I am not done with divisibility yet. Not by a long shot. In my last post, I discussed how divisibility was basically making groups and subtracting them out. To understand the basics of what I am going to talk about now, just imagine the numbers as balls. When I say 7, imagine 7 balls in a group.

Let's start easy: Is 7 divisible by 3? Before answering, think of the following figure:

To divide by 3, I have to split 7 into groups of 3. I can make two groups of 3 and then 1 ball is leftover. Hence, when I divide 7 by 3, quotient is 2 and remainder, the ball that could not be put in a group, is 1.

Similarly, what happens when I divide 12 by 3?

I get four groups and nothing leftover. (Is this then the diagrammatic representation of 12 = 3*4? Sure. After all 12/3 is 4, the quotient) Then when I divide 12 by 4, I should be able to get 3 groups with nothing leftover.

I guess you get the picture. Now, a question:

I have a number that when divided by 6, leaves a remainder of 2. What will be the remainder when the number is divided by 3?

So here, I do not know what the number is, but I know that when I make groups of 6, 2 is leftover. Logically, it follows that when I split each of the groups of 6 into two groups of 3 each, I will still have the same remainder of 2.

Then the answer is 2. you don't have to make the diagram in the exam of course! Just imagine it and you will see the answer.

A slight twist on the question above: I have a number that when divided by 6, leaves a remainder of 4. What will be the remainder when the number is divided by 3?

Imagine the picture. Just like above but with groups of 6, there are 4 balls leftover. You divide the groups of 6 into groups of 3, no issues, but now, you can make another group of 3 from the 4 balls that are leftover. Therefore, only one ball will be leftover giving you the remainder of 1. Answer is 1.

Another little twist: A number when divided by 3 gives a remainder of 1. How many distinct values can the remainder take when the same number is divided by 9?

Now imagine that there are lots of groups of 3 and 1 ball leftover. You have to make groups of 9 out of these so you start combining three groups of 3s to make groups of 9. Let's see what the possibilities at the end are:

1. All groups of 3s get used to make groups of 9 and the 1 ball from before is again leftover.

2. One group of 3 is leftover and the 1 ball from before, giving you a total of 4 balls leftover.

3. Two groups of 3 are leftover and the 1 ball from before, giving you a total of 7 balls leftover.

(Three or more groups of 3 cannot be leftover because then, we will be able to make another group of 9 out of them)

Therefore, you can have the remainder in three distinct ways: 1, 4 and 7.

That was easy, wasn't it? I will take more advanced remainder questions in the near future, after discussing factors and multiples.

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