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Quantitative Aptitude

Fermat Theorem

There exists ...

Question

There exists positive integers $a, b$ and $c$ with no common factor greater than $2$ such that $a {log}_{200} 5 + b {log}_{200} 2 = c$ , then the value of $\frac{a + b + c}{3 c}$ can be

1

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Solution

The correct option is B $2$
$a l o g_{200} 5 + b l o g_{200} 2 = c$
$l o g_{200} 5^{a} + b l o g_{200} 2^{b} = c$
$l o g_{200} 5^{a} {.2}^{b} = c$
$(200)^{c} = 5^{a} {.2}^{b} \Rightarrow$
$5^{2 c} 2^{3 c} = 5^{a} {.2}^{b} \Rightarrow 2 c = a$ $3 c = b$
$\Rightarrow \frac{2 c + 3 c + c}{3} = 2 c$

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There exists positive integers a,b and c with no common factor greater than 2 such that alog2005+blog2002=c, then the value of a+b+c3c can be

The correct option is B 2alog2005+blog2002=clog2005a+blog2002b=clog2005a.2b=c(200)c=5a.2b⇒52c23c=5a.2b⇒2c=a 3c=b⇒2c+3c+c3=2c

There exists positive integers $a, b$ and $c$ with no common factor greater than $2$ such that $a {log}_{200} 5 + b {log}_{200} 2 = c$ , then the value of $\frac{a + b + c}{3 c}$ can be

The correct option is B $2$
$a l o g_{200} 5 + b l o g_{200} 2 = c$
$l o g_{200} 5^{a} + b l o g_{200} 2^{b} = c$
$l o g_{200} 5^{a} {.2}^{b} = c$
$(200)^{c} = 5^{a} {.2}^{b} \Rightarrow$
$5^{2 c} 2^{3 c} = 5^{a} {.2}^{b} \Rightarrow 2 c = a$ $3 c = b$
$\Rightarrow \frac{2 c + 3 c + c}{3} = 2 c$