**Example 1**. The sum of squares of the digits of a three-digit positive number is 146, while the sum of the two digits in the unit’s and the ten’s place is 4 times the digit in the hundred’s place. Further, when the number is written in the reverse order; it is increased by 297. Find the number.

**Example 2**. First 100 natural numbers are written on a black board. Two persons *A* and *B* are playing a game of putting + and – sign one by one between any two consecutive integers out of these 100 natural numbers. Both *A* and *B* are free to put any sign (+ or –) anywhere provided there is no sign placed already between the two natural numbers. At the end, when the signs are put between all such two consecutive natural numbers, result is calculated. If the result is even, then *A* wins and if the result is odd, then *B* wins. Who will win?

**Example 3**. Let *ABC* be a triangle and *D* be a point on the segment *BC* such that *DC* = 2*BD*. Let *E* be the mid-point of *AC*. Let *AD* and *BE* intersect in *P*. Determine the ratios *BP*/*PE* and *AP*/*PD*.

Of course, you can find the solution just below, but it is highly recommended that, you first try to solve it on your own.

Just remember the words of Paul Halmos, who says “**the only way to learn mathematics is to do mathematics**”.

**Solution 1**. Let the number be *abc* i.e., 100*a* + 10*b* + *c* whose digits are *a*, *b* and *c*.

According to the symmetry of the question,

*a*^{2}+*b*^{2}+*c*^{2}= 146; (**1**)*b*+*c*= 4*a*; (**2**)- and 100
*c*+ 10*b*+*a*– 100*a*– 10*b*–*c*= 297. (**3**)

From relation (3), we get

- 99(
*c*–*a*) = 297 - or,
*c*–*a*= 3 - or, c =
*a*+ 3 (**4**)

Now using (4) in (2), we find that

*b*+*a*+ 3 = 4*a*- or,
*b*= 3*a*– 3 (**5**)

Finally, substituting the values from (4) and (5) in (1), we have

*a*^{2}+ (3*a*– 3)^{2}+ (*a*+ 3)^{2}= 146- or,
*a*^{2}+ 9*a*^{2}– 18*a*+ 9 +*a*^{2}+ 6*a*+ 9 = 146 - or, 11
*a*^{2}– 12*a*– 128 = 0

Only acceptable value *a* from here is 4.

Hence, *b* = 3*a* – 3 = 3 × 4 – 3 = 9 and *c* = *a* + 3 = 4 + 3 = 7.

Therefore, the required number is 497.

**Solution 2**. Among 100 natural numbers, there are 50 odd and 50 even numbers.

So, 100 consecutive natural numbers = set of 50 odd and 50 even numbers.

Whatever sign we put between two odd numbers, resultant of 50 odd numbers = Even, and similarly, whatever sign we put between two even numbers, resultant of 50 even numbers = Even.

Hence the net result is either Even + Even or Even – Even.

In any case, net result = Even. Hence *A* will win.

We can also see here that + sign or – sign does not matter here. Irrespective of the signs put by *A* or *B*, *A* is always going to win.

**Solution 3**. Let *F* be the midpoint of *DC*, so that *D*, *F* are points of trisection of *BC*.

Now in triangle *CAD*, *F* is the mid-point of *CD* and *E* is that of *CA*.

Hence *CF*/*FD* = 1 = *CE*/*EA*.

Thus *EF* || *AD*. Hence we find that *EF* || *PD*.

Hence *BP*/*PE* = *BD*/*DF*. But *BD* = *DF*. We obtain *BP*/*PE* = 1.

In triangle *ACD*, since *EF* || *AD* we get *EF*/*AD* = *CF*/*CD* = 1/2.

Thus *AD* = 2*EF*. But *PD*/*EF* = *BD*/*BF* = 1/2.

Hence *EF* = 2*PD*. Therefore, *AD* = 2*EF* = 4*PD*.

This gives *AP* = *AD* − *PD* = 3*PD*.

We obtain *AP*/*PD* = 3.