Number System - Questions & Discussions

MBA CAT 2022
844
teacher age is 65

65 is multiple of 13

so c)13

what approach did u applied?
Did u see the options,got their multiples and checked with those the given conditions?
845

What is the highest power of 12 that can divide 5^36 - 1 ?
Sorry don't have OA.

846

Highest power shud be 2 coz 4^3 is not able to divide 5^36 -1...

847
What is the highest power of 12 that can divide 5^36 - 1 ?
Sorry don't have OA.

highest power of 2 is 4 and 3 is 3..
so answer is 3
848
highest power of 2 is 4 and 3 is 3..
so answer is 3


limit shud be checked with 4 and not by 2....For 4,highest power will be 2..Hence highest power of 12 shud be 2...
849
What is the highest power of 12 that can divide 5^36 - 1 ?
Sorry don't have OA.



answer should be 2 for this one !
850
MangoMan Says
Is this possible? I doubt .... please provide an example if you have one ...

Abhinav90 Says
this thing cannot exist...



questions such as these are based on the identity concept.....if a n degree equation has more than n roots...then it becomes an identity and changes from being an n degree equation to being a n degree expression....which means that for any value besides it's roots...the expression becomes true...... the answer in such cases would be true for all real values of x ....
851
barclays_boss Says
questions such as these are based on the identity concept.....if a n degree equation has more than n roots...then it becomes an identity and changes from being an n degree equation to being a n degree expression....which means that for any value besides it's roots...the expression becomes true...... the answer in such cases would be true for all real values of x ....


are you talking about trivial eqn(or it is called non-trivial ?) that have more than one solutions(more than the degree)?

like this one: -4x^2 + 4x^2 =0 ?

Plz elucidate !
852

@abhinav....no these are not trivial eqn....although such expressions might have a trivial soln..but not always...actually they cease to be called as equations...they turn into identities and so are called expressions....for eg if you have a question as...for how many real values of 'a' will the equation (a-2) x^2 + 4a+2) x + 1 have three roots....also given that 'a' is a non negative integer....then the solution set is a =[0,infinity)

853
barclays_boss Says
@abhinav....no these are not trivial eqn....although such expressions might have a trivial soln..but not always...actually they cease to be called as equations...they turn into identities and so are called expressions....for eg if you have a question as...for how many real values of 'a' will the equation (a-2) x^2 + 4a+2) x + 1 have three roots....also given that 'a' is a non negative integer....then the solution set is a = but how will it have 3 roots?

For if we take a=3(in interval),x^2 + 14x+1 =0 will have two roots only...how come three ?
854

The age of Confusius' teacher was a number divisible by
a. 7
b. 9
c. 13
d. 15

=> 10a + b + 10b + a is a perfect square
= 11(a+b) a perfect square, hence a+b = 11 ---> 1
=> 10a + b - 10b - a is a perfect square
= 9(a-b) a perfect square, hence a-b can be 1 or 9, as per eqn 1 a-b shall be 1 so a=6 b=5

The age of Confusius' teacher was a number divisible by 13, as 65 is her age
855
S is a set of 10 consecutive two-digit integers such that the product of these 10 integers has the highest power of 2 contained in it. How many such sets S are possible?

a. 4
b. 8
c. 10
d. 24

Good question.

In order to maximize the power of 2 we will consider the set which has highest power of 2.
I.e., we will include multiples of 8

S={64,65,66,67,68,69,70,71,72,73}

S={63,64,65,66,67,68,69,70,71,72}

S={56,57,58,59,60,61,62,63,64,65}

S={55,56,57,58,59,60,61,62,63,64}

Only 4 possible sets.


why not 10?? pls xplain
S=
64,65,66,67,68,69,70,71,72,73
63,64,65,66,67,68,69,70,71,72
62,63,64,65,66,67,68,69,70,71
61,62,63,64,65,66,67,68,69,70
60,61,62,63,64,65,66,67,68,69
59,60,61,62,63,64,65,66,67,68
58,59,60,61,62,63,64,65,66,67
57,58,59,60,61,62,63,64,65,66
56,57,58,59,60,61,62,63,64,65
55,56,57,58,59,60,61,62,63,64
856
viveknitw Says
(7777....300 times )^300 % 19.

7^300 mod 19 * (100^150 + 100^149 + ...+1) mod 19

1*(5^150 + 5^149 + .. + 1) mod 19

(5^151 - 1) mod 76 => 72 mod 19 = 15

i dont understand my solution and i cant find any fault with my approach
..here goes
777777....300 times
when 7777...i times is divided by 19 the remainders are 7,1,17,6,10,12,13,4,,9,....
there are set of 18 remainders (including 0) . so 7777...300 times has the same remainder as
777....12 times.
that comes out to 11.

now 11^300 divided by 19 remainder be r
11^3 = 19k+1
so (19k+1)^100 divided by 19 will have a remainder of 1.
...
where is the fault.

isnt the process a lengthy one!
is there anyway to find 11 directly ?(euler nnumber etc...........)


