100-2-2

=98-2

=96

100-2-2=98-2=96

100-2-2

=98-2

=96

bn nhe

tk nhe

=6 nka k mk mk k lại         

\(10-2-2\)

\(=10-\left(2+2\right)\)

\(=10-4=6\)

= 6 quá dễ

10-1-2

=-9-2

=7

tk nhe

\(10-1-2\)

\(=9-2\)

\(=7\)

10-1-2=7 ban nhe nho k cho minh nhe

10-1-1-1

=9-1-1

=8-1

=7

tk nhé

10-1-1-1= 10-(1+1+1)= 10-3=7

Vậy 10-1-1-1=7

cau tra loi la 7 ban nhe .

vi 10-1-1-1

=9-1

=8-1

=7

100 + 10 + 1 = 110 +1 = 111

100 + 10 + 1 = 111

100+10+1=111

100 -10 - 1 = 89

100-10-1=89

100-10-1

=90-1

=89

tk nhe

trao doi!

1000-100-10-1

=900-10-1

=890-1

=889

tk nhe

1000-100-10-1

=900-10-1

=890-1

=889

1 000 - 100 - 10 - 1

= 900 -10 - 1

= 890 - 1

= 889

tk r đó tk nha 

program abcdef;

uses Crt;

var

k, lowerLimit, upperLimit, i, j, reversed, temp, remainder: integer;

isPrime, isPalindrome: boolean; 

begin

clrscr;

write('Nhap so chu so k (1<=k<=9): ');

readln(k);

 lowerLimit := 1;

for i := 1 to k - 1 do

lowerLimit := lowerLimit * 10; 

upperLimit := lowerLimit * 10 - 1; 

writeln('Cac so nguyen to doi xung co ', k, ' chu so la:');

for i := lowerLimit to upperLimit do

begin

_{// Kiểm tra số nguyên tố}

isPrime := True;

if i < 2 then

isPrime := False

else

for j := 2 to trunc(sqrt(i)) do

if i mod j = 0 then

beginisPrime := False;break;end;

_{// Kiểm tra số đối xứng}

if isPrime then

begin

reversed := 0;

temp := i;

while temp <> 0 dobeginremainder := temp mod 10;

reversed := reversed * 10 + remainder;

temp := temp div 10;

end;

isPalindrome := (i = reversed);

 if isPalindrome then

writeln(i);

end;

end; 

readln;

end.