Quant by Arun Sharma

MBA CAT 2024
2378



10. what is the highest power of 3 available on the expression 58!-38!
a.17
b.18
c.19
d.none of these



58!/3=19
58!/9=6
58/27=2

58! can be written as 58!=3^27*x

similarly 38!=3^17*y

58!-38!=3^27*x-3^17*y
=3^17(3^10*x-y)

so 3^17 is the highest common factor of 3!!
2379

HI Guys,

Please help me in this series :

1) 1 + 1/3 +1/6 + 1/10 + 1/15 +...............

a) 2 b) 2.25 c) c) 3 d) 4


2) 1/2 + 1/6 + 1/12 + 1/20 + .........+ 1/156 + 1/182

a) 12/13 b) 13/14 c) 14/12 d) 13/12

2380

3) 1/2 + 1/6 + 1/12 + 1/20 + .........+ ......infinity

a) 12/13 b) 1.1 c) 14/12 d) 1

2381
the highest power on 990 that will exactly divide 1090! is
a.101
b.100
c.108
d.109
e.110



my ans is 108. .. . .
2382

hi puys....
plz help me in solving following questions...

1) What is the total no. of divisors of the no. 12^33*34^23*2^47 ?

(a) 4658
(b) 9316
(c) 2744
(d) None of these


2) What is the remainder when (1!)^3+(2!)^3+(3!)^3+(4!)^3+.......(1152!)^3 is divided by 1152 ?

(a) 125
(b) 225
(c) 325
(d) 205


3) How many two-digit numbers less than or equal to 50 have the product of the factorials of their digits less than orequal to the sum of the factorials of their digits?

(a) 17
(b) 16
(c) 15
(d) None of these


4) M is a two digit no. which has the property that the product of factorials of its digits>sum of factorials of its digits. How many values of M exist?

(a) 56
(b) 64
(c) 63
(d) None of these

Thanks in advance 😃

2383

1) What is the total no. of divisors of the no. 12^33*34^23*2^47 ?
(a) 4658
(b) 9316
(c) 2744
(d) None of these

Answer

12^33==>2^66.3^33
34^23==>2^23.17^23
2^47 as it is
Therefore 2^136.3^33.17^23
meaning no of factors (137)(34)(24)=111792

I think it should be none of these

2384

Please help me with this:

Find the max of 'n' such that
42*57*92*91*52*62*63*64*65*66*67
is perfectly divisible by 42^n.

a)4
b)3
c)5
d)6

2385

i think it should b none of these........
it should be 137*34*24=111792....

2386

@saurav chetri...

i think d answer is (b)3....

becoz to find out n...we mst first factorise 42 which gives us 2,3,7.
now make out d pairs of 2,3,7 frm d series by factorising each number............

2387
Please help me with this:

Find the max of 'n' such that
42*57*92*91*52*62*63*64*65*66*67
is perfectly divisible by 42^n.

a)4
b)3
c)5
d)6


Taking out only 2's, 3's and 7's

7*2*3*3*2*2*7*2*2*2*3*3*7*2^6*2*3

2^13*7^3*3^5

42=2*3*7

So, 3 42's are present. So, option (B)
2388
Please help me with this:

Find the max of 'n' such that
42*57*92*91*52*62*63*64*65*66*67
is perfectly divisible by 42^n.

a)4
b)3
c)5
d)6

look dude...the answer wud b 3..... frm the expression u can clearly see that one power is already present in the expression n for other powers . our approach wud b as follows... since 42= 2x3x7 n since in the expression we see that the factor 7 appears for the least number of times, hence answer= no. of 7s in the expression...
2389
hi puys....
plz help me in solving following questions...


2) What is the remainder when (1!)^3+(2!)^3+(3!)^3+(4!)^3+.......(1152!)^3 is divided by 1152 ?

(a) 125
(b) 225
(c) 325
(d) 205

since 1152 = 2^7 x 3^2 and every term in this expression, except the first three terms is completely divisible by 1152 thereby resulting in zero remainder, hence the remainder in this case wud be because of first 3 terms i.e 1!)^3+(2!)^3+(3!)^3=225
so answer is (b) 225
2390
hi puys....
plz help me in solving following questions...

1) What is the total no. of divisors of the no. 12^33*34^23*2^47 ?

(a) 4658
(b) 9316
(c) 2744
(d) None of these


2) What is the remainder when (1!)^3+(2!)^3+(3!)^3+(4!)^3+.......(1152!)^3 is divided by 1152 ?

(a) 125
(b) 225
(c) 325
(d) 205


3) How many two-digit numbers less than or equal to 50 have the product of the factorials of their digits less than orequal to the sum of the factorials of their digits?

(a) 17
(b) 16
(c) 15
(d) None of these


4) M is a two digit no. which has the property that the product of factorials of its digits>sum of factorials of its digits. How many values of M exist?

