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A bomb squad consisting of three members, person C had to defuse the bomb while other members (A,B) had the task to detect the bomb. The bomb squad was summoned to the bank to defuse 2 bombs located inside the two of the 50 rooms of the bank.

The bank was divided into two sections, room number 1-25 were in section A and 26-50 were located in section B. For efficiency, person A went to the section A and person B went to the section B to detect the bomb. They detected the three possible rooms with the bomb and came back to report the room numbers to the person C. BUT, on the way back they forgot the room numbers of three rooms. However, person A knows that the multiplication of all three room numbers is 84, and person B knows that sum of all three room numbers is 92.

Person C knows that the sum of the room numbers in which the bomb is located is 53. Because of lack of time, Person C can only check 4 rooms of the whole bank (consider negligible time is taken during traversing the bank).

Can the person C defuse the bomb in time? If not, how many rooms do he need to check in order to defuse the bombs?

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Since B's sum is 92, one or all the terms is/are even, and at least one is less than 31:

$(25+d)+(25+e)+(25+f)=92$
$d+e+f=17$

The greatest room no could be 39 at most and 32 at least.

Based on C's sum, the rooms must be 26 and 27 unless one is from A and the other from B. If the former is the case, B's third possible number must be 39. Otherwise, $(26-g)+(25+h)=53$, making $h-g=2$. Since the possible factors of 84 less than 26 allowing $g$ to be less than 24 are 3, 4, 6, 7, 12, 14 and 21, $25+h$ can be 50, 49, 47, 46, 41, 39 or 32, but due to what we learned from B, the duos can only be (32, 21), (39, 14) or (26, 27). Just search one room from each possible duo until you find a bomb, and then go to the room corresponding to the counterpart of the bomb room.

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Three different numbers that multiply to 84, none of which is more than 25, are 1,4,21; 1,7,12; 1,6,14; 2,3,14; 2,6,7; or 3,4,7. So the room number is one of 1,2,3,4,6,7,12,14,21. The other is one of 53 minus each of those, viz 32,39,41,46,47,49,50. (1 and 2 are thus impossible.) The sum of 92 includes at least 26+27=53 besides the correct room so the latter is at most 39. So the possibilities are 39,14; 32,21. But reject 39,14 because there's no room to make them sum to 92. Thus, the bombs would seem to be in 21 and 32.

But

that assumes a bomb is in a room below 26. It's possible the bombs are in 26 and 27, and B found the possibilities 26,27,39. Thus, checking three rooms will suffice: 26, 32, and one more. But I recommend firing A and B, and hiring replacements who will remember the room numbers better.

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    $\begingroup$ I think you cannot reject 39,14 possibility because 26+27+39=92 so it is possible for B to have these as the possible rooms. $\endgroup$
    – I'm Nobody
    Commented Feb 27, 2022 at 12:52
  • $\begingroup$ Ah, right, @I'mNobody $\endgroup$
    – msh210
    Commented Feb 27, 2022 at 14:28

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