Topic - Quant Mixed Bag

Solved ? - Yes

Source - ]]>

Topic - Quant Mixed Bag

Solved ? - Yes

Source - ]]>

f (1 – x) = 4^(1 – x) / [4^(1 – x ) + 2]

= 4 / (4 + 2 * 4^x)

= 2 / (2 + 4^x)

f(x) + f(1 – x) = [4^x/(4^x + 2)] + [ 2/(4^x + 2)] = 1

f(1/100) + f(1 – 1/100) = f(1) + f(99/100) = 1

f (2/100) + f( 1 – 2/100) = f(2) + f(98/100) = 1

Similarly f(49/100) + f(50/100) = 1

Hence f(1/100) + f(2/100) + f(3/100) ... f(99/100) = 49

f ’(x) = 2ax + b.

2ax + b = 0.

x = -b / 2a

At x = 5/4

b = -5, a = 2.

At x = 1,

a + b + c = 1.

2 – 5 + c = 1.

c = 4.

f(x) = 2x^2 - 5x + 4.

f(5) = 2(5)^2 – 5(5) + 4 = 29

f(x^5) = x^20 + x^15 + x^10 + x^5 + 1.

Let’s take x = 2.

f (x) = f(2) = 31.

f (x^5) = f(32) = 2^20 + 2^15 + 2^10 + 2^5 + 1

f(32) mod f(2) = (2^20 + 2^15 + 2^10 + 2^5 + 1) mod 31

= (2^5)^4 + (2^5)^3 + (2^5)^2 + (2^5) + 1 mod 31

= 1 + 1 + 1 + 1 + 1

= 5 ]]>

f ’(x) = 6x – 4

6x – 4 = 0

x = 2/3.

f(2/3) = 17/3. ]]>

Highest power = [n/p]+[n/p^2]+[n/p^3]……..

52 = [n/7] + [n/7^2] + [n/7^3]

The only things strike in our mind to split 52.

1 + 7 + 49 (not equal to 52)

6 + 42 (not equal)

7 + 49 (not equal)

6 + 42 < 52 < 7 + 49

6 + 43 < 52

6 + 46 = 52.

[n/7] = 46

n > 7 * 46.

n > 322.

[n/49] = 6.

n > 49 * 6.

n > 294.

For least possible number n = 322. ]]>

52 = [ n/13] + [n/169]

Let’s think , how 52 can be split.

3 + 39 < 52 < 4 + 52

3 + 49 = 52.

[n/13 ] = 49.

n > 13 * 49

n > 637.

[n/169] = 3

n = 169 * 3

n > 507.

For least value, n = 637. ]]>

Highest power of 2 = [100/2] + [100/2^2] + [100/2^3] + [100/2^4] + [100/2^5] + [100/2^6]

= 50 + 25 + 12 + 6 + 3 + 1

= 97.

But power of 2 is 3 here, so 97/3 = 32.

Highest power of 3 = [100/3] + [100/3^2] + [100/3^3] + [ 100/3^4]

= 33 + 11 + 3 + 1

= 48.

But power of 3 is 2 here, so 48/2 = 24.

24 < 32.

Power of 72 in 100! Is 24.

Power of 2 in 50! = [50/2] + [50/2^2] + [50/2^3] +[50/2^4]+[50/2^5]

= 25+12+6+3+1

= 47.

Power of 2 is 3 here.

47 / 3 = 15.

Power of 3 in 50!

[50/3]+[50/3^2]+[3^3]

= 16+5+1

= 21.

15 < 21.

Power of 24 in 50! is 15.

Highest power of 2 = [17/2] + [ 17/2^2] + [ 17/2^3] + [17/2^4]

8+4+2+1 = 17.

Highest power of 3 = [17/3] + [17/3^2]

5+1 = 6.

Highest power of 5 = [17/5] + [17/5^2] = 3.

Highest power of 7 = [17/7] + [17/7^2] = 2.

Highest power of 11 = [17/11] = 1.

Highest power of 13 = [17/13] = 1.

Highest power of 17 = [17/17] = 1.

17! = 2^17 x 3^6 x 5^3 x 7^2 x 11 x 13 x 17

Factors = (17+1)(6+1)(3+1)(2+1)(1+1)(1+1)(1+1)

= 18 x 7 x 4 x 3 x 2 x 2 x 2

= 126 x 96

= 12096. ]]>

= (100 x 99 x 98 ... 1) x (99 x 98 x 97... 1) x (98 x 97 x 96 ... 1) ... 1

= 100! x 99! x 98! x 97! ... 1!

= 1! x 2! ... 100!

1! to 4! = 0.

5! to 9! = 1 x 5 = 5.

10! to 14! = 2 x 5 = 10.

15! to 19! = 3 x 5 = 15.

20! to 24! = 4 x 5 = 20.

1! to 24! = 5(0+1+2+3+4) = 50.

25! To 49! = 5(6+7+8+9+10) = 200.

50! to 74! = 5(12+13+14+15+16)=350.

75! to 99! = 5(18+19+20+21+22)=500.

100! = 24.

Total = 50 + 200 + 350 + 500 + 24 = 1124.

= 5^101 (150 !) / (49! )

No of 5’s in 150! = [150 / 5] + [ 150 / 5^2 ] + [ 150 / 5^3 ]

= 30 + 6 + 1 = 37.

Total no of 5’s in numerator = 101 + 37 = 138.

No of 2’s in 150!

= [150 / 2] + [150 /2^2 ] + [ 150 /2^3 ] + [ 150/2^4 ] + [ 150 / 2^5 ] + [ 150 / 2^6 ] + [150 / 2^7]

= 75 + 37 + 18 + 9 + 4 + 2 + 1

= 146.

No of 2’s > No of 5’s

Hence no of zeroes at numerator = no of 5’s = 137.

Number of zeroes at denominator = [ 49 / 5 ] + [ 49 / 5^2 ]

= 9 + 1

= 10.

Total no of zeroes = 137 – 10 = 127.

]]>k+2 = 25n+1.

k+2 = 26,51,76,101,126,151,176,201.

But for 125, no of zeroes increase by 3.

So we need to remove 126.

k+2 = 26,51,76,101,151,176,201

k = 24,49,74,99,149,174,199.

Total possible values = 7. ]]>