The Baffling Bishop

Let the three departing clergymen's ages are $x$ , $y$ , and $z$ satisying $1 \leq x \leq y \leq z$ (because the permutation of the ages is not important).

There are $20$ integer solutions of the set of $x$ , $y$ , and $z$ (I used WolframAlpha here) : $\{ 1 , 1 , 2450 \}$ , $\{ 1 , 2 , 1225 \}$ , $\{ 1 , 5 , 490 \}$ , $\{ 1 , 7 , 350 \}$ , $\{ 1 , 10 , 245 \}$ , $\{ 1 , 14 , 175 \}$ , $\{ 1 , 25 , 98 \}$ , $\{ 1 , 35 , 70 \}$ , $\{ 1 , 49 , 50 \}$ , $\{ 2 , 5 , 245 \}$ , $\{ 2 , 7 , 175 \}$ , $\{ 2 , 25 , 49 \}$ , $\{ 2 , 35 , 35 \}$ , $\{ 5 , 5 , 98 \}$ , $\{ 5 , 7 , 70 \}$ , $\{ 5 , 10 , 49 \}$ , $\{ 5 , 14 , 35 \}$ , $\{ 7 , 7 , 50 \}$ , $\{ 7 , 10 , 35 \}$ , and $\{ 7 , 14 , 25 \}$ .

Take a look that $x + y + z$ is twice of the vicar's age. The vicar (of course) knows his age so he knows the exact value of $x + y + z$ . Note that in here, the vicar hadn't known the values of $x$ , $y$ , and $z$ . It proves that there are more than one possibility of $x$ , $y$ , and $z$ .

Luckily, there is exactly only one pair of set of $x$ , $y$ , and $z$ satisfy the conditions : $\{ 5 , 10 , 49 \}$ and $\{ 7 , 7 , 50 \}$ . The sum of them is $64$ which means the age of the vicar is $32$ .

Next, the bishop said that he was older than the three ages . There are 3 cases of bishop's age $p$ :

• if $p < 50$ , then there is no valid set of $x$ , $y$ , and $z$ satisfies the condition.
• if $p = 50$ , then there is exactly one set of $x$ , $y$ , and $z$ satisfies the condition : $\{ 5 , 10 , 49 \}$ .
• if $p > 50$ , then the two sets are satisfying the condition.

In here, bishop had given all the conditions to get the exact values of $x$ , $y$ , and $z$ . It has to be only one possibility of their values. Therefore, the value of $p$ must be $50$ so the values of $x$ , $y$ , and $z$ are $5$ , $10$ , and $49$ .

The answer of this question is the value of $p$ = $\boxed{50}$ .

In order for the vicar to not know the answer for the first time, there should be at least two distinct (a,b,c) combinations that sum up to the same number. If every combination is worked out, one can see such two sets:

(50, 7, 7) and (49, 10, 5) sets both have 2450 as their multiplication and 64 as their summation.

Since the bishop gave additional information to distinguish this set, we know that he's older than 49 and not older than 50. Therefore, bishop is 50

Let x,y,z be age of worshipers and v and b age of vicar and bishop respectively. Assume vicar knows his age :)

Vicar should be able to figure out the age unless there are two sets of solutions for x+y+z = 2v and xyz=2450. Enumerate all the possible solutions of xyz=250 and find out if there are multiple sets of solutions which results in same value for x+y+z.

Here we found (50,7,7) and (10,5,49) which adds up to 64. This is what confused the vicar so he needed a second clue. The second clue would help if the pope's age is 50.

NB: You need to enumerate the entire set of solutions for xyz=50 since the last clue of bishop helped reach a definite answer. So the max age of the worshiper should be consecutive number.

simple : xyz= 2450 , and x+y+z is twice the vicar's age , now 2450=[5x5x7x7x2] , ideally if we start grouping [since there are more variables] its like 30 permutations, but it reduces 3 trial and error steps, first group is 5x49x10 then 10x14x35 then 49x25x2 , now the bishop is elder than all the three so he must be atleast 50 , so answer is 50.

ok so , let's say that the three worshipers are a b c . And if we look at 2450 we obviously know that 50 goes in to 2450 . So I did 2450 / 50 = 49 . so I now knew that 49 x b x c = 2450 . so the b x c must make 50 . b = 5 and c = 10 . so now we know the age of the worshipers are 49 , 10 , 5 . Since 49 is the largest number . The number after 49 is 50 . So the bishop can be 50 or up . But guessing that he's the youngest possible so this makes him 50

The bishop is older than 49.

50 is not the only answer.

Factor of 2450 is 2 5 5 7 7, now let us consider the maximum case so the age of one person is 7 7=49, the second person age is 5 5=25 or 5*2=10 and third one is 2 or 5.

The reasonable age combinations :

Ages Sum

7 14 25 2450 46 49 10 5 2450 64 7 7 50 2450 64 35 35 2 2450 72 49 2 25 2450 76 7 70 5 2450 82 49 50 1 2450 100 35 70 1 2450 106 98 5 5 2450 108 98 25 1 2450 124

The fact that the vicar was initially stumped (inspite of knowing his own age) shows that there are at least 2 combinations yielding the same sum , that is 64 , i.e 49, 10 and 5 and 7,7 and 50. The bishop's statement that he is older than each of them shows that 49, 10 and 5 is the right answer since each would imply that their ages are unique. So the bishop's age being 50 is the minimal age (Bishops too are human in that they may not want to grow too old!!! :-) ) that is required to fulfil conditions as well as provide a kind of red herring for those proceeding purely verbally

Let's call the ages of the worshippers a,b,c

Let's call the Vicar's age V

From the given clues we know the following:

abc = 2 x 5 x 5 x 7 x 7 = 2450

a + b + c = 2V

So, a+b+c must be even

There's only one factor of two in the prime decomposition of 2450. It follows that only one of the three worshippers will have an even age, the other two will be odds. Keeping this constraint in mind we can write down all the possible different a+b+c values

2+25+49 = 76

10+5+49 = 64

50+7+7 = 64

2+35+35 = 72

10+7+35 = 52

98+5+5 = 108

2+175+7 = 184

14+25+7 = 46

70+5+7 = 82

2+245+5 = 252

14+35+5 = 54

Let's assume that the Vicar knows his age: in order for the vicar to not know the answer there should be at least two distinct (a,b,c) combinations that sum up to the same number and these are

10+5+49 = 64

50+7+7 = 64

At this point we use the same "trick" (and assume the Bishop knows his own age, just as a reminder let's call it B). If B were any value greater than 50 then the Bishop couldn't have cried "You should be able to figure out their ages now" because both of the two cases would have been acceptable. So, we just showed that B is less or equal (not greater) than 50. But just looking at the two possible cases we know that B is greater than 49, leaving B = 50 as the only option.

Call it beginner's luck, but I just multiplied 49*5 to get 245 and assumed that the third worshiper was 10 and the others were 49 and 5 years old respectively. Then, I just arbitrarily added one year to the oldest worshiper and scored the correct answer. Rather bizarre to me.

After looking at other solutions to this problem, I found that I interpreted the problem differently and still arrived at the same solution. I assumed that this was a trick question, and the vicar and bishop (and one other person) were observing their receding reflections in a mirror or other reflective surface. In that case:

bishop's age = b

vicar's age = v

third person's age = x

$\frac{b+v+x}{2} = v$

and

$b \times v \times x = 2540$

and

b > "each of them", referring to the vicar and the third person

Prime factorize 2540, and you get 1,2,5,5,7,7. The only possibility that works is 1,49,50 .