AMC 10 Step by Step

2016 AMC 10A Problem 2

P1-10~1 minPrint

For what value xx does 10x⋅1002x=1000510^{x}\cdot 100^{2x}=1000^{5} ?

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Key insight

Write 100 and 1000 as powers of 10. The left side becomes 10 to the power 5x and the right side 10 to the power 15, so 5x = 15.

Solution

The numbers 1010, 100100 and 10001000 are all powers of 1010, so rewrite everything with base 1010.

  1. 100=102100 = 10^2, so 1002x=(102)2x100^{2x} = (10^2)^{2x}. A power of a power multiplies the exponents, so this is 104x10^{4x}.
  2. 1000=1031000 = 10^3, so 10005=(103)5=10151000^5 = (10^3)^5 = 10^{15}.
  3. The left side is now 10x⋅104x10^x \cdot 10^{4x}. Multiplying powers of the same base adds the exponents, so it is 105x10^{5x}.
  4. The equation reads 105x=101510^{5x} = 10^{15}. Two powers of 1010 are equal only when their exponents are equal, so 5x=155x = 15 and x=3x = 3.

Check: 103⋅1006=103⋅1012=101510^3 \cdot 100^6 = 10^3 \cdot 10^{12} = 10^{15}, and 10005=10151000^5 = 10^{15} too.

The answer is (C) 3\boxed{\textbf{(C)}\ 3}.

Why this works

When every number in the equation is a power of the same base, rewrite them all with that base; then you only need to compare exponents. Keep the two rules apart: a power raised to a power multiplies the exponents, and two powers multiplied together add them.

Alternative approach

Try the answer choices. With x=3x = 3 the left side is 103⋅1006=1000⋅1012=101510^3 \cdot 100^6 = 1000 \cdot 10^{12} = 10^{15}, which equals 100051000^5. It works on the first try.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Express everything with base 1010: 100=102100 = 10^2 and 1000=1031000 = 10^3, so
1002x=(102)2x=104x,10005=(103)5=1015.100^{2x} = (10^2)^{2x} = 10^{4x}, \qquad 1000^5 = (10^3)^5 = 10^{15}.

The left side becomes 10x⋅104x=105x10^x \cdot 10^{4x} = 10^{5x}. Setting 105x=101510^{5x} = 10^{15} gives 5x=155x = 15, so x=3x = 3.

The answer is (C) 3\boxed{\textbf{(C)}\ 3}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.