CSAT Solved Papers/ 2025/Q58

2025 CSAT — Q58

Quant Number theory 2.5 marks Hard

The 55-digit number PQRSTPQRST (all distinct digits) is such that T≠0T \ne 0. PP is thrice TT. SS is greater than QQ by 44, while QQ is greater than RR by 33. How many such 55-digit numbers are possible?

  1. A 3
  2. B 4 Answer
  3. C 5
  4. D 6

Worked rationale

Translate the conditions, then enumerate the few feasible chains.

P=3T,Q=R+3,S=Q+4=R+7.P = 3T,\qquad Q = R + 3,\qquad S = Q + 4 = R + 7.

Bound TT: T≠0T \ne 0 and P=3T≤9⇒T∈{1,2,3}P = 3T \le 9 \Rightarrow T \in \{1,2,3\}, giving P∈{3,6,9}P \in \{3,6,9\}.

Bound RR: S=R+7≤9⇒R≤2S = R + 7 \le 9 \Rightarrow R \le 2, so R∈{0,1,2}R \in \{0,1,2\}, giving the chains

(R,Q,S)=(0,3,7), (1,4,8), (2,5,9).(R,Q,S) = (0,3,7),\ (1,4,8),\ (2,5,9).

Now pair each (T,P)(T,P) with each (R,Q,S)(R,Q,S) and keep only all-distinct 55-tuples:

(R,Q,S)(R,Q,S)(T,P)=(1,3)(T,P)=(1,3)(T,P)=(2,6)(T,P)=(2,6)(T,P)=(3,9)(T,P)=(3,9)
(0,3,7)(0,3,7)P=3=QP=3=Q ✗{6,3,0,7,2}\{6,3,0,7,2\} ✓T=3=QT=3=Q ✗
(1,4,8)(1,4,8)T=1=RT=1=R ✗{6,4,1,8,2}\{6,4,1,8,2\} ✓{9,4,1,8,3}\{9,4,1,8,3\} ✓
(2,5,9)(2,5,9){3,5,2,9,1}\{3,5,2,9,1\} ✓T=2=RT=2=R ✗S=9=PS=9=P ✗

Four surviving numbers: 63072, 64182, 94183, 3529163072,\ 64182,\ 94183,\ 35291.

Answer: (b) 4.

Why the other options miss

  • A
    missed a case: misses one valid chain (commonly the (1,4,8)(1,4,8) with (T,P)=(2,6)(T,P)=(2,6) branch), under-counting.
  • C
    missed a case: admits one clashing tuple (e.g. keeps (0,3,7)(0,3,7) with (T,P)=(1,3)(T,P)=(1,3) where P=QP=Q) without enforcing distinctness.
  • D
    solved the wrong question: counts all 3×3=93 \times 3 = 9 chains minus a couple, ignoring most of the distinct-digit collisions.

Specialist insight

The whole item is constraint-driven pruning: the relations collapse five free digits to two short ladders — T∈{1,2,3}T\in\{1,2,3\} and R∈{0,1,2}R\in\{0,1,2\} — leaving only 3×33\times3 candidate combinations. Marks are won or lost entirely on the distinctness check: five of the nine combinations die on a digit collision (P=QP=Q, T=RT=R, T=QT=Q, S=PS=P). Lay the 3×33\times3 grid out explicitly and test each cell for repeats rather than computing in your head; that disciplined sweep yields exactly four.

The trap, in one line

The relations force T∈{1,2,3}T\in\{1,2,3\} and R∈{0,1,2}R\in\{0,1,2\} (99 combinations); enforcing all-distinct digits kills five, leaving 44.

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