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Short-cut Techniques to solve Mathematics of level 10+2


This contains 5 problems each with an unique solving method.
These problems are collected from OLYMPIAD problems.
Problem-1. Show that 10000. . . . . 001 with 1961 zeros is composite(not prime).
Sol- The number = 10^(1962) +1 = {10^(654)}^3 + 1
Now by induction,we know
 a^n + b^n is divisible by a+b,where n is odd integers.
So, the given no is divisible by (10^(654) + 1).

Problem-2.  A printer numbers the pages of a book starting with 1 & uses 3189 digit in all.Find the number of pages of the book.
Sol:- No of digits used for numbering pages 1 to 9 is 9.
Similarly 10 to 99 is= 90*2=180, 100 to 999 is=900*3=2700.
Number of digits will remain after using 9+180+2700(=2889) digits =3189 - 2899 =300.
The digits can be used for numbering 300/4=75 pages, i.e.,from 1000 to 1074. Hence the book has 1074 pages.

Problem-3. The four digit number aabb is a square. Find it.
Sol:- let aabb= n^2
Then n^2 = 1100a + 11b=11(100a + b)
Since, n^2 is divisible by 11^2 ,we see that 11 divides (a+b).
Let us assume a+b=11. Since,n^2 is a perfect square so b can't be 0,1,2,3,5,7,8. Checking the remaining we see that 7744 = 88^2.

Problem-4. Find the number of solutions(positive integers) of the equation
 3x + 5y=1008.
Sol:- x,y belongs to the set of natural numbers(N). Then 3|5y => 3|y , i.e. y=3k for all k belongs to N.
Thus 3x+15k=1008
=>  x+5k = 336
=>  5k ≤ 335
=> k ≤ 67
Ans is=67.

Problem-5. If a,b,c are positive integers s.t. abc+ab+bc+ca+a+b+c=1000. Find the value of a+b+c.
Sol:- abc+ab+bc+ca+a+b+c=1000.
=> a(bc+b+c)+a+b+c+bc+1=1001
=> a(bc+b+c+1)+(bc+b+c+1)=1001
=> (a+1)(b+1)(c+1)=13.7.11
=> a=12, b=6, c=10
=> a+b+c=28.

Note:- There are several techniques to solve these kinds of problems.I have tried to give a time-consuming method to solve these.Hope you will like it.And you can apply these techniques to solve some more problems in exams.keep visiting for the rests.