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Number Theory Level 2

If both 11^2 and 3^3 are factors of the number a * 4^3 * 6^2 * 13^11, then what is the smallest possible value of a?


The answer is 363.

Anand Shah by Anand Shah , 22, India

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2 solutions

Of all the numbers with their exponents being multiplied, any one of them or all of them together had to be divisible by 11^2 and 3^3. Of the numbers, 6^2 satisfies 3^2. So what is left over is 11^2 and 3. Thus, a=11^2*3=363

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363

Srinivas Nani - 7 years, 2 months ago

a × 4 3 × 6 2 × 1 3 11 1 1 2 × 3 3 = a × 4 3 × 3 2 × 1 3 11 1 1 2 × 3 \frac{a \times 4^3 \times 6^2 \times 13^{11}}{11^2\times3^3} = \frac{a \times 4^3 \times 3^2 \times 13^{11}}{11^2\times3}

as 11 and 3 is coprime to {4,3,13} so a = 1 1 2 × 3 a = 11^2\times 3

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a = ((11^2) * 3) and not ((11^3)*3)

Sahil Gohan - 7 years, 2 months ago

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