If a certain positive integer is divided by 9, the remainder

but don't we need to test values that satisfy both equations?

Statement A openly tells us that the number is a multiple of 5. So regardless of what the number is, when we divide that number by 5, the remainder is going to be 0. We do not have to find the number itself.

I realize my mistake, I misinterpreted the question. Thanks!

Hi Bunuel,

One question. With statement 1, are the following inferences all valid?
- N is divisible by 5 since --> n = 5 (9q + 6)
- N is also divisible by 3 since --> n = 3 (15q + 10)
- N is therefore also divisible by 15 since --> n = 15 (3q + 2)

I know this goes beyond the scope of answering this question. I just wanna check if my reasoning is correct for future problems such as this one.

Thanks,

@bunnel
do we really need x = 9q+3 to get an answer ?

Forget the conventional way to solve DS questions.

We will solve this DS question using the variable approach.

DS question with 2 variables and 1 equation: Let the original condition in a DS question contain 2 variables. Now, 2 variable and 1 equation would generally require 1 more equation for us to be able to solve for the value of the variable. We know that each condition would usually give us an equation and Since we need 1 equation to match the numbers of variables and equations in the original condition, the logical answer is D. The answer could be A, B, or D, but the default answer will be D.

To master the Variable Approach, visit https://www.mathrevolution.com and check our lessons and proven techniques to score high in DS questions.

Let’s apply the 3 steps suggested previously. [Watch lessons on our website to master these 3 steps]

Step 1 of the Variable Approach: Modifying and rechecking the original condition and the question.

=> Certain positive integer is divided by 9, the remainder is 3: Let the positive integer be 'x'. Then, x = 9p + 3 , where 'p' is the quotient.

We have to find remainder when 'x' is divided by '5'.

Second and the third step of Variable Approach: From the original condition, we have 2 variables (x and p) and 1 equation(x = 9p + 3).To match the number of variables with the number of equations, we need 1 more equation. Since conditions (1) and (2) will provide 1 equation each, D would most likely be the answer.

Let’s take a look at each condition.

Condition(1) tells us that If the integer is divided by 45, the remainder is 30.

=> x = 45p + 30

=> x = 5 (9p + 6).

=> Therefore, 'x' being multiple of '5', the remainder when divided by '5' will be '0'.

Since the answer is unique , condition(1) is sufficient by CMT 2.

Condition(2) tells us that 'x' is divisible by '2' - that means 'x' is an even number.

=> For x = 18: \(\frac{18 }{ 5}\) - remainder= 3

=> For x = 20: \(\frac{20 }{ 5}\) - remainder= 0

Since the answer is not unique , condition(2) is not sufficient by CMT 2.


Condition(1) is alone sufficient by CMT 2.

So, A is the correct answer.

Answer: A

avigutman KarishmaB

How can I solve it logically for statement 2 without listing value of n (3,12,21,30...) and concluding that remainder could 2 (12) or 0 (30)?

Thank you for your help!

While experts provide their reply, sharing my 2 cents if that helps.

Given: A certain positive integer is divided by 9, the remainder is 3.

Let's assume that number is x, therefore we can represent x as

x = 9*q + 3

In this representation, q is the quotient when x is divided by 9.

Statement 2 tells us "The integer is divisible by 2"

Hence, we can conclude that x is even.

Therefore 9q + 3 is even.

As 9q + 3 is an even integer we can reason out that 9q is odd (odd + odd = even) ⇒ which in turn tells us ⇒ q is odd.

Having done the analysis, let's move back to our equation

x = 9*q + 3

We know that q is odd, so q can be 5 (as 5 is odd) and if it were the case, the remainder, when 9q + 3 is divided by 5, would be 3.

Now, the other possibility is 9q is not divisible by 5. In that case, the remainder of \(\frac{9q}{5}\) will be between 1 and 4, both inclusive. While 9q is not divisible by 5, if the remainder is 2 then the net sum (9q+3) is divisible by 5.

So we can have a scenario in which 9q+3 is divisible by 5, i.e. when \(\frac{9q}{5}\) leaves a remainder = 2 .

As we're getting two different answers, Statement 2 is insufficient on its own.

Well, Sneha2021, let's try this: if I tell you that positive integer n is divisible by 7, can you tell me whether it's odd or even?
What if I tell you that positive integer k has a remainder of 3 when divided by 7, a remainder of 1 when divided by 5, and a remainder of 0 when divided by 3. Now can you tell me whether k is odd or even?
The answer to both questions above is no. We can't tell, because the terms odd and even exist only in the world of divisibility by 2, and the divisor '2' doesn't have any common factors with the divisors '3', '5', or '7'. These worlds of divisibility are completely separate from one another, so no inferences can be made about remainders going from one world to another.
By the way, my description of k is going to fit numbers that are (3*5*7 = ) 105 apart on the number line. So, if we find possible value of k, adding 105 over and over again will keep yielding possible values of k. And, of course, each time we add 105, we change the value of k back and forth between odd and even.
k could be 66, 171, 276, 381, 486, etc.
Back to your original question, Sneha2021: yes, we can evaluate statement (2) logically, without testing cases, and without pen and paper. The information we have about the divisibility of the "certain integer" is for divisors of 9 and 2, but the question has a divisor of 5, which has no common factors with 9 or 2. Therefore, we can't infer anything about the remainder in the world of divisibility by 5.

Using the divisibility concept, we know that n leaves remainder 3 upon dividing by 9. So imagine groups of 9 and a lone group of 3 leftover.
If n is divisible by 2, it means it is an even number. So we will have odd number of groups of 9 (so that when we add them to 3, we get an even number).
So we will have 1 group of 9 or 3 groups of 9 etc.

When we divide this by 5, 4 is leftover from each group fo 9. So we will have 1 group of 4 or 3 groups of 4 etc and we will still have that 3 leftover.
So we will have 7 leftover (remainder 2) or 15 leftover (remainder 0) etc.

Hence statement 2 is not sufficient alone. (Since its not sufficient alone, you will need to make multiple cases so you cannot avoid that)

hi avigutman HAPPY new years !

(1) do you do S1 w/o pen and paper too ? if so , what was your logic for S1

(2) could you perhaps link a video where the yellow topic is being exercised (perhaps in another OG problem or an AMA or perhaps your book).. the yellow is NOT "intuitive" for me unfortunately.

(3) just curious, how did you come up with "66" from the list in the green.

Yes, jabhatta2. The free info is a distraction when evaluating S1. A remainder of 30 when dividing some number by 45 means that the number is a multiple of 5 (consider the simpler case in which dividing by 10 yields an even remainder - since both 10 and the even remainder are each divisible by 2, the original number must be divisible by 2).

https://youtu.be/_uwR9AZas9M?t=1813

We know that k is a positive integer that has a remainder of 3 when divided by 7, a remainder of 1 when divided by 5, and a remainder of 0 when divided by 3.
So, I just looked at numbers that are 3 above a multiple of 7, that happen to also have a units digit of 1 or 6 (a.k.a. a remainder of 1 when divided by 5). And, I know that such numbers will be LCM(5,7) = 35 units away from one another on the number line. So I quickly wrote down 3, 10, 17, 24, 31 (I continued listing them until I got one with a units digit of 1 or 6). Once I found 31, I checked whether the sum of the digits is divisible by 3. It isn't, so I added 35 and checked again (and I would've kept adding 35 over and over until I got to a multiple of 3). Bingo. That's how I identified 66 as a number that fits the description of k.

thank you so much avigutman

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