Let's learn a simple yet a powerful concept before we answer this query.
Let R be the remainder when N is divided by D1.
Now the remainder when N is divided by D2, where D2 is a factor of D1 is either
1. R ( if R <D2) or
2. R1, where R1 is the remainder obtained when R is divided by D2 ( if R>D2)
Remember the converse is also true.
Let's look at an example here. ( let me give the converse example)
Let 23 be the remainder obtained when a number N is divded by 30.
Now the remainder obtained when the number is divided 90 must be either 23 or 30×1+23 or 30×2+23 i.e the remainder will be of the form 30p+23 where p =0 or 1 or 2( I think this is clear to everyone)
Let's apply this rule to the follwoing question.
1. Last two digits of N = 3^999.
All of us know that when a number is
- Divided by 10 the remainder is the unit digit
- Divided by 100 the remainder is the last two digits of the number
- Divided by 1000 the remainder is the last four digits of the number and so on.
3999 % 2= (27333) % 25 = (2333)%25 = [(102433) x 23] % 25
Remainder obtained when 1024 is divided by 25 is 24 but we can take it as -1 also (always take the smallest remainder irrespective of sign)
So finally the problem becomes -1 x 23 %25= -8 =17 ( finally give the positive remainder)
So when our number is divided by 25 the remainder is 17
So when it is divided by 100 the remainder R must be of the form 25p+17
Similarly when N is divided by 4 the remainder is 3.I am taking 4 here because the other factor of 100 is 4.
Hence 25p+17 must leave a remainder 3 when divided by 4. i.e p+1 must leave a remainder 3 when divided by 4. Hence the value of p is 2.
Therefore R = 25p+17 = 67. Hence the number N ends with 67.
That's it folks. This example reveals the power of this simple rule.
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