Qu@ntose

When the natural numbers 1,2,3,4,5,.......,500 then the digit 3 is used n times in this way. The value of n is- (a)100 (b)200 (c)300 (d)400 SOLUTION:-   We have 3 places where 3 can come, unit digit, tens digit and hundred digit.        _   _    _ Now, if we fix 3 at unit digit place,    _  _  3 At hundred digit place, there can be a total of 5 numbers(0,1,2,3,4) and at tens digit place there can be any of the 10 numbers(0-9) Total ways = 5*10 = 50------(1) Similarly when we fix 3 at tens digit place,  _  3  _  Again we will have 50 numbers (5*10)------(2) Similarly fixing 3 at hundred digit place, 3  _  _ Now we will have 10*10 = 100 ----------(3) Adding these 3 equations we get, 50+50+100 = 200 times ANS ## Counting in such manner had worked because we have to count the number of times 3 appeared, fixing 3 at the unit place will also give 333, fixing 3 at tens place will also give 333 and we will also get 333 when we fix 3 at hundred digits place. We have counted 333 three times whi

Arithmetic Progression:-   Arithmetic Progression is a sequence in which common difference is constant. e.g:- 1,2,3,4,5,6,7 Here the common difference is 1. Consider a series 4,8,12,16,20,.............84 First term = 4 Last term = 84 Common difference = 4 General term of AP is defined as, first term + (number of terms - 1)*Common Difference Tn = a + (n - 1)*d From the above sequence, a = 4 Tn = 84 d = 4 Let's find the number of terms in the above sequence 84 = 4 +(n - 1)*4 4*n = 84 n = 21 Sum of AP:- [[[        Sn = n/2(2a +(n - 1)*d)       ]]] Now, Sn = n*(a + [(n-1)/2]*d) [[[       Sn = n*Middle term      ]]] Arithmetic Mean(AM):- if the numbers are in AP then Arithmetic mean is nothing but the middle term(if the number of terms is odd). On the other hand, arithmetic mean is nothing but the average of the numbers. i.e, Sum of numbers/Total numbers e.g:- 1,3,5,7,9 AM = (1+3+5+7+9)/5 = 5 On the other hand  if we look it as an AP then we can directly say that AM is 5. Question- Is

The digits of a three-digit number are in G.P. when the digits of this number are reversed and this resultant number is subtracted from original number the difference comes out to be 792. The actual number is- (a)842 (b)961 (c)421 (d)931 SOLUTION- Approach-1:- Let the number be xyz. Now,according to the question, (100x+10y+z) -  (100z+10y+x) = 792 99x-99z=792 (x - z) = 8 --------(1)  Now we can use options to solve this equation  As, (x - z) = 8 only option (b) and (d) satisfies the criteria and among these options only option (d) have numbers in G.P. So, option (d) will be the answer. Approach-2:- As the numbers are in G.P., only options (a),(c) and (d) satisfies the criteria. Option (a), 842 - 248 = 594 Option (c), 421 - 124 = 297 Option (d), 931 - 139 = 792 So option (d) will be the answer.

If  x = (9+4√5)^48 = [x] + f, where [x] is defined as integral part of x and f is a fraction, then find the value of x(1-f)? SOLUTION- x = (9+4√5)^48 x = 48c0*(9)^48 + 48c1*(9)^47 *(4√5) + 48c2*(9)^46 *(4√5)^2+........+48c47*9*(4√5)^47 +                  48c48*(4√5)^48----------------(i) All the terms, that have even power of (4√5) will be integers and the other terms will be irrational numbers. Now, x = [x] + f Let x' = (9 - 4√5)^48 Now (9 - 4√5) < 1 Hence, (9-4√5)^48 < 1 x' = (9-4√5)^48 = f' x' = 48c0*(9)^48 - 48c1*(9)^47 *(4√5) + 48c2*(9)^46 *(4√5)^2 - 48c3*(9)^45 *(4√5)^3+.......+             48c48*(4√5)^48------------(ii) Adding eqn (i) and (ii), we get x + x' = [x] + f +f'           =2*[48c0*(9)^48 + 48c2*(9)^46 *(4√5)^2 + .................+ 48c48*(4√5)^48] From above result we can see that [x] + f + f' is an Integer but f and f' are fractions, Hence, the only possible value of (f + f') = 1 Now, f' = (1 - f) = (9 - 4√5)^48 ∴ x(1 - f)