Minimising Difference Between Numbers | Number Properties

Not every question of a standardised test can be solved using pure algebra, especially at higher levels. There will be questions which will need logic and quite a bit of thinking on your part. These questions tend to throw test-takers off. Often, the complaints are – From where do I start? Thinking through takes too much time… etc. Unfortunately, there is no getting away from such questions so let’s practie how to deal with them.

Question: ${N}$ and ${M}$ are each 3-digit integers. Each of the numbers 1, 2, 3, 6, 7, and 8 is a digit of either ${N}$ or ${M}$ . What is the smallest possible positive difference between ${N}$ and ${M}$ ?

(A) 29

(B) 49

(D) 113

(E) 131

This is not a simple algebra question where we make equations and solve them.

We are given 6 digits: 1, 2, 3, 6, 7, 8

Each digit needs to be included to make two 3 digit numbers. This means that we will use each of the digits only once and in only one of the two numbers. We need to minimise the difference between the two numbers so they should be as close as possible to each other. Obviously, since the numbers cannot share any digits, they cannot be equal and hence, the smaller number needs to be as large as possible and the greater number needs to be as small as possible to be close to each other. Think of the numbers of a number line. You need to reduce the difference between them. Then, under the given constraints, push the smaller number to the right on the number line and the greater number to the left to bring them as close as possible to each other.

STEP 1:

The first digit (hundreds digit) of both numbers should be consecutive integers i.e. the difference between 1** and 2** can be made much less than the difference between 1** and 3** (the difference between these two will certainly be more than 100).

We get lots of options for hundreds digits: (1** and 2**) or (2** and 3**) or (6** and 7**) or (7** and 8**). All of them could satisfy our purpose.

STEP 2:

Now let’s think about the next digit (the tens digit). To minimise the difference between the numbers, the tens digit of the greater number should be as small as possible (1 if possible) and the tens digit of the smaller number should be as large as possible (8 if possible). So let’s not use 1 and 8 in the hundreds places and reserve them for the tens places since we have lots of other options (which are equivalent) for the hundreds places. Now what are the options?

Let us try to make a pair of numbers of the form 2** and 3**. Make the 2** number as large as possible and make the 3** number as small as possible. As discussed above, the tens digit of the smaller number should be 8 and the tens digit of the greater number should be 1. We get 28* and 31*

STEP 3:

Now we use the same logic for the units digit. Make the units digit of the smaller number as large as possible and the units digit of the greater number as small as possible. We have only two digits leftover – 6 and 7.

So the two numbers could be 287 and 316. The difference between them is 29.

Let’s try the same logic on another pair of hundreds digits.

Let’s make a pair of numbers of the form 6** and 7**. Make the 6** number as large as possible and make the 7** number as small as possible. Using the same logic as above, we get 683 and 712. The difference between them is 29 again.

The smallest of the given options is 29 so we need to think no more.

Answer must be (A).

Note that even if you express the numbers algebraically,

$N = 100a + 10b + c$

$M = 100d + 10e + f$

Still a lot of thought will be needed and there is no process which can be followed.

Assuming $N$ is the greater number, we need to minimise $N - M$ .

$N - M = 100 (a - d) + 10( b - e) + (c - f)$

Since $a$ and $d$ cannot be the same, the minimum value $a - d$ can take is 1.

$(a - d)$ cannot be negative because we have assumed that $N$ is greater.

So $a$ and $d$ must be consecutive (2 and 1 or 3 and 2 or 7 and 6 etc). This is another way of arriving at STEP 1 above.

Next, $(b - e)$ should take the minimum value. From the available digits, 1 and 8 are the farthest and can give a difference of -7. Then, $b = 1$ and $e = 8$ . This leaves the consecutive pairs of 2, 3 and 6, 7 for hundreds digits. This takes care of our STEP 2 above.

$(c - f)$ should be minimum too. We have only one pair of digits leftover and they are consecutive. So minimum value of $(c - f)$ is -1. If the hundreds digits are 3 and 2, then $c = 6$ and $f = 7$ . This is our STEP 3.