RMO Practice

Binary numbers from 000000 - 111111 represent 0 to 63 in base 10. Exactly $^{6}C_3 = 20$ numbers have three 0's and three 1's. Being a binary situation in respect of 0 and 1 there are exactly half of 20, i.e. 10 pairs that add - up to binary 111111 ( $63_{10}$ ). Hence they all add up to 630.

$We\quad are\quad interested\quad in\quad the\quad numbers\quad less\quad than\quad 64.\\ Notice\quad that\quad 64={ 2 }^{ 6 }\\ Therefore,\quad all\quad the\quad numbers\quad less\quad than\quad 64\quad have\quad utmost\quad 6\quad digits\quad in\quad their\quad binary\quad representation.\\ Now,\quad we\quad have\quad to\quad place\quad three\quad ones\quad within\quad six\quad places.\\ There\quad are\quad _{ 3 }^{ 6 }{ C } = \frac { 6! }{ 3!.3! } = 20\quad ways\quad of\quad doing\quad so.\\ Therefore,\quad there\quad are\quad 20\quad numbers\quad less\quad than\quad 64\quad which\quad have\quad exactly\quad three\quad ones\quad in\quad their\quad binary\quad representation.\\ Now,\quad if\quad we\quad fix\quad one\quad 1\quad in\quad the\quad leftmost\quad place\quad i.e.\quad 1\quad \_ \quad \_ \quad \_ \quad \_ \quad \_ \quad ,\quad we\quad have\quad to\quad fill\quad the\quad remaining\quad two\quad ones\quad in\quad five\quad places.\\ There\quad are\quad _{ 2 }^{ 5 }{ C }=\frac { 5! }{ 2!.3! } =10\quad ways\\ So\quad we\quad can\quad conclude\quad that\quad there\quad are\quad ten\quad numbers\quad with\quad 1\quad as\quad the\quad leftmost\quad digit.\\ Now\quad that\quad we\quad need\quad the\quad sum,\quad the\quad sum\quad is\quad equal\quad to\quad 111111\times 1010\\ Note\quad :\quad { 1010 }_{ 2 }\quad =\quad { 10 }_{ 10 }\\ For\quad the\quad multiplication\quad part,\quad { 1\quad 1\quad 1\quad 1\quad 1\quad 1 }_{ 2 }\quad \\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \times \quad \quad { 1\quad 0\quad 1\quad 0 }_{ 2 }\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad ----------------\\ \qquad \qquad \qquad \qquad \qquad \quad { 1\quad 0\quad 0\quad 1\quad 1\quad 1\quad 0\quad 1\quad 1\quad 0 }_{ 2 }\\ Now,\quad we\quad have\quad to\quad convert\quad the\quad result\quad to\quad base\quad 10.\\ \quad { 1001110110 }_{ 2 }=\boxed{630}\\$