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# How many multiples of $3$ are in between $201$ and $427$?(a) $54$(b) $65$(c) $76$(d) $82$ Verified
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Hint: The given problem can be solved by using the concept of arithmetic progression (sequence of numbers in order such that the difference between the consecutive terms is constant). since $201$ is the first multiple of $3$ in the range, and $426$ is the last multiple of 3 in the range. Then the arithmetic progression can be written as: $201,204,207...,423,426.$ It can be solved by using the formula ${T_n} = a + (n - 1)d$ in which the last term is $426$. We have to find the value of n that will give the number of multiples of $3$ from $201$ through $427$.

Complete step-by-step solution:
To find the solution of the given problem we can use the formula of arithmetic progression which is given by
${\text{Last term = first term + }}\left( {{\text{n - 1}}} \right){\times{ \text common difference}}$
It can be written as ${T_n} = a + (n - 1)d$
By taking the first term $a = 201$
To calculate the common difference $d = 204 - 201 = 3$
${T_n}$ is the last term in the progression and is given =$426$
On substitution we get
$426 = 201 + (n - 1) \times 3$
$\Rightarrow 426 - 201 = (n - 1) \times 3$
$\Rightarrow \dfrac{{426 - 201}}{3} = (n - 1)$
$\Rightarrow \dfrac{{225}}{3} = (n - 1)$
$\Rightarrow \dfrac{{225}}{3} + 1 = n$
$\Rightarrow \dfrac{{225 + 3}}{3} = n$
$\Rightarrow \dfrac{{228}}{3} = n$
$\Rightarrow n = 76$
Therefore, there are $76$ multiples of $3$ from $201$ through $427$.

Note: In the above problem $427$ is not a multiple of $3$. The easy way to check this is to find the sum of its digits $(4 + 2 + 7 = 13)$. Since $13$ is not divisible by $3$, the number whose sum of digits is $13$ is also not divisible by $3$. We therefore must take the largest number in the range which is divisible by, which is $426$. Its sum of digits $(4 + 2 + 6 = 12)$ which is divisible by $3$.