Topic - Quant Mixed Bag

Solved ? - Yes

Source - Elite's Grid Prep Forum ]]>

Topic - Quant Mixed Bag

Solved ? - Yes

Source - Elite's Grid Prep Forum ]]>

a) 23

b) 14

c) 5

d) 9 ]]>

1001 = 10^3+1

so make group of 3 and add with alternate sign from right side

so 000-096+abc-687+355

(-428 + abc) mod 1001 = 0

so abc = 428

so sum of digit = 4 + 2 + 8 = 14 ]]>

1001=10^3+1

so make group of 3

and add with alternate sign from right side

000 - abc + 674 - 307 + 1 = -abc+368 so abc=368 ]]>

first non zero digits from right side

34! = 5 * 6+4

so 2^6 * 6! * 4!

4 * 2 * 4 = 2

so a = 2

now number is div by 7,11,13 so div by 1001 so make group 3 and add with alternate sign from right side

000-000+5ab-643+609-618+847-140+604-cd9+799-232+295=2041-cd9

so 2041-cd9 mod 1001=0

so 39-cd9 mod 1001=0

so c=0 and d=3 ]]>

291 + 70 + 40 + b + 10a + 3

404 + ab/99

8 + ab/99

where ab is a 2 digit number

clearly ab can only be 91 , so only 1 pair ]]>

like in base 10, if a number is div. by 11 then sum of its digits(sign changed alternatively) is div. by 11 and vice-versa.

In base 7 k = 7, k+1 = 8. 4 is the divisor of 8. The alternating sum of the digits of 123abc231bca312cab is 1-2+3-a+b-c+2-3+1-b+c-a+3-1+2-c+a-b = 6 -(a +b +c) is divisible by 4

Now (a+b+c) can be 2,6,10,14, or 18

Solutions in each case is 6,28,36,15,1

total solution's = 6+28+36+15+1 = 86 ]]>

any number written p - 1 times div by p where p is prime

so only last 5 digits will be dere so 44444 mod 103=51 ]]>

now 2014^2015 mod 121 = (-43)^35 mod121

= -43 * (43 * 43)^17 mod 121

= -43 * (34)^17 mod 121

= -43 * 34 * (34 * 34)^8 mod 121

= -43 * 34 * (-54 * -54)^4

= -43 * 34 * (12)^4 mod 121

= -43 * 34 * 144 * 23 mod 121

= -43 * 34 * 23^2 mod 121

= -43 * 34 * 45 Mod 121

= -87 mod 121 = 34.

OR

78^2015 mod 121

= (1+77)^2015 mod 121

= 77C0 + 2015C1 * 77 + rest of the terms divisible by 11

= (1 + 2015 * 77 ) mod 121

= 34

N! = 2^(N-1)*k

N! mod 2^N = 2^(N-1) { k mod 2} =2^(N-1) ]]>

a. 9

b. 8

c. 10

d. 11 ]]>

OR

Just find number of coprime less than 30 = 30 * 1/2 * 2/3 * 4/5 = 8

]]>I. an odd number.

II. an even number.

III. a prime number.

IV. a perfect square.

a) II only

b) III and IV only

c) I and IV only

d) I only

e) None of these

= > x is even and x+1 is odd

so 1 is true and 2nd is false

now to decide where n is prime or not

try for first four natural numbers which give a product of 24 and therefore n =25, therefore n is not prime but could be prefect square

OR

n= (a-1) * (a) * (a+1)(a+2)

n= (a^2+a-1)(a^2+a-1)

so n is perfect square

OR

x = 2 * 3 * 4 * 5.

then 120 + 1 = 121 = 11^2

so option 3.

]]>3n+1 = 121 = 11^2

40 mod 40 = 0 ]]>

(a) 22

(b) 33

(c) 18

(d) 26

(e) none ]]>

Maximum occurs when the numbers are 7, 5, 4, 3, 2, 2.

To achieve min LCM look for max number of hcf among the 6C2 pairs.

So we need number in form of coprime.

Thus, M = 414 = 2 * 3^2 * 23, M has 2 * 3 * 2 = 12 divisors

M^2 has 3 * 5 * 3 = 45 divisors.

so 45 - 12 = 33 divisors

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