Question

How many multiples of \[3\] are in between \[201\] and \[427\]?
(a) \[54\]
(b) \[65\]
(c) \[76\]
(d) \[82\]

Answer 1

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 \[3\].