Source: Official Guide Revised GRE 1st Ed. Arithmetic Exercises; #15

100

When the positive integer n is divided

When the positive integer n is divided by 3, the remainder is 2 and when n is divided by 5, the remainder is 1. What is the least possible value of n?

6 Explanations

1

Habtamu Anteneh

First we divide n'/3=2, cross multiplying them gives, n'=6. when we divide n''/5=1 cross multiplying gives n''=5. finally we add n'+n''= 6+5=11.

hope this helps!

Nov 16, 2017 • Comment

1

Habtamu Anteneh

we have a ratio of 3:2. we have to add them so we have 5, now we multiply 2 (ratio of cats) by 20 (the total number), which gives us 40, finally we divide 40 by 5 (3+2)= 8. similarly, we can multiply 3 by 20/5= 12.

Nov 16, 2017 • Comment

2

Aditya Om

Is there a general method other that this trial and error method?

Mar 30, 2017 • Comment

Cydney Seigerman, Magoosh Tutor

Hi Aditya,

Excellent question! Yes, we can derive a general formula to approach this question :) First, use the statements to write two equations in the form

dividend = (integer quotient)*(divisor) + remainder
n = 3p + 2
n = 5q + 1

where p and q are the quotients.

Now, we can plug in the values of p and q into the equations, starting with 0:

n = 3p + 2
p = 0 --> n = 3*0 + 2 = 2
p = 1 --> n = 3*1 + 2 = 5
p = 3 --> n = 3*2 + 2 = 8
p = 4 --> n = 3*3 + 2 = 11
etc.

n = 5q + 1
q = 0 --> n = 5*0 + 1 = 1
q = 1 --> n = 5*1 + 1 = 6
q = 2 --> n = 5*2 + 1 = 11

We can stop here, since we have found a common value for n, n = 11. Therefore, the least possible value for n for which both statements is true is n = 11 :)

Hope this helps!

Apr 1, 2017 • Reply

Md Habib Ullah Khan

Thanks

Nov 20, 2017 • Reply

3

Samia Aljuwayed

I've question about the substitution part(y = 0, 1, or 2) is it randomly?

Jan 7, 2017 • Comment

Sam Kinsman, Magoosh Tutor

Hi Samia,

What Priyanka was saying is that since x = (5y - 1)/3, where x and y are whole numbers, we know that 5y - 1 must be divisible by 3.

We want to find the smallest possible value of 5y-1. This value will be small when y is small. So we can start with the smallest possible value of y (which is 0), and move up from there until we find a value of y which will make 5y-1 divisible by 3.

If we plug in y=0, we get 5(0)-1 = -1, which is not divisible by 3. So this doesn't work.

If we plug in y=1, we get 5(1)-1 = 4, which is not divisible by 3. So this doesn't work.

If we plug in y=2, we get 5(2)-1 = 9, which is not divisible by 3. So this works.

If y = 2, then n = 11 (because, as Priyanka said, 5y + 1 = n).

Jan 11, 2017 • Reply

Shubhangi Kukreti

I don't really understand why 5y-1 is divisible by 3

Jul 29, 2017 • Reply

Cydney Seigerman, Magoosh Tutor

Hi Shubhangi,

Happy to clarify :) This idea comes from the solution posted by Priyanka:

"We have,
3x + 2 = n
5y + 1 = n (x & y being their respective quotient)

Therefore, 3x + 2 = 5y + 1
3x = 5y - 1
x = (5y - 1)/3"

When we solve for x in terms of y, we find that

x = (5y - 1)/3

Since x is an integer, we know that the right side, which is equal to x, must be an integer. So, (5y - 1) must be divisible by 3. On the other hand, if the expression (5y - 1) were not divisible by 3, then the expression (5y - 1)/3 would not be an integer. In that case, it could not be equal to x, since we know x is an integer.

Does that make sense? I hope this clears things up for you! :)

Jul 30, 2017 • Reply

23

khorshid Azarnoush

Thank you

Nov 5, 2015 • Comment

32

Priyanka Shetty

We have,
3x + 2 = n
5y + 1 = n (x & y being their respective quotient)

Therefore, 3x + 2 = 5y + 1
3x = 5y - 1
x = (5y - 1)/3

Now we know that, 5y - 1 is divisible by 3, so lets substitute y = 0, 1, or 2

Trying these values we find y = 2

3x = 5(2) - 1
3x = 9
x = 3

n = 3x + 2 = 9 + 2 = 11
n = 5y + 1 = 10 + 1 = 11

Thus 11 is the least possible value of n

Oct 8, 2013 • Comment

MANOJ KHAIRNAR

(Y)

Apr 2, 2015 • Reply

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