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What is the sum of the digits of the number of integers multiples of 3 between 1 and 200?​

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GREATNICLAUSIN

✒️[tex]\large{\mathcal{ANSWER}}[/tex]

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The sum of the digits of the number of whole numbers between 1 and 200 that are multiples of 3 is 12.

  • The multiples of 3 are: 3, 6, 9, 12, 15...

This sequence forms an arithmetic progression (AP) of ratio 3, and first term 3 and last term 198 (since we are considering a discrete interval).

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Thus, we have that the number of numbers multiples of 3 between 1 and 200 is 66.

[tex] a_{n} = a_{1} + (n - 1)r[/tex]

[tex]198 = 3 + (n - 1)3[/tex]

[tex] 3_{n} = 198[/tex]

[tex]n = \frac{198}{3} = 66[/tex]

The sum of the digits for this number is 12:

  • [tex]6 + 6 = 12[/tex]

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KANTOINEDOIXIN

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[tex] \large\underline{\mathbb{PROBLEM}:} [/tex]

  • What is the sum of the digits of the number of integers multiples of 3 between 1 and 200?

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[tex] \large\underline{\mathbb{ANSWER}:} [/tex]

[tex] \qquad \Large \:\rm{The \: sum \: is \: 12} [/tex]

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[tex] \large\underline{\mathbb{SOLUTION}:} [/tex]

» Find the first and the last numbers that are divisible by 3 between 1 and 200 which are 3 and 198.

  • [tex] a_1 = 3 \:\: and \:\: a_n = 198 [/tex]

» Find the number of terms using the arithmetic sequence formula. since it's divisible by 3, then the common difference will be 3.

[tex] \begin{align} & \bold{Formula:} \\ & \quad \boxed{\rm a_n = a_1 + d(n-1)} \end{align} [/tex]

  • [tex] 198 = 3 + 3(n-1) [/tex]

  • [tex] 198 = 3 + 3n-3 [/tex]

  • [tex] 198 = 3n [/tex]

  • [tex] \frac{\,198\,}{3} = \frac{\,3n\,}{3}\\ [/tex]

  • [tex] n = 66 [/tex]

[tex] \therefore [/tex] There are 66 numbers that are divisible by 3 between 1 and 200. Now find the sum of the digits of 66.

  • [tex] 6 + 6 = 12 [/tex]

[tex] \therefore [/tex] The sum is 12.

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