2022 CSAT — Q5

Quant Number theory 2.5 marks Hard

An Identity Card has the number $ABCDEFG$ , not necessarily in that order, where each letter represents a distinct digit (only $1, 2, 4, 5, 7, 8, 9$ ). The number is divisible by $9$ . After deleting the first digit from the right, the resulting number is divisible by $6$ . After deleting two digits from the right, the resulting number is divisible by $5$ . After deleting three digits from the right, the resulting number is divisible by $4$ . After deleting four digits from the right, the resulting number is divisible by $3$ . After deleting five digits from the right, the resulting number is divisible by $2$ . Which of the following is a possible value for the sum of the middle three digits of the number?

A 8 Answer
B 9
C 11
D 12

Worked rationale

All seven distinct digits come from $\{1,2,4,5,7,8,9\}$ — so all seven are used and their sum is $1+2+4+5+7+8+9 = 36$ , already divisible by $9$ . So ” $ABCDEFG$ divisible by $9$ ” is automatic.

Now decode each deletion (deleting from the right keeps the leading prefix):

$ABCDE$ divisible by $5$ $\Rightarrow E \in \{5\}$ (no $0$ in the set), so $\boxed{E = 5}$ .
$AB$ divisible by $2$ $\Rightarrow B$ even.
$ABCD$ divisible by $4$ $\Rightarrow$ last two digits $CD$ divisible by $4$ $\Rightarrow D$ even.
$ABCDEF$ divisible by $6$ $\Rightarrow F$ even (and the prefix digit-sum divisible by $3$ ).
$ABC$ divisible by $3$ $\Rightarrow A+B+C \equiv 0 \pmod 3$ .

The even digits available are $\{2,4,8\}$ — exactly three of them, and $B, D, F$ must all be even, so $\{B,D,F\} = \{2,4,8\}$ . That leaves the odd digits $\{1,7,9\}$ for $A, C, G$ .

$ABCDEF$ divisible by $3$ : $A+B+C+D+E+F = 36 - C\text{-odd partners} \dots$ more directly, since the full sum is $36$ , $A+B+C+D+E+F = 36 - G$ , which is $\equiv 0 \pmod 3$ iff $G \equiv 0 \pmod 3$ . Among $\{1,7,9\}$ only $G = 9$ works, so $\boxed{G = 9}$ , leaving $\{A,C\} = \{1,7\}$ .

$CD$ divisible by $4$ with $D$ even and $C \in \{1,7\}$ : test $D \in \{2,4,8\}$ — $72/4 = 18$ ✓ ( $C=7, D=2$ ) and $12/4 = 3$ ✓ ( $C=1, D=2$ ). Both need $D = 2$ , so $\{B,F\} = \{4,8\}$ .

$ABC$ divisible by $3$ : $A + B + C = (1+7) + B = 8 + B$ , need $\equiv 0 \pmod 3 \Rightarrow B = 4$ (gives $12$ ), so $F = 8$ .

Two valid cards: $7412589$ ( $A{=}7,C{=}1$ ) and $1472589$ ( $A{=}1,C{=}7$ ). The middle three digits are $C, D, E$ (positions 3–4–5):

$7412589 \Rightarrow C{+}D{+}E = 1 + 2 + 5 = 8$ ;
$1472589 \Rightarrow 7 + 2 + 5 = 14$ .

$14$ is not offered; $8$ is a possible value.

Answer: (a) 8.

Why the other options miss

B

invented an assignment that doesn’t survive the checks: never actually holds against all six divisibility filters simultaneously.
C

an arithmetic slip: mis-pins $E$ or sums the wrong three positions (e.g. $D+E+F$ ).
D

solved the wrong question: reads “middle three” as $B,C,D$ or chases the always-divisible total ( $36$ ) instead of the positional digits.

Specialist insight

The lever is forced positions before search: divisibility by $5$ pins $E=5$ , the three even slots $B,D,F$ exhaust $\{2,4,8\}$ , and the divisible-by- $3$ chain forces $G=9$ . Only then is the search over $\{A,C\}$ trivially small. Note the question asks for a possible value, not the value — the configuration is not unique ( $8$ or $14$ ), and the examiner deliberately offers only one of the two reachable sums.

The trap, in one line

$E=5$ , evens $\{B,D,F\}=\{2,4,8\}$ , $G=9$ forced; valid cards give middle-sum $8$ or $14$ , and only $8$ is offered $\Rightarrow$ (a).

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