Problem Solving

[theanother]

There is a two-digit number whose digits are the same, and has got the following property: When squared, it produces a four-digit number, whose first two digits are the same and equal to the original’s minus one, and whose last two digits are the same and equal to the half of the original’s. Find that number.

[Ghost]

Im too lazy to think

Lol

[kyoshiro029]

Okay so there's 2 numbers in the beginning and let just say that as "X" so the X when squared will produce 4 number and that number let say XX will become XY hence the lesser 1 so from XX it turmed into XY, or if you say that the lesse 1 would turn the equation to YY then that original number would be half of the last 2 digits in the sqaured number

The answer is 88

88x88 = 7744 hence 77 is lesser by 1 in the original number and the last 2 digit is half of the original number

:3

[kyoshiro029]

actually xD i didn't use any of those things i said... what i did was just take numbers and randomly use process of elimination

if its a number that can produce 4 digits when squared then.... by elimination 11 and 22 are out since they can only produce a 3 digit number when square

and if the last 2 digits of the squared number are exactly the same.... then 33, 44, 55, 66, 77, and 99 are out... so whats left are 0 and 8... since the value of 00 is still 0 xD i excluded that cause you said the first 2 digits are lesser numbers of the original number... then all that's left is 88

of course i checked it with the last thing you said.... 44 is half of 88 :3

hence i concluded that the answer to your problem is 88

*nice nice xD i like a few brain teaser