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

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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If m and n are positive integers, is 10^n+m divisible by 3?  [#permalink]

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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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"Only $149 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Manager Joined: 17 May 2015 Posts: 249 If m and n are positive integers, is 10^n+m divisible by 3? [#permalink] Show Tags 31 Dec 2016, 01:00 1 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) Thanks. Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6981 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: If m and n are positive integers, is 10^n+m divisible by 3? [#permalink] Show Tags 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 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$149 for 3 month Online Course"
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Re: If m and n are positive integers, is 10^n+m divisible by 3?  [#permalink]

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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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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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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.
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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