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පුළුවන්නම් විසඳන්න... ඔට්ට්යි බෑ!! Challenge...
- Thread starter BarakOsama
- Start date
[SIZE=+1]Kris found a treasure box. To open the box, she had to put four numbers in their correct order, using the following clues:[/SIZE]
*The first and second numbers are prime.
*The third number is a multiple of the second number.
*The second number is even.
*The third number is more than nine and less than twenty.
*The first number is less than ten.
*The fourth number is the least common multiple of the first two numbers.
*The fourth number is the largest of all four numbers.
*The first number is more than three times the second number.
*The third number is one more than the sum of the first two numbers.
[SIZE=+1]She opened the box and found a treasure equal to 1000 times the sum of the four numbers. How much was the treasure worth? [/SIZE]
*The first and second numbers are prime.
*The third number is a multiple of the second number.
*The second number is even.
*The third number is more than nine and less than twenty.
*The first number is less than ten.
*The fourth number is the least common multiple of the first two numbers.
*The fourth number is the largest of all four numbers.
*The first number is more than three times the second number.
*The third number is one more than the sum of the first two numbers.
[SIZE=+1]She opened the box and found a treasure equal to 1000 times the sum of the four numbers. How much was the treasure worth? [/SIZE]
ABCDE*4=EDCBA
මෙම අකුරු 0-9 දක්වා එක් සංඛ්යාවක් පමණක් නිරූපණය කරයි නම් A,B,C,D,E වල අගයන් සොයන්න.
මෙම අකුරු 0-9 දක්වා එක් සංඛ්යාවක් පමණක් නිරූපණය කරයි නම් A,B,C,D,E වල අගයන් සොයන්න.
21978*4=87912ABCDE*4=EDCBA
මෙම අකුරු 0-9 දක්වා එක් සංඛ්යාවක් පමණක් නිරූපණය කරයි නම් A,B,C,D,E වල අගයන් සොයන්න.
Because there is no carry for A x 4, we know A = 2, 1, or 0. Now,A=0 implies E= 1, 2, or 3 (due to carry from B x 4).
But, A=0 implies E=5 since E x 4 ending in 0. Thus, A must be 1 or 2.
But, A must be even by E x 4, which implies A=2 and E is either 3 or 8.
But E=3 cannot work with A x 4 where A=2. Thus, E=8.
We note that there is no carry from B x 4, which implies B = 0 or 1.
But, B must equal 1 as (D x 4) + 3-carry is odd...which implies D = 7.
Finally, a carry of 3 is necessary to make D=7 for B x 4, which implies C=9
[SIZE=+1]21978 x 4 = 87912
[/SIZE]
sarala deyak thiyenne 
සිංහළයා;10141362 said:උඹලට අපි දාන ප්රශ්න ගින්නක් නෑ කියල හිතෙනවනම් මේ ප්රශ්න ටිකට උත්තර හොයපල්ලා.මේවට තාම උත්තර හොයාගෙන නෑ:lol
http://en.wikipedia.org/wiki/Hilbert's_problems
http://en.wikipedia.org/wiki/List_of_unsolved_problems_in_mathematics
mata owata uththara kiyanna thibba English walin thibba nam .... koheda me sinhala mata therennne naa ne......
apitath amba one......දැන් මම යනවා අඹ කඩන්න... කෙක්ක ගහේ එල්ලල ඇවිල්ලා දැන් පැයකටත් වැඩී...