Solutions

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?

 

Difficulty Level

Hard

Solution

Option A is the correct answer.

Option Analysis

Let’s Consider N = (100X1 + 10Y1+ Z1)

Consider M = (100X2 + 10Y2+ Z2)

N – M = 100(X1-X2) + 10(Y1-Y2) + (Z1-Z2)

Let’s analyze these terms:-

100(X1-X2) = 100; we need to keep it minimum at 100 (i.e. X1-X2=1 with pair of consecutive numbers). We do not want it over 200 as it will increase the overall value.

10(Y1-Y2) = -70; to offset 100 from above, we should minimize this term to lowest possible negative value. Pick extreme numbers as 1 & 8 -> 10(1-8)= -70

(Z1-Z2) = -1; Excluding (1,8) taken by (Y2,Y2) and pair of consecutive numbers taken by (X1,X2) -> we are left with 1 pair of consecutive numbers -> Minimize it to -1;

Finally, N−M=100(X1−X2)+10(Y1−Y2)+(Z1−Z2)=100−70−1=29.