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

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by Brent@GMATPrepNow » Tue Nov 13, 2018 5:43 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

OA A

Source: Veritas Prep
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

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Brent
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by fskilnik@GMATH » Wed Nov 14, 2018 3:24 am
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
\[?\,\,\,\,:\,\,\,\,\# \,\,N\,\,,\,\,\,10 \leqslant N \leqslant 99\,\,\,{\text{with}}\,\,\,\left( * \right)\]
\[\left( * \right)\,\,\,\,\left. \begin{gathered}
N = 4M + 1\,\,\,\,\left( {M \geqslant 3\,\,\operatorname{int} } \right)\,\,\,\,\,\, \hfill \\
N = 14K + 1\,\,\,\left( {K \geqslant 1\,\,\operatorname{int} } \right) \hfill \\
\end{gathered} \right\}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,N - 1\,\,{\text{multiple}}\,\,{\text{of}}\,\,LCM\left( {4,14} \right) = 28\]
\[\left\{ \begin{gathered}
\,\,10 \leqslant N \leqslant 99 \hfill \\
\,\,N = 28L + 1\,\,\,\,\left( {L \geqslant 1\,\,\operatorname{int} } \right) \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,L = 1,2\,\,{\text{or}}\,\,3\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 3\]

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

Regards,
Fabio.
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by GMATGuruNY » Wed Nov 14, 2018 5:12 am
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
Let x = the two-digit number.

Since dividing by 4 leaves a remainder of 1, x is equal to one more than a multiple of 4:
x = 4k + 1
Since dividing by 14 leaves a remainder of 1, x is equal to one more than a multiple of 14:
x = 14m + 1

Since the two expressions in blue must be equal, we get:
4a + 1 = 14b + 1
4a = 14b
2a = 7b
a = (7/2)b

If b=2, then a=7.
If b=4, then a=14.
If b=6, then a=21.
The resulting values for a indicate that a must be a POSITIVE MULTIPLE OF 7.
21 is the largest multiple of 7 such that 4a+1 will yield a two-digit integer.
Thus, there are 3 options for a: 7, 14, 21.
Since there are 3 options for a, there are 3 possible values for x.

The correct answer is A.
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by Scott@TargetTestPrep » Sun Jan 27, 2019 6:34 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
The smallest positive integer that is divisible by both 4 and 14 is their LCM, which is 28. Thus 28 + 1 = 29 will yield a remainder of 1 when divided by both 4 and 14. We can add 28 to each previous dividend to obtain additional dividends:
29 + 28 = 57

57 + 28 = 85

Since the next number will be 85 + 28 = 113, which is a 3-digit number, we can stop at 85. Thus, we have three 2-digit numbers (29, 57, and 85) that will yield a remainder of 1 when divided by both 4 and 14.

Answer: A

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