{2}+3x+5\)
\((3033)_{y}=3y^{3}+0y^{2}+3y+3\)
| Option | (1235)x | (3033)y | Equal? |
|---|---|---|---|
| 2: x=7, y=5 | 343+98+21+5 = 467 | 375+15+3 = 393 | ✗ |
| 3: x=8, y=6 | 512+128+24+5 = 669 | 648+18+3 = 669 | ✓ |
| 4: x=9, y=7 | 729+162+27+5 = 923 | 1029+21+3 = 1053 | ✗ |
So x = 8 and y = 6, both sides representing the decimal number 669 — option 3.
Working option 3 out in full, since it is the answer:
\((1235)_{8}=1\times512+2\times64+3\times8+5\times1=512+128+24+5=669\)
\((3033)_{6}=3\times216+0\times36+3\times6+3\times1=648+0+18+3=669\)
A useful shortcut for problems of this kind. The two numbers have the same digit count, and the left one begins with 1 while the right begins with 3 — so the right-hand base must be appreciably smaller, since three of its highest power equals roughly one of the other's. Indeed \(3y^{3}\approx x^{3}\) gives \(x\approx1.44y\), and 8 against 6 fits that ratio far better than 7 against 5 or 9 against 7.
The underlying principle is positional notation: a digit's contribution is the digit multiplied by the base raised to its position index, counting from zero at the right. The same number has many representations, but its value is invariant — which is exactly what the question is testing.
Hence, x = 8 and y = 6.
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