You need to check 2 more cases.
CASE-IV: 9(11a+b) = 10x9
11a+b = 10 ⇒ a=1 and b=-1 Not PossibleCASE-V: 9(11a+b) = 19x18
11a+b = 38 ⇒ a=3 and b=5
a+b+c = 19
c = 19 - (a+b) = 19 - (3+5) = 11 Not Possible
I always miss one thing or the other..........
]]>CASE-IV: 9(11a+b) = 10x9
11a+b = 10 ⇒ a=1 and b=-1 Not Possible
CASE-V: 9(11a+b) = 19x18
11a+b = 38 ⇒ a=3 and b=5
a+b+c = 19
c = 19 - (a+b) = 19 - (3+5) = 11 Not Possible
(100a+10b+c) = (a+b+c)[sup]2[/sup]
⇒ 99a+9b = (a+b+c)[sup]2[/sup] - (a+b+c)
⇒ 9(11a+b) = (a+b+c)(a+b+c-1)
Since LHS of the equation is a Multiple of 9 and that the Maximum Value of RHS (a+b+c) can be (9+9+9) = 27, we have the following three cases...
CASE-I: 9(11a+b) = 9×8
11a+b = 8 ⇒ a=1 and b=-3 Not Possilbe (b can't be -ve)
CASE-II: 9(11a+b) = 18×17
11a+b = 34 ⇒ a=3 and b=1
a+b+c = 18
c = 18 - (a+b) = 18 - (3+1) = 14 Not Possilbe (c=14 is not possible)
CASE-III: 9(11a+b) = 27×26
11a+b = 78 ⇒ a=7 and b=1
a+b+c = 27
c = 27 - (a+b) = 27 - (7+1) = 19 Not Possilbe (c=19 is not possible)
Hence, we can conclude that NO SUCH THREE DIGIT NUMBER IS POSSIBLE!
I think it'd be much easier to show this for higher digit numbers!
]]>Consider for example..
Find a number which is square of its last digit (digit in unit's place)?
Ans (quite obvious) is 25 & 36 and these can be determined using almost the same logic as above, for solving the equation!
The required number can only be 'one/two digit' number can also be shown..
An 'n digit' number 'a' would satisfy the above question only if "square of the sum of the digits of the GREATEST 'n digit' number EQUALS/EXCEEDS the number'!
I don't agree with this.
For example, take 199. 1+9+9 = 19 and 19² = 361 > 199, so even though the highest 3-digit number is greater than the sum of its digits, there are still some 3-digit numbers that are less (which means there's no reason why there couldn't be some that are equal).
Testing exhaustively shows that there aren't any, but you couldn't say that without trying them all.
However, you could use a similar argument to prove there aren't any 5-digit (or higher) numbers with that property though.
The highest sum-of-digits-squared number would be made by 99999.
5x9 = 45, and 45² = 2025.
The highest sum-of-digits-squared number is lower than the lowest 5-digit number (10000), and so there can't be any 5-digit numbers equal to the squared sum of their digits.
Greatest 'one digit' number is 9..
9^2 = 81 > 9... Satisfies!
Greatest 'two digit' number is 99..
9+9 = 18
18^2 = 324 > 99... Satisfies!
Greatest 'three digit' number is 999..
9+9+9 = 27
27^2 = 729 < 999.. Doesn't Satisfy!!
& so on...
Comments are most welcome!!
Regards...
I think i've found a nice Mathematical way to crack the above problem but i'll find it a bit difficult to post it coz i don't know how to code for equations n i'm surfing through my Nokia6600 rather than computer..
Here it goes...
0 = 0 (yes)
1 = 1 (yes)
4 = 4 (no)
9 = 9 (no)
1 + 6 = 7 (no)
2 + 5 = 7 (no)
3 + 6 = 9 (no)
4 + 9 = 13 (no)
6 + 4 = 10 (no)
8 + 1 = 9 (yes)
1 + 0 + 0 = 1 (no)
1 + 2 + 1 = 4 (no)
.
.
.
etc etc etc