Ta có :
abc = 100a + 10b + c = n2 - 1 (1)
cba = 100c + 10b + c = (n - 2)2 (2)
Lấy (2) trừ (1) ta được:
99(a - c) = 4n - 5
=> 4n - 5 \(⋮\)99
Lại có: 100 \(\le\)n2 - 1 \(\le\)999
<=> 101 \(\le\)n2 \(\le\)1000
<=> 11 \(\le\)n \(\le\)31
<=> 44 \(\le\)4n \(\le\)124
<=> 39 \(\le\)4n - 5 \(\le\)119
Mà 4n - 5 \(⋮\)99
=> 4n - 5 = 99
=> n = 26
=> abc = 262 - 1 = 675
Vậy abc = 675
Tim tat ca cac so tu nhien co 3 chu so abc sao cho abc = n2 - 1 va cba= (n- 2)2
tim tat ca cac so tu nhien co 3 chu so abc sao cho abc=\(n^2-1\) va cba=\(\left(n-2\right)^2\)
tim tat ca cac so tu nhien co 3 chu so abc sao cho abc =n^2-1 va cba=(n-2)^2
tim tat ca cac so tu nhien co 3 chu so abc sao cho abc = n2 - 1 va cba = (n - 2)2
tim tat ca cac so tu nhien co ba chu co abc sao cho abc=n^2-1 va cba=(n-2)^2
Tim tat ca cac so tu nhien co 3 chu so abc sao cho abc = n2 - 1 va cba = ( n-2 )2
tim tat ca cac so tu nhien co 3 chu so abc sao cho abc =n2 -1 va cba =(n-2)2
Tim tat ca cac so tu nhien co 3 chu so abc sao cho abc = (n-2)2
tim so tu nhien n sao cho ton tai so co 3 chu so abc mma abc= n mu 2 -1 va cba = (n-2) ^2
