Albert, Bernard and Cheryl again

There is no problem with this question. Amazing question by the way! Did it take you long to create it?

Anyway here's the solution:

Based on the first statement, $\textbf{Albert tells Bernard, "I don't know her birth year."}$ , it implies that all the numbers with a unique number of factors could not be Cheryl's birth year. This excludes numbers: $64,1,36$ , we would call this set ${P}$ . For the set: ${64+48,1+48,36+48}$ ,we would define it as set ${S}$ for the second part.

Based on the second statement, $\textbf{Bernard tells Albert, "I don't know her birth year."}$ , it implies that any number $+48$ that has a unique number of factors are out. This eliminates numbers: $$ . This second statement also implies that any number $+48$ that shares a number of factors with $n$ other numbers, of which $n-1$ of these numbers are inside set ${S}$ . This eliminates numbers for Cheryl's birth year: ${16, 33, 52,96,32,73 }$ , we would call this set ${C}$

Now, based on the third statement, $\textbf{Albert tells Bernard, "Now I know her birth year."}$ . This implies that of the numbers possible based solely from the number Albert was given, all of the numbers except $1$ has been eliminated. So what we have to do is to find the factors of the eliminated numbers. We do not have to find the factors of the numbers in set ${P}$ since these numbers have a unique number of factors. $\text{Factors of set }{C}:\text{ }{5, 4, 6, 12, 6, 2}$

Checking the factors one by one, we remain with the number $\boxed{81}$ , with $5$ factors.

Here is the link for the number of factors from numbers $1$ to $150$ :

1 to 50

51 to 100

101 to 150

Here is my C++ code

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#include<stdio.h>
#include<vector>
vector<int> v1[25];
vector<int> v2[25];

int num_of_factor(int n)
{
         int count = 0;
    for(int i = 1 ; i<n;i++)
    {
        if(n%i==0)
            count++;
    }
    return count;
}

void printf1(int n)
{
    for(int i = 0;i < v1[n].size();i++)
    {
        printf("%d  ",v1[n][i]);
    }
    printf("\n");
}

void printf2(int n)
{
    for(int i = 0;i < v2[n].size();i++)
    {
        printf("%d  ",v2[n][i]);
    }
    printf("\n");
}

int main()
{
    int temp;
    scanf("%d",&temp);
    int i;
    for(i=0;i<100;i++)
    {
        int temp = num_of_factor(i);
        v1[temp].push_back(i);
    }
    for(i=0;i<100;i++)
    {
        int temp = num_of_factor((i+48)%100);
        v2[temp].push_back(i);
    }
    for(i=0;i<25;i++)
    {
        printf("%d : ",i);
        printf1(i);
    }
    for(i=0;i<25;i++)
    {
        printf("%d : ",i);
        printf2(i);
    }
}

In figure the initial are number of factor and the rest following are the number having total that number of factor.

First 1 to 25 for Albert and next 1 to 25 for Bernald and as from the code 48 is added.

Now, as per the question as soon as Bernald tell Albert that he doesn't know solution Albert comes to know the answer

So the number cheryl give to Albert have two number

As from the figure the possible is (16,81) and (48,80)

Bernald says he doesn't know the answer so so answer can be 80,48,81

Albert now knows the answer so Cheryl must have told Albert number 4.

So, Albert comes to know that answer is 81 i.e. 1981

Only two numbers below 100 have four factors: 16 and 81 (both fourth powers) 16+48=2^6, the only number with 6 factors, so it's not 16.

Great question