Question:hard

A six-digit code is to be formed using 6 distinct numbers. The number in the first place is the square of a prime number in the third place. The numbers in the \(4^{\text{th}}\), \(6^{\text{th}}\), \(2^{\text{nd}}\) and \(1^{\text{st}}\) place are consecutive numbers. If all odd digits except 1 are present in the code, what is the sum of all the digits?

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Find which single-digit values can be a prime's square, then use the consecutive-number clue to fix four of the six digits.
Updated On: Jul 21, 2026
  • 17
  • 21
  • 34
  • 38
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: List the possible prime-square pairs.
The first-place digit equals the square of a prime digit placed third, and both must be single digits.
Only \(2^2=4\) and \(3^2=9\) give single-digit squares, so the pair (third place, first place) is either (2, 4) or (3, 9).

Step 2: Translate the consecutive-place clue into an equation.
The 4th, 6th, 2nd and 1st place digits increase by one each, in that reading order.
Writing the 4th place value as \(k\), the other three become \(k+1\), \(k+2\) and \(k+3\) for the 6th, 2nd and 1st places.

Step 3: Check the pair (2, 4).
Here the 1st place is 4, so \(k+3=4\) gives \(k=1\), placing 1, 2, 3 at the 4th, 6th and 2nd places.
The 6th place becomes 2, but the third place is also 2 in this pair, so a digit repeats and the code cannot be distinct.

Step 4: Check the pair (3, 9).
Here the 1st place is 9, so \(k+3=9\) gives \(k=6\), placing 6, 7, 8 at the 4th, 6th and 2nd places.
Together with the third place 3, the five fixed digits are 6, 7, 8, 3 and 9, all different, so this pair is valid.

Step 5: Apply the odd-digit condition to the last open slot.
Every odd digit other than 1, namely 3, 5, 7 and 9, must appear in the code.
Among these, 3, 7 and 9 are already used, so 5 is the only one missing and it must fill the fifth place.

Final Answer:
The six digits are 6, 7, 8, 3, 5 and 9 placed as worked out above, and their total is \(6+7+8+3+5+9=38\). \[ \boxed{38} \]
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