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How many positive two-digit numbers yield a remainder of 1

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by BTGmoderatorDC » Wed Sep 12, 2018 5:12 pm
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How many positive two-digit numbers yield a remainder of 1 when divided by 4 and also yield a remainder of 1 when divided by 14?

A. 3
B. 4
C. 5
D. 6
E. 7

OA A

Source: Veritas Prep
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Source: — Problem Solving |

by Brent@GMATPrepNow » Wed Sep 12, 2018 5:39 pm
BTGmoderatorDC wrote:How many positive two-digit numbers yield a remainder of 1 when divided by 4 and also yield a remainder of 1 when divided by 14?

A. 3
B. 4
C. 5
D. 6
E. 7
When it comes to remainders, we have a nice rule that says:
If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.
For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.

Two-digit number yields a remainder of 1 when divided by 14.
So, the possible values are: 15, 29, 43, 57, 71, 85 and 99
At this point, we have 7 possible values

Two-digit number yields a remainder of 1 when divided by 4.
Examine each of the 7 values and determine which ones yield a remainder of 1 when divided by 4
They are: 15, 29, 43, 57, 71, 85 and 99

So, there are 3 values that satisfy BOTH conditions.

Answer: A

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by fskilnik@GMATH » Wed Sep 12, 2018 6:19 pm
BTGmoderatorDC wrote:How many positive two-digit numbers yield a remainder of 1 when divided by 4 and also yield a remainder of 1 when divided by 14?

A. 3
B. 4
C. 5
D. 6
E. 7

Source: Veritas Prep
Let´s use some powerful "divisibility tools" carefully explained in our course.

(Our students perform all operations below in approximately 2min with no rush!)

\[N = 4Q + 1 = 14J + 1\,\,\,\,\,\,\,\,\left( {Q,J\,\,{\text{ints}}} \right)\]
\[\left. \begin{gathered}
N - 1 = 4Q \hfill \\
N - 1 = 14J\, \hfill \\
\end{gathered} \right\}\,\,\,\,\mathop \Rightarrow \limits^{LCM\,\left( {4,14} \right)\, = \,\,{2^2} \cdot 7} N - 1 = 28K\,\,\,\,\, \Rightarrow \,\,\,\,N = 28K + 1\,\,,\,\,\,K\,\,\operatorname{int} \]
\[10 \leqslant N \leqslant 99\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,10 \leqslant 28K + 1 \leqslant 99\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,9 \leqslant 28K \leqslant 98\]
\[9 \leqslant 28K \leqslant 98\,\,\,\, \Rightarrow \,\,\,\,1 \cdot 28 \leqslant 28K \leqslant 3 \cdot 28\,\,\,\,\, \Rightarrow \,\,\,\,1 \leqslant K \leqslant 3\,\,\,\,\, \Rightarrow \,\,\,\,? = 3\,\,\,\,\,\,\,\]

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
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by Scott@TargetTestPrep » Mon Sep 17, 2018 5:26 pm
BTGmoderatorDC wrote:How many positive two-digit numbers yield a remainder of 1 when divided by 4 and also yield a remainder of 1 when divided by 14?

A. 3
B. 4
C. 5
D. 6
E. 7
We can use the fact that if a number n leaves a remainder of r when divided by a and b (a ≠ b), then n leaves a remainder of r when divided by the least common multiple (LCM) of a and b.

Since the LCM of 4 and 14 is 28, the smallest such two-digit number is 29 (notice that 29/4 = 7 R 1 and 29/14 = 2 R 1). The next two-digit number is 29 + LCM = 29 + 28 = 57. The next one is 57 + 28 = 85. Since the next one will be more than 100, we can stop at 85. Therefore, we see that there are 3 such integers.

Answer: A

Scott Woodbury-Stewart
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