CSAT Solved Papers/ 2023/Q59

2023 CSAT — Q59

Quant Data sufficiency 2.5 marks Hard

Consider a 33-digit number.

Question: What is the number?

Statement-1: The sum of the digits of the number is equal to the product of the digits.

Statement-2: The number is divisible by the sum of the digits of the number.

Which one of the following is correct in respect of the above Question and the Statements?

  1. A The Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone
  2. B The Question can be answered by using either Statement alone
  3. C The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone
  4. D The Question cannot be answered even by using both the Statements together Answer

Worked rationale

Statement-1 alone (digit-sum == digit-product): for digits a,b,ca,b,c (leading a≥1a\ge 1), a+b+c=abca+b+c = abc. A 00 digit forces product 0≠0 \ne a positive sum, so all digits are ≥1\ge 1; the only solution is the multiset {1,2,3}\{1,2,3\} (1+2+3=6=1⋅2⋅31+2+3 = 6 = 1\cdot 2\cdot 3). Its permutations 123,132,213,231,312,321123, 132, 213, 231, 312, 321 all qualify — not unique. Insufficient.

Statement-2 alone (number divisible by digit-sum): vast numbers of 33-digit Harshad numbers qualify. Insufficient.

Both together: restrict to permutations of {1,2,3}\{1,2,3\} (digit-sum 66) that are divisible by 66. Test:

  • 123/6=20.5123/6 = 20.5 ✗,   132/6=22\;132/6 = 22 ✓,   213/6=35.5\;213/6 = 35.5 ✗,
  • 231/6=38.5231/6 = 38.5 ✗,   312/6=52\;312/6 = 52 ✓,   321/6=53.5\;321/6 = 53.5 ✗.

Two survivors, 132132 and 312312 — still not unique even together.

Answer: (d) The Question cannot be answered even by using both the Statements together.

Why the other options miss

  • A
    thought it was enough when it wasn’t: imagines St-1’s a+b+c=abca+b+c = abc pins a single number, forgetting the six permutations of {1,2,3}\{1,2,3\}.
  • B
    thought either statement was enough when neither is: treats either condition as uniquely identifying the number.
  • C
    stopped at the first solution: spots that the number is a permutation of 1,2,31,2,3 divisible by 66 but stops at the first such number (132132), missing the second (312312).

Specialist insight

Two layers of non-uniqueness. St-1 narrows to the permutations of {1,2,3}\{1,2,3\} (the unique multiset with sum == product), but order is free. St-2 (divisible by digit-sum 66 ⇒\Rightarrow divisible by 66, i.e. even) keeps the even ones: 132132 and 312312. Because two numbers survive both statements, the answer is (d). The discipline is to enumerate every survivor of the combined conditions before declaring sufficiency — stopping at one (132132) is the engineered (c) trap.

The trap, in one line

St-1 ⇒\Rightarrow a permutation of {1,2,3}\{1,2,3\}; even ones (÷66) are 132132 and 312312 — two survivors, so not even together == (d).

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