Example 1: Let \(A\) be the set of prime numbers less than 100. We multiply all the elements of \(A\) to obtain a number \(B\). With how many consecutive zeros will \(B\) end?

Solution: From 1 to 100, there are 2 prime numbers 2, 3, 5, 7, 11,…..97. Out of these numbers only one number is even and only one is containing power of 5, hence total number of zeroes at the end of the product is 1

Example 2: Let \(X\) be the set of integers {1, 7, 13, 19, 25, 31,…. 361, 367} and \(Y\) be a subset of \(X\) such that the sum  of no two elements of y is 368. Find the maximum possible number of elements in \(Y\).

Solution: The numbers in the brackets are in \(AP\), where sum of  first and last is 368, sum of second and second last term = 368,  Out of first and last only one can be selected Out of second and second last only one can be selected \(X\) contains 61 terms, hence total number of term in \(Y = 31\)

Example 3: Let \(p, q, r\) are 3 natural numbers such that \(p + q + r = 3t + 7\), where \(t\) is a positive integer. For a given \(t\), find the minimum possible value of \({p^2} + {\rm{ }}{q^2} + {\rm{ }}{r^2}\) 

Solution: We know that if \(a + b\) is given, their product is maximum when both the numbers are equal and for the same condition, \({a^2} + {b^2}\) will be minimum.

Now in this case, \(p, q\) and \(r\) can not be equal , we will take \(p , q\) and \(r\) as close as possible.

\(p = t + 2, q = t + 2 \) and \(r = t + 3\)

Now \({p^2} + {q^2} + {r^2} = {\rm{ }}3{t^2} + {\rm{ }}14t + 17\)

Example 4: "DELHI SWEET HOUSE" sells laddus in boxes of different sizes. The laddus are priced at Rs.5 per laddu up to 400 laddus. For every additional 10 laddus. The price of the whole lot goes down by 10 paise per laddu. What should be the size of the box that would maximize the revenue?

Solution: Suppose there are \(400 +10x\) laddus, so the price of each laddu will be  \(\left( {5 - \frac{x}{{10}}} \right)\) Rs. Total price of the whole lot =\(\left( {5 - \frac{x}{{10}}} \right)\left( {400 + 10x} \right)\) 

\(= (50 - x)(40 + x)\)

Now sum of \((50 - x)\) and \((40 + x)\) is 90, hence their product is maximum when both are same. \(50 - x\) 

\(= 40 + x\) or \(x = 5\)

Total number of laddus = 450

Example 5: When a number \(p\) is divided by \(q\), remainder is 21, when \(2p\) is divided by \(q\), remainder is 3. Find the value of \(p\).

Solution: When \(p\) is divided by \(q\), remainder is 21. Hence when \(2p\) is divided by \(q\), remainder will be 42. But according to the question, remainder is 3, hence \(q\) is either 39 or 13. Also \(q\) can not be 13 because remainder is 21, hence \(q  = 39\).  

Example 6: Product of three numbers is 8960 and HCF of any two numbers is 4. Find LCM of three numbers.

Solution: Suppose the three numbers are \(4a\), \(4b\) and \(4c\), where \(a, b\) and \(c\) are co prime to each other. Given that \(4a.4b.4c = 8960\) \( \Rightarrow abc = {\rm{ }}140\)

Again LCM of the three numbers \(= 4abc\)

\( = {\rm{ }}4 \times 140{\rm{ }} = {\rm{ }}560.\)

Example 7: If two numbers \(M\) and \(N\) are defined as \(M = (2000)!\) and 

\(N{\rm{ }} = {\rm{ }}2002 \times 2003 \times 2004 \times 2005 \times 2006 \times 2007 \times 2008\). Find LCM of \(M\)  and \(N\).

Solution: The numbers 2002, 2004, 2005,2006 and 2008 are clearly composite, hence these numbers are already contained in and 2000!.

The number 2007 is multiple of 9, hence it is also contained in \(2000!\). The number 2003 is a prime number, hence LCM of the numbers \(M\) and \(N\) will be \(2003 \times 2000!\)

Example 8: Find the number of digits in the number \({1002^{2000}}\).

Solution: The number can be written as

\({(1000 + 2)^{2000}} = {1000^{2000}}{\left[ {1 + \frac{2}{{1000}}} \right]^{2000}}\)

= \({10^{6000}}{\left[ {1 + \frac{1}{{500}}} \right]^{500 \times 4}}\)

As we know that \({\left[ {1 + \frac{1}{n}} \right]^n}\) is very close to 2.7, when \(n\) is a large number, thus the approximate value of \({10^{6000}}{\left[ {1 + \frac{1}{{500}}} \right]^{500 \times 4}}\)will be \({10^{6000}}{(2.7)^4} \approx 50 \times {10^{6000}}\), which contains 6002 digits. Thus the number \({1002^{2000}}\) contains 6002 digits.

Example 9: Find the difference in hundreds and tens digit of the three digit number \(abc\) if \(\left[ {\frac{{100a + 10b + c}}{{a + b + c}}} \right]\)is minimum.

Solution:  Since the number is a three digit number, hence \(100a + 10b + c > 100 and a + b + c < 27\), hence to minimize the number \(\left[ {\frac{{100a + 10b + c}}{{a + b + c}}} \right]\), \(c\) must be maximum, because on increasing \(c\), percentage change in \(a + b + c\) is more than that in \(100a + 10b + c\). 

Thus \(c = 9\). Also a should be minimum, because on increasing \(a\), the number \(100a + 10b + c\) increases rapidly. Thus \(a = 1\).

Now \(\left[ {\frac{{100a + 10b + c}}{{a + b + c}}} \right]\)= \(\left[ {\frac{{109 + 10b}}{{10 + b}}} \right]\).

Now increasing \(b\) will increase denominator more rapidly as compared to numerator as

\(\frac{{10}}{{109}} < \frac{1}{{10}}\), thus b must be largest.

Hence \(a = 1, b = 9\) and \(c = 9\) 

Example 10: If \(a, b\) and \(c\) are three single digit numbers such that \((a + b + c)! = 90(abc)\), where \(abc\) is the product of \(a, b\) and \(c\).

Solution: From the given condition it is clear that \((a + b + c)!\) is a multiple of 90. 

Thus  \((a + b + c) = 6\) or \(7\) 

as \(6! = 720\) and \(7! = 5040\).

But if we take \(a + b + c = 7\), then

\(abc = \frac{{5040}}{{90}} = 56\), which is not possible. 

Thus \(a + b + c = 6\) and \(abc = \frac{{6!}}{{90}} = 8\)

So there is only one possibility \((a,\,b,\,c) = (2,\,2,\,2)\)