CSAT Solved Papers/ 2021/Q19

2021 CSAT — Q19

Quant Number theory 2.5 marks Medium

Integers are listed from 700700 to 10001000. In how many integers is the sum of the digits 1010?

  1. A 6
  2. B 7
  3. C 8
  4. D 9 Answer

Worked rationale

Every integer in [700,1000][700,1000] except 10001000 (digit sum 11) is a 33-digit number h t u‾\overline{h\,t\,u} with h∈{7,8,9}h\in\{7,8,9\}. Require h+t+u=10h+t+u=10, i.e. t+u=10−ht+u = 10-h, with t,u∈{0,…,9}t,u\in\{0,\dots,9\}.

  • h=7h=7: t+u=3⇒(0,3),(1,2),(2,1),(3,0)=4t+u=3 \Rightarrow (0,3),(1,2),(2,1),(3,0) = 4 ways.
  • h=8h=8: t+u=2⇒(0,2),(1,1),(2,0)=3t+u=2 \Rightarrow (0,2),(1,1),(2,0) = 3 ways.
  • h=9h=9: t+u=1⇒(0,1),(1,0)=2t+u=1 \Rightarrow (0,1),(1,0) = 2 ways.

Total =4+3+2=9= 4+3+2 = 9.

Answer: (d) 9.

Why the other options miss

  • A
    missed a case: handles only the h=7h=7 band or stops counting pairs once the obvious ones are found.
  • B
    off by one: drops a boundary pair (e.g. omits (3,0)(3,0) or (2,0)(2,0) where one digit is 00).
  • C
    off by one: misses exactly one ordered pair, often forgetting (0,1)(0,1) or (1,0)(1,0) in the h=9h=9 band.

Specialist insight

For “digit sum =S=S in a range” the move is to fix the leading digit and count ordered pairs for the remaining two — the number of solutions of t+u=kt+u=k with 0≤t,u≤90\le t,u\le 9 is just k+1k+1 (when k≤9k\le 9). So the counts are (10−h)+1(10-h)+1: 4,3,24,3,2. This formula avoids hand-listing and the classic fencepost slip of dropping the pair with a 00. Remember 10001000 is in the list but contributes nothing.

The trap, in one line

Pairs of t+u=10−ht+u=10-h number (10−h)+1(10-h)+1: 4+3+2=9⇒4+3+2=9 \Rightarrow (d).

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