Quant Boosters - Kamal Lohia - Set 6

[sumit agarwal]

@kamal_lohia

Here the expression has odd number of factors.
Which means it is a perfect square.

Now if N is even expression is odd(e+e+o =o)
And if n is odd expression is still odd(o+o+o=o)
Thus, the square is of an odd positive number.

Putting expression equal to 11^2 satisfies it
And n comes out to be 9
As n(n+3)=108

But ,how to ensure that there is no other solution besides this for higher squares????

[sumit agarwal]

@kamal_lohia

S =2^2=4
R=2×3=6
Q =2^2×3=12
P=2^2×3×5=60

SO The sum of digits is 6.

[sumit agarwal]

@kamal_lohia

the gcd cant be greater than the difference of the two consecutive terms of the sequence

Diff= 2n+1
Equating to 41 gives n =20 , nd the numbers are 410 and 451...whose gcd is 41..

59 on equating doesnt give two numbers which hv gcd as 59

Hence 41.

[sumit agarwal]

@kamal_lohia

X=[10^20000/10^100+2]
X= 10^20000-2^200/ 10^100+2
X=2^200(A-1)/2(B+1)
X=2^199(some expression)

Hence highest power is 199

Is my solution right?

[sumit agarwal]

@kamal_lohia

Remainder of (P-3)!/p=(p-1)/2
Remainder of (p-4)!×(p-3)/p =(p-1)/2
Rem(p-4)!/p = (1-p)/6

Hence rem( 6*(p-4)!/p)= rem((1-p)/p)= 1

[sumit agarwal]

@kamal_lohia

12100 =2^2 ×5^2 ×11^2
110 =2×5×11

Number of factors of 12100 less than 110 =
3×3×3-1/2= 13
Numbr of factors of 110 less than 110 =7

Hence total factors of 12100 which are less than 110 but do not divide it are 13-7 =6

[sumit agarwal]

@kamal_lohia

What is the logic to be used for tgis prob? Could anyone guide?

[sumit agarwal]

@kamal_lohia

It will satisfy for 11th digit x=0,7

[sumit agarwal]

@kamal_lohia

Could anyone suggest how to estimate that the series like
10^2 -> 19^2 gives 1, 2, 3
20^2 -> 29^2 gives 4, 5, 6, 7, 8

3 + 5 + 7 + 11 ....... which gives 100.
So It continues like this but fails for higher squares...which gives answer as 76 ...because the series of integers generated is not continuous.
So what is the right way to solve it?

[zabeer]

@sumit-agarwal

In such scenarios like finding distinct values of [x^2/n] where x can be from 1, 2, 3 ... n
[1^2/n], [2^2/n] ... [(n/2)^2/n] will yield all numbers from 0 to [n/4] (means [n/4] + 1 distinct integers)
Then the next set (from [(n/2 + 1)^2/n] till [n^2/n] will be all different integers (means [n/2] distinct integers)
So the number of distinct integers would be [n/2] + [n/4] + 1

if n = 100,
number of distinct integers would be [100/2] + [100/4] + 1 = 76

if n = 2014,
number of distinct integers would be [2014/2] + [2014/4] + 1 = 1511

if n = 13
number of distinct integers would be [13/2] + [13/4] + 1 = 10

Just trying to generalize a solution shared by Kamal sir (Quant Boosters - Set 1 - Q2).

You can try out with various numbers (may be smaller numbers) so that this can be verified.