When a positive integer m is divided by 17, the remainder is 3. Which

Question

When a positive integer m is divided by 17, the remainder is 3. Which integer could be m?

I. 258
II. 274
III. 292

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III

BrentGMATPrepNow · Answer

When a positive integer m is divided by 17, the remainder is 3.  
In other words,   m is 3 greater than some multiple of 17

Now let's find some multiples of 17 that are close to the given values...

(15)(17) = 255
(16)(17) = 272
(17)(17) = 289

Since 255 is a multiple of 17, we can see that 258   is 3 greater than some multiple of 17  
So,  m could equal 255. 
So, we can eliminate answer choices B and C

Since 272 is a multiple of 17, we can see that 275   is 3 greater than some multiple of 17  
So,  m CANNOT equal 274 
So, we can eliminate answer choice E

Since 289 is a multiple of 17, we can see that 292   is 3 greater than some multiple of 17  
So,  m could equal 292. 
So, we can eliminate answer choice A

Answer: D

Karmoker · Answer

Let, m=20
When we devide m by 17 i.e., 20/17, we get 3 as remainder.
Hence,the options are divided by 17 with a remainder of 3, are our desired options.

Lets check out the options:
I. 258 (258/17) gives a remainder of 3
II.274 (274/17)gives a remainder of 2
III.292 (292/17) gives a remainder of 3

So, options I & III are the options that are devided by 17 with a remainder of 3.

Hence, D is our desired choice

Tusher Karmoker 🙂
Love from Bangladesh🇧🇩

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CrackverbalGMAT · Answer

In a could be true question on GMAT we are trying to prove the statements true.

When, a positive integer m is divided by 17, the remainder is 3,we can infer that if we reduce the number (given as choices) by 3,the resulting numbers must be divisible by 17.

What are the numbers that we have when we reduce the given number by 3?
I. 258 - 3 = 255

II. 274 - 3= 271

III. 292 - 3 =  289

I occurs thrice, II twice and III thrice. It makes sense to start either with I or III to eliminate maximum choices.

I. 255 should be divisible by 17 - It is. Eliminate B,C.

III.289 we know is the square of 17. Divisible. Eliminate A

If you check II.271 iS NOT DIVISIBLE by 17 as 17 * 16 =272. Eliminate E.

(option d)

D.S
GMAT SME

Sunoonazeez · Answer

The number should n]be of the form 17X+3
so when 3 is subtracted from the given numbers , the result should be a multiple of 17
among the options, (I) 258 and (III)292 satisfy this condition 
so the answer is D

100mitra · Answer

Correct Option D

Reminder formula : Dividend = Divisor * Quotient + Reminder

\frac{(Dividend - Reminder) }{ Divisor}  = Positive integer (Quotient)

Option 1 =  \frac{(258 - 3) }{ 17}  = 15 - Correct

Optio. 2 =  \frac{(274 - 3) }{ 17}  = integer - Wrong

Option 3 =  \frac{(292-3)}{17}  = 17 - Correct

Correct Answer D (I and III)

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BrushMyQuant · Answer

When a positive integer m is divided by 17, the remainder is 3

Theory: Dividend = Divisor*Quotient + Remainder

m -> Dividend
17 -> Divisor
a -> Quotient (Assume)
3 -> Remainders
=> m = 17*a + 3 = 17a + 3

Which integer could be m

m = 17a + 3
=> m - 3 = 17a
=> m - 3 should be a multiple of 17.

We will take each option choice and subtract 3 and check if it is a multiple of 17 or not

I. 258  
258 - 3 = 255 and we know that 255 IS a multiple of 17 =>   POSSIBLE

II. 274  
274 - 3 = 271 and we know that 271 is NOT a multiple of 17 =>   NOT POSSIBLE

III. 292  
292 - 3 = 289 and we know that 289 IS a multiple of 17 =>   POSSIBLE

So,  Answer will be D 
Hope it helps!

Watch the following video to learn the Basics of Remainders

https://www.youtube.com/watch?v=P0S6j2uTaj4

When a positive integer m is divided by 17, the remainder is 3. Which

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