CSAT Solved Papers/ 2024/Q48

2024 CSAT — Q48

Quant Number theory 2.5 marks Medium

If the sum of the two-digit numbers AB and CD is the three-digit number 1CE, where the letters A, B, C, D, E denote distinct digits, then what is the value of A?

A 9 Answer
B 8
C 7
D Cannot be determined due to insufficient data

Worked rationale

Write the addition column by column. $\overline{AB} + \overline{CD} = \overline{1CE}$ means

(10A + B) + (10C + D) = 100 + 10C + E.

Cancel $10C$ from both sides:

10A + B + D = 100 + E.

Now use carries. Units column: $B + D = E + 10c_1$ with carry $c_1 \in \{0,1\}$ . Tens column: $A + C + c_1 = C + 10c_2$ , i.e. $A + c_1 = 10c_2$ . Hundreds: the result’s hundreds digit is $1$ , so $c_2 = 1$ .

Then $A + c_1 = 10$ . Since $A \le 9$ , we need $c_1 = 1$ , giving $A = 9$ . (A units carry is required: $B + D \ge 10$ .)

A consistent assignment exists, e.g. $98 + 27 = 125$ ( $A{=}9, B{=}8, C{=}2, D{=}7, E{=}5$ , all distinct), so $A$ is determined.

Answer: (a) 9.

Why the other options miss

B

missed a case: allows $c_1 = 2$ or mishandles the tens carry, wrongly admitting $A = 8$ .
C

an arithmetic slip: a carry slip — forgets the hundreds digit forces $c_2 = 1$ , so the tens equation is misread and $A$ comes out too small.
D

solved the wrong question: thinks the five free letters leave $A$ open, missing that the place-value structure alone pins $A$ regardless of the other digits.

Specialist insight

The deciding move is reading the tens column with the matching $C$ : $A + C + c_1$ must produce a tens digit of $C$ again, so $A + c_1$ is a clean multiple of $10$ . With the hundreds digit fixed at $1$ ( $c_2 = 1$ ), this forces $A + c_1 = 10$ , and since $A \le 9$ the units must carry ( $c_1 = 1$ ), pinning $A = 9$ uniquely — independent of $B, D, E$ . Cryptarithms are solved by carries, not by guessing digits; the repeated $C$ in column two is the structural key here.

The trap, in one line

The repeated $C$ in the tens place forces $A + c_1 = 10$ ; with the hundreds digit $1$ ( $c_2=1$ ) and $A\le9$ , the units must carry, so $A=9$ — uniquely, whatever $B,D,E$ are.

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