Can a number be partitioned into parts satisfying this condition? Let $a,b,c,d,e,f$ be distinct decimal digits, with $ab,cd,ef$ being two-digit numbers formed from these digits (so that, for example, $ab=10a+b$).
How may I prove that the expression $ab+cd+ef$ assumes all integer values between $69$ and $255$? Thanks.
 A: I am not sure that the question statement is accurate. Here is the best I could come up with.
Let K be a number satisfying:
$$K=ab+cd+ef$$
We want to prove:
$$K_{min}  <= K <= K_{max}$$
given that a,b,c,d,e,f are non zero, positive, non-duplicate. Also, ab,cd and ef are two digit numbers that is none of them is $<9$ (however, $10,11$, and $12$) is not valid because $a$, $c$ and $e$ would be equal. 
$$K=(10a+b)+(10c+d)+(10e+f)$$
$$K=10(a+c+e)+(b+d+f)$$
Since a,b,c,d,e and f are different, positive, the minimum value for K would use {a,b,c,d,e,f}={1,2,3,4,5,6,7} in some order to get the minimum value of K.
To have a minimum value of K, we need the value of (a+c+e) to be minimum since it gets multiplied by 10. We choose $a=1,c=2,e=3$. We can choose the numbers b, d, f to be less than 9 and as small as possible. 
We choose $b=4, d=5, f=0$ to get
$$K_{min}=10(1+2+3)+4+5+0=69$$
Please note that here we made the assumption that one of the 2nd digits could be zero (as in the case for 10+24+35 -See comment below).
To find the maximum value of $K$, we need to choose values of a,c and e as large as possible since their sum is multiplied by 10. We use $a=7, b=8, and c=9$. We choose $b, d , f$ as large as possible, so $b=6, d=5, f=4$.
$$K_{max}= 10(7+8+9)+6+5+4=255.$$
$$69<=K<=255$$
****Note:Thanks for user sulspatz for correcting an error in my answer****
