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

Question

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

ccooley · Accepted Answer

I'd start with the ones that have a remainder of 1 when divided by 14. There'll be fewer of those, so it'll be much quicker to list them out.

14 + 1 = 15
15 + 14 = 29
29 + 14 = 43
43 + 14 = 57
57 + 14 = 71
71 + 14 = 85
85 + 14 = 99

Notice that I'm not taking the slow route of doing '14*2+1, 14*3+1, 14*4+1...' etc. I find that tends to be slower unless you happen to have the multiples of 14 memorized, since multiplication is a bit slower than addition. A nice shortcut for listing numbers with a particular remainder, is to start with the remainder (1), then count up by the divisor (in this case 14) repeatedly.

Okay, we have our list. How many of those have a remainder of 1 when divided by 4? The first thing I notice is that 15 has a remainder of 3 when divided by 4, but 29 has a remainder of 1. Why? Well, it looks like when the 14 is multiplied by 2, it becomes a multiple of 4 (28). So, adding 1 to it will give you a number that also has a remainder of 1 when you divide it by 4. On the other hand, if the 14 isn't multiplied by 2, it isn't a multiple of 4. So, adding 1 to it won't give you a 'good' number.

That means every other value on our list will work.

14 + 1 = 15
15 + 14 =  29 
29 + 14 = 43
43 + 14 =  57 
57 + 14 = 71
71 + 14 =  85 
85 + 14 = 99

The answer is (A) 3.

AMahmoodi · Answer

Hi,
an alternative solution would be:

the LCM of 4 & 14 is 28 therefore the numbers are 28+1, 56+1, 84+1.

thanks.

NandishSS · Answer

Hi,

Take the numbers which are multiple of 14-28,42,56,70,84,98,112,126--Out of which only 28,56,84 are divisible by both 14 & 4 hence if you add 1 to these numbers. It will give you remainder of 1 when divided by 4 and remainder of 1 when divided by 14. Hence 3

KrishnakumarKA1 · Answer

First we need to find the numbers which are divisible by 4 and 14. the numbers are oing to be LCM of the 4 and 14 which will be 28, 56, 84.

hence the number leaving 1 as remanider will be 28+1, 56+1 and 84+1
Option A

KrishnakumarKA1 · Answer

First we need to find the numbers which are divisible by 4 and 14. the numbers are oing to be LCM of the 4 and 14 which will be 28, 56, 84.

hence the number leaving 1 as remainder will be 28+1, 56+1 and 84+1
Option A

gracie · Answer

least value of n=1
1+ lcm 2*14=29=next least value of n
let x=number of such two digit numbers
29+(x-1)28<100
x<3.6
x=3
A

SidKan · Answer

Hi,
an alternative solution would be:

the LCM of 4 & 14 is 28 therefore the numbers are 28+1, 56+1, 84+1.

thanks.

BrentGMATPrepNow · Answer

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

fskilnik · Answer

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\,\,{	ext{ints}}} ight)

\left. \begin{gathered}
 N - 1 = 4Q \hfill \
 N - 1 = 14J\, \hfill \ 
\end{gathered} ight\}\,\,\,\,\mathop \Rightarrow \limits^{LCM\,\left( {4,14} ight)\, = \,\,{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.

ScottTargetTestPrep · Answer

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

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

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

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