The first six positive integers are used once to form two three-digit

Question

The first six positive integers are used once to form two three-digit numbers. What is the minimum possible difference between the two numbers formed?

A. 87
B. 78
C. 69
D. 60
E. 47

chetan2u · Accepted Answer

Since we are looking for least possible difference and the numbers used are 1 to 6, let us see whether the hundreds digit should be 5&6 or 4&3 or something else.  
 Hundreds digit: we should pick up two numbers that are consecutive, so it could be 5&6 or 1&2 or 2&3 etc. 
 Tens digit:  We have to give maximum to the smaller hundreds digit and smallest to larger hundred digit. That is a6b and (a+1)1d
 units digit:  Same as 2. From the remaining 4 digits, let us give maximum to the smaller hundreds digit and smallest to larger hundred digit. so 5 and 2 are the options NES with maximum difference. => a65 and (a+1)12

Two remaking numbers a and a+1, which should be 3&4. 
Clearly a is 3, so numbers are 365 and 412.

Difference = 412-365 = 47

sanjitscorps18 · Answer

The answer should be 47.

The numbers would be 412 and 365.

Option E

PyjamaScientist · Answer

Good one  sanjitscorps18 . Rarely do we see someone beating Chetan to the correct answer.
Could you share the method you used to solve this? Or was this through number intuition/ Trial and error?

sanjitscorps18 · Answer

Th trick that I employed was as below.
1. To keep the first digits as close as possible, usually the middle numbers (took 3 and 4)
2. Maximize the last two digits on the number starting with the lower digit (in this case it was 3), and using 65 rather than 56
3. Minimize the last two digits on the number starting with the higher digit (in this case it was 4), and using 12 rather than 21

Hope this would help !!

samarth1222 · Answer

hi 
In this question what if we take 421 and not 412 and the minimum possible number will be 46 421-365= 46 will it be the correct method to do?

sanjitscorps18 · Answer

Please check the difference. It would be 56 and not 46.

Posted from my mobile device

sayan640 · Answer

KarishmaB  Would you like to provide your explanation on this one?

Posted from my mobile device

surya.nair · Answer

Let the two numbers be abc and xyz.
So , abc => 100*a + 10*b+c
xyz = 100*x + 10*y + z
abc - xyz = 100(a-x) + 10(b-y) + c-z

We just need to substitue the numbers and find what brings this value to minimum
let a = 4 and x = 3, then a-x = 1
let b = 1 and y = 6, then b-y = -5
let c = 2 and z = 5, then c-z = -3
I am not randomly assigning the number, I think of the differences first and then assign the order to get the minimum value. This way we have (100 * 1) - (10 * 5) - 3 = 47

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The first six positive integers are used once to form two three-digit

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