Giải:
Ta có:
\(\overline{abc}=100a+10b+c=n^2-1\) (1)
\(\overline{cba}=100c+10b+a=n^2-4n+4\) (2)
Lấy (1) trừ (2) ta được:
\(99\left(a-c\right)=4n-5\)
\(\Rightarrow4n-5⋮99\)
Vì \(100\le\overline{abc}\le999\) nên:
\(100\le n^2-1\le999\)
\(\Rightarrow101\le n^2\le1000\)
\(\Rightarrow11\le31\)
\(\Rightarrow39\le4n-5\le119\)
Vì \(4n-5⋮99\)
\(\Rightarrow4n-5=99\)
\(\Rightarrow n=26\)
\(\Rightarrow\overline{abc}=675\)
Vậy \(\overline{abc}=675\)
Giải:
Ta có:
abc¯¯¯¯¯¯¯=100a+10b+c=n2−1abc¯=100a+10b+c=n2−1 (1)
cba¯¯¯¯¯¯¯=100c+10b+a=n2−4n+4cba¯=100c+10b+a=n2−4n+4 (2)
Lấy (1) trừ (2) ta được:
99(a−c)=4n−599(a−c)=4n−5
⇒4n−5⋮99⇒4n−5⋮99
Vì 100≤abc¯¯¯¯¯¯¯≤999100≤abc¯≤999 nên:
100≤n2−1≤999100≤n2−1≤999
⇒101≤n2≤1000⇒101≤n2≤1000
⇒11≤31⇒11≤31
⇒39≤4n−5≤119⇒39≤4n−5≤119
Vì 4n−5⋮994n−5⋮99
⇒4n−5=99⇒4n−5=99
⇒n=26⇒n=26
⇒abc¯¯¯¯¯¯¯=675⇒abc¯=675
Vậy abc¯¯¯¯¯¯¯=675abc¯=675
Chúc bn hk tốt!!