= 77(10^298 + 10^296 + 10^294 + ................+ 10^2 + 1) % 19
= 1(5^149 + 5^148 + 5^147 +......................+ 5 + 1) %19, bold in GP with a = 1, r =5 and n = 150, hence the summation reduced to (5^150-1)/4
= ((5^150 -1)/4)%19
= (5^150 -1)%76, E(76) = 36
= (5^6 -1)%76
= (45-1)%76
= 44/4
= 11
I'm getting difference ans, pls check.. any observations kindly let me know..
857
= 77(10^298 + 10^296 + 10^294 + ................+ 10^2 + 1) % 19
= 1(5^149 + 5^148 + 5^147 +......................+ 5 + 1) %19, bold in GP with a = 1, r =5 and n = 150, hence the summation reduced to (5^150-1)/4
= ((5^150 -1)/4)%19
= (5^150 -1)%76, E(76) = 36
= (5^6 -1)%76
= (45-1)%76
= 44
I'm getting difference ans, pls check.. any observations kindly let me know..

could you please explain what you mean by E(76) = 36.....

also, there is a mistake in your first step..... 7777...300 times is 77(10^148+10^146+10^144+.....+10^2+1) {77(10^2+1) will give be 7777 so i need to do only till 10^148}

also checked the repetitiveness of divisibility by 19 in the case of 777777 (tried till 777777...20 times 😛 :p) and remainder is coming to 11 using that logic.......
858
viveknitw Says
61^59 % 59 ?

61^59 mod 59 =
2^59 mod 59 =
32* (2^6)^9 mod 59 = 32 * 125^3 mod 59 =
32 * 343 mod 59 = 2
remainder = 2


= 61^59 % 59
= 2^59 % 59, e(59) = 58
= 2. 2^58 % 59
= 2
859
= 77(10^298 + 10^296 + 10^294 + ................+ 10^2 + 1) % 19
= 1(5^149 + 5^148 + 5^147 +......................+ 5 + 1) %19, bold in GP with a = 1, r =5 and n = 150, hence the summation reduced to (5^150-1)/4
= ((5^150 -1)/4)%19
= (5^150 -1)%76, E(76) = 36
= (5^6 -1)%76
= (45-1)%76
= 44
I'm getting difference ans, pls check.. any observations kindly let me know..

could you please explain what you mean by E(76) = 36.....

also, there is a mistake in your first step..... 7777...300 times is 77(10^148+10^146+10^144+.....+10^2+1) {77(10^2+1) will give be 7777 so i need to do only till 10^148}

also checked the repetitiveness of divisibility by 19 in the case of 777777 (tried till 777777...20 times 😛 :p) and remainder is coming to 11 using that logic.......


1. Euler(76) = 76(1-1/2)(1-1/19) = 36
2. Pls check,
77 * 10^6 = 77000000
77 * 10^4 = 770000
77 * 10^2 = 7700
77 * 10^0 = 77
-------------------------
= 77777777
-------------------------
77(10^298 + 10^296 + 10^294 + ................+ 10^2 + 1)
= 777777....300 times
as u can c from the above ex 10^6 gives 8 (7s), 10^4 gives 6 (7s) likewise 10^298 gives 300 (7s)

puys pls clarify whether,
((5^150 -1)/4)%19 = (5^150 -1)%76
I have a doubt only in this step of my soln
860


puys pls clarify whether,
((5^150 -1)/4)%19 = (5^150 -1)%76
I have a doubt only in this step of my soln


No... (5^150-1)%76 will be 4 times of %19....
861

Plz solve this:
if 2x+3y=10 maximise x^5.y^6

862
1. Euler(76) = 76(1-1/2)(1-1/19) = 36
2. Pls check,
77 * 10^6 = 77000000
77 * 10^4 = 770000
77 * 10^2 = 7700
77 * 10^0 = 77
-------------------------
= 77777777
-------------------------
77(10^298 + 10^296 + 10^294 + ................+ 10^2 + 1)
= 777777....300 times
as u can c from the above ex 10^6 gives 8 (7s), 10^4 gives 6 (7s) likewise 10^298 gives 300 (7s)

puys pls clarify whether,
((5^150 -1)/4)%19 = (5^150 -1)%76
I have a doubt only in this step of my soln

apologies for the 10^148 thingie......also cud u explain the Euler concept....not familiar with it (commerce background : :)

also am not sure but wont x mod 19 (where x is some number) give a remainder between 0 and 18, while x mod 76 will give a remainder between 0 and 75.....how can i say then that (x/4) mod 19 = x mod 76?

mebbe what i can do is find the remainder of x mod 76 and divide by 4 to get (x/4) mod 19..... just guessing.....in this case that will give me the answer....

x being 5^150-1

x mod 76 = 44

meaning x = 76p +44
or x/4 = 19p + 11 giving a remainder of 11......

@Sahana Kavitha: request you to explain your solution step by step......am confused how u made the following steps....
= (5^150 -1)%76, E(76) = 36
= (5^6 -1)%76
= (45-1)%76

please do explain......
863

Guys pls help me i dont understand how to approach quant section my verbal is good, di is also fine but everytime i switchover to quant,n i feel like im on a war with my small n tiny head....pls help.suggest some good way out