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If m and n are positive integers, is 10^n+m divisible by 3?

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If m and n are positive integers, is 10^n+m divisible by 3?  [#permalink]

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New post 31 Dec 2016, 00:37
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If m and n are positive integers, is \(10^n+m\) divisible by 3?

1) n=1
2) m=2

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If m and n are positive integers, is 10^n+m divisible by 3?  [#permalink]

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New post 31 Dec 2016, 01:00
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MathRevolution wrote:
If m and n are positive integers, is \(10^n+m\) divisible by 3?

1) n=1
2) m=2


Statement 1: n= 1 and m =1, remainder = 2; n=1 and m=2, remainder = 0, no unique answer.

Statement2: Given m = 2, for any value of n \(10^n + m\) will be divisible by 3. Hence, remainder = 0, unique answer. Answer: (b)

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Re: If m and n are positive integers, is 10^n+m divisible by 3?  [#permalink]

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New post 01 Jan 2017, 17:40
==> If you modify the original condition and the question, “Is \(10^n\)+m=3t? (t=any positive integer)” is equal to “Is the sum of all the digits of \(10^n\)+m divisible by 3?” Then, as from con 2), if you know m=2, from \(10^n\)+2 => 12, 102, 1002…., the sum of the digits always become 3, which is divisible by 3, and hence yes, it is sufficient.

Therefore, the answer is B.
Answer: B
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Re: If m and n are positive integers, is 10^n+m divisible by 3?  [#permalink]

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New post 30 Aug 2018, 09:04
MathRevolution wrote:
If m and n are positive integers, is \(10^n+m\) divisible by 3?

1) n=1
2) m=2


\(10^n\) will invariably always lead a remainder 1 upon division by 3 (try dividing 10, 100 for example). Therefore value of n(+ve) has no bearing on the divisibility of \(10^n+m\) by 3

\(10^n+m\)- is only divisible by 3 if m has a value equal to 2 or m leaves a remainder of 2 upon division by 3, since (1 + 2) is divisible by 3

Therefore option B is sufficient to know. Our winner option B
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If m and n are positive integers, is 10^n+m divisible by 3?  [#permalink]

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New post 31 Aug 2018, 10:11
MathRevolution wrote:
If m and n are positive integers, is \(10^n+m\) divisible by 3?

1) n=1
2) m=2

\(m,n\,\,\, \geqslant 1\,\,\,{\text{ints}}\)

\(\frac{{{\text{1}}{{\text{0}}^n} + m}}{3}\,\,\,\mathop = \limits^? \,\,\operatorname{int}\)

To prove (1) is insufficient, let us present what our method calls a BIFURCATION (ALGEBRAIC, in this case):

\(\left( 1 \right)\,\,\,n = 1\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,m = 1\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\, \hfill \\
{\text{Take}}\,\,m = 2\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\
\end{gathered} \right.\)
\(\left( 2 \right)\,\,\,\left\{ \begin{gathered}
m = 2 \hfill \\
{10^{n\,\,\,}} = \left( {{{10}^n} - 1} \right) + 1 \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\frac{{{\text{1}}{{\text{0}}^{n\,}} + m}}{3}\,\,\, = \,\,\frac{{\left( {{{10}^n} - 1} \right) + 1 + 2}}{3}\,\,\,\,\mathop { = \,}\limits^{n\,\, \geqslant \,\,1} \,\,\,\,\frac{{\overbrace {\left( {{\text{1}}{{\text{0}}^n} - 1} \right)}^{\left\langle {9 \ldots 9} \right\rangle }}}{3} + 1\,\,\,\, = \,\,\,\,\left\langle {3 \ldots 3} \right\rangle + 1\,\,\,\, = \,\,\,\,\operatorname{int} \,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle\)


The above follows the notations and rationale taught in the GMATH method.
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Re: If m and n are positive integers, is 10^n+m divisible by 3?  [#permalink]

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New post 04 Sep 2018, 17:07
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If m and n are positive integers, is \(10^n+m\) divisible by 3?

1) n=1
2) m=2

Ans B.

\(10^n+m\) = \((1+9)^n+m\), from the binomial theorem, we know that for (\(1+9)^n\), all the terms will have a factor of 9 except the first term (\(1)^n\).

So the remainder will always be 1 for (\(10)^n\). That means the divisibility now depends on only the value of m.

Statement1: n=1, no value of m, insufficient.

Statement2: m=2, so the expression is divisible by 3, sufficient.
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Re: If m and n are positive integers, is 10^n+m divisible by 3?   [#permalink] 04 Sep 2018, 17:07
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