(a) 56
(b) 64
(c) 63
(d) None of these

Thanks in advance :)


1.d

12^33*34^23*2^47 = (2^2 * 3)^33 * (2*17)^23 * 2^47
= 2^(66+23+47) * 3^33 * 17 ^23 = 2^ 136 * 3^33 * 17 ^23

hence N = (136+1) *(33+1) *(23+1)

i.e none of these

2.b:225
1152= 4*4*4*2*9

hence deciding factor must be first 3 terms after 3 terms all terms are divisible by 1152

(1!)^3+(2!)^3+(3!)^3 =225
rem(225/1152) = 225

3.d

Let two digit number in format of ab

a! * b!
No for this condition satisfied
10 - 19 ---- i.e 10
20,20,40,50 --- i.e 4
21,31,41 ----- i.e 3
22 ----- i.e 1
total 18

4.c : 63

condition is a! * b! > a! + b! which is against of a! * b!
for a! * b! 10- 19 ---------10
20,30,40 ......,80,90 i.e 8
21,31,41,......81,91 i.e 8
22----------------1
total 10+8+8+1 = 27

10-99 --- i.e 90 numbers

hence 90-27=63
2391
HI Guys,

Please help me in this series :

1) 1 + 1/3 +1/6 + 1/10 + 1/15 +...............

a) 2 b) 2.25 c) c) 3 d) 4


2) 1/2 + 1/6 + 1/12 + 1/20 + .........+ 1/156 + 1/182

a) 12/13 b) 13/14 c) 14/12 d) 13/12


1. a : 2

1 + 1/3 +1/6 + 1/10 + 1/15 +...............
sn = 1 +3 +6 +10 +15 ........tn
sn = 1 +3 +6 +10 ...........tn-1 + tn

subtract to each others
tn = 1+ 2+3 +4+5 .......n = n*(n+1)/2

thus tn of terms = 1/n(n+1)/2 = 2/n(n+1) = 2 * [(n+1 - n)/n(n+1)]
= 2 * [ 1/n - 1/n+1]

take
t1 = 2 * [ 1/1 -1/2]
t2 = 2*[1/2-1/3]
t3= 2*[1/3 -1/4]
:
:
tn= 2*[1/n -1/n+1]

sum = t1 + t2 +t3 +...tn = 2*[1-1/n+1]

series in infinity

sn = 2*[1- 1/infinty] = 2*[1-0] =2


2.b :: 13/14

1/2 + 1/6 + 1/12 + 1/20 + .........+ 1/156 + 1/182
1/2 + 1/6 + 1/12 + 1/20 + .........+ 1/2*tn
= 1/2[1+1/3+1/6+10+....+1/tn] -----------same as in question 1
here tn = 91
n=13 :::: 91 -78 = 13
hence tn = n(n+1)/2

tno = 2 * tn = n(n+1)

1/tno = 1/n(n+1) = 1/n - 1/n+1
1/t1 = 1-1/2
1/t2 = 1/2 - 1/3
:
:
1/t13= 1/13-1/14

sn = 1/t1 +1/t2 + 1/t3.....1/t13 =1-1/14 = 13/14
2392
3) 1/2 + 1/6 + 1/12 + 1/20 + .........+ ......infinity

a) 12/13 b) 1.1 c) 14/12 d) 1


1. d : 1

already posted the approach,kindly follow that
2393
Please explain how to solve:
1.The remainder when no 123456789101112..........484950 is divided by 16 is-
a.3 b.4c.5d.6
ans-d

2. 333^555 + 555^333 is divisible by
a.2 b.3 c.37 d.11 e.all of these
ans-d

1=>the last 4 digit shuld divisible by 16.
4950mod 16=6
2=>the no. is divisible by2,3,37..
plz check your ans.
2394

Please help me out with this one:

Distance bet Varanasi and Lucknow is 220.
2 buses start from these towns towards each other.
They can meet halfway if first bus starts 2 hrs earlier than the second.
If they start simultaneously, they meet in 4 hrs.
Find their speeds?

2395
Please help me out with this one:

Distance bet Varanasi and Lucknow is 220.
2 buses start from these towns towards each other.
They can meet halfway if first bus starts 2 hrs earlier than the second.
If they start simultaneously, they meet in 4 hrs.
Find their speeds?


Let speed of buses are V1 and V2

Case1 : If they start simultaneously, they meet in 4 hrs.

4(V1+V2) = 220
V1+ V2 = 55 ------------1

case2 :They can meet halfway if first bus starts 2 hrs earlier than the second

110/V1 - 110/v2 = 2
1/V1 - 1/V2 = 1/55 ---------2

V2 = 55/2 * (root5 -1)
V1= 55/2 * (3- root5)
2396

hi puys. . . .pls help me. . ..

1.Three Mangoes,four guavas and five watermelons cost Rs. 750. Ten watermelons, six mangoes and 9 guavas cost Rs. 1580. What is the
cost of six mangoes, ten watermelons and four guavas?
a) 1280
b) 1180
c) 1080
d) Cannot be determined

2.Iqbal dealt some cards to Mushtaq and himself from a full pack of
playing cards and laid the rest aside. Iqbal then said to Mushtaq
"If you give me a certain number of your cards, I will have four times
as many cards as you will have. If I give you the same number of
cards, I will have thrice as many cards as you will have". Of the given
choices, which could represent the number of cards with iqbal?
a)9
b)31
c)12
d)35
3. Find the general term of the GP with the third term 1 and the seventh
term 8.
a)(2^3/4)^n-3
b)(2^3/2)^n-3
c)(2^3/4)^3-n
d)(2^3/4)^2-n
e) None of these

2397

Please help me out with this one:

Set of 15 different words is given.
In how many ways is it possible to choose a subset of not more than 5 words?

a) 4944
b) 4^15
c) 15^4
d) 4943