Let $\overline{ABCD}$ be a $4$-digit number. What is the smallest possible positive value of $\overline{ABCD}- \overline{DCBA}$?
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Tags: number theory, Digits
GammaZero
14.09.2021 22:21
Do different letters represent different numbers?
parmenides51
14.09.2021 22:23
not necessarily
parmenides51
14.09.2021 22:27
the result has 2 digits, it asks for smallest positive
AkramElOmrani
14.09.2021 22:28
no 1119 - 9111 = -7 992 it's smaller
AkramElOmrani
14.09.2021 22:31
9999 - 9999 = 0 the smallest possible number no ? Is it positive or strictly positive ?
parmenides51
14.09.2021 22:36
positive means greater than zero
fuzimiao2013
14.09.2021 22:52
90. Simply expand, and you get that it's equal to 999(A-D) + 90(B-C), and since B-C < 10, and we can't have A-D > 0, thus A=D and B-C = 1.
pinkpig
14.09.2021 23:01
We can easily find that $$\overline{ABCD}-\overline{DCBA}=(1000A+100B+10C+D)-(1000D+100C+10B+A)=$$$$999A+90B-90C-999D=999(A-D)+90(B-C).$$To minimize this, we let $A-D=1,0,-1.$ These three cases give us a minimum of $189,90,-189.$ We conclude that the answer is $\boxed{90}.$
kmuruganandam
14.09.2021 23:08
fuzimiao2013 wrote: 90. Simply expand, and you get that it's equal to 999(A-D) + 90(B-C), and since B-C < 10, we can't have A-D > 0, thus A=D and B-C = 1. Sorry, could you explain why A-D must be greater than 0 if B-C<10.
OlympusHero
15.09.2021 05:10
yuh
Rewrite as $1000A+100B+10C+D-1000D-100C-10B-A = 999A+90B-90C-999D = 9(111(A-D)+10(B-C))$. The minimum possible positive value is attained when $A-D=0$ and $B-C=1$, which gives a result of $\boxed{90}$. This is attainable; one way is $A=1, B=2, C=1, D=1$.
kmuruganandam
15.09.2021 18:05
Yeah, I figured it out using the "common sense" approach that we are looking for the smallest positive integer but I still don't understand @fuzimiao2013's solution