2023 CSAT — Q29

Sum the digit-sums of every integer from $10$ to $100$. Handle $10$–$99$ by place value, then add $100$ separately.

Tens digits ($10$–$99$): the tens digit runs $1, 2, \dots, 9$, each holding for a block of $10$ numbers, so each appears $10$ times:

Units digits ($10$–$99$): in each tens-block the units run $0$–$9$; there are $9$ blocks, so each unit digit $0$–$9$ appears $9$ times:

So digits over $10$–$99$ sum to $450 + 405 = 855$.

Add $100$: its digits are $1, 0, 0$, contributing $1$.

Answer: (b) 856.

Split the count by place value: tens digits appear $10$ times each ($450$), units digits appear $9$ times each across the nine blocks ($405$). The single most common leak is the boundary — the range is ”$10$ to $100$” inclusive, so $100$ contributes its $1$, turning $855$ into $856$. The deadliest distractor is exactly $855$, engineered for the candidate who does the place-value work perfectly but ignores the endpoint.