Consider the numbers 1 2 100 on the blackboard Consider the

Consider the numbers 1, 2, ..., 100 on the blackboard. Consider the following rule: We select two numbers, say, x, y from the ones written on the blackboard, we delete them and add the number y on the blackboard. If we apply the rule 99 times then we will be left with only one number on the blackboard. What are all the possible numbers we might end up with? If we apply the rule 98 times then we will be left with two numbers on the blackboard. What are all the possible pairs of numbers we might end up with?

Solution

A.

In this case when we have numbers

1, 2, 3, 4,......... 100, when we remove two number and replace it with the multiple of those two numbers the product of all these numbers will remain constant.

which means the last on the board will be:

= 100*99*98*97*96*......4*3*2*1 = 100!

B.

Last two remaining digits on the board will be the two digits which can give us the factorial of 100, because we know that final digit will give us 100!.

which means possible pairs will be:

= (99!, 100) or (100!, 1), or (98!, 99*100) or (97!, 96*95*94) or (66!, 67*68*69*70*....99*100) .... and many more.

Here ! denotes factorial function.

Comment below if you have any doubt.

To under this question better, take 5 consecutive numbers.

1, 2, 3, 4, 5

S1: remove 1 and 2, then replace with 1*2 = 2

2, 3, 4, 5

S2: remove 2 and 3, then replace with 2*3 = 6

6, 4, 5

S3: remove 4 and 5 replace with 4*5 = 20

6, 20

S4: remove 6 and 20, then replace with 6*20 = 120

final number will be = 120 = 5!

for B part possible pairs will be in this case:

(6, 20) Or (5, 24) Or (120, 1) or (60, 2) Or (30, 4)

which means number which can give us 5! after multiplication.