Number System - Questions & Discussions

MBA CAT 2022
121


2.define a no: k such thtit is the sum of the squires of the first m natural nos where m


the numbers of the form 8n - 1 ,8n satisfies the condition

so number of values = 6*2 = 12
122

Originally Posted by janvats View Post
find the remainder when (38^16!)^1777 is divided by 17

a. 1
b. 16
c. 8
d. 13

E(17) = 16.
(38^16!)^1777 mod17 = (38^16k)^1777 mod 17 = 1


Using remainder theroem
(38^16!)^1777 % 17
=> (4^16!)1777 % 17
=> (-1)^((16!^1777)/2) % 17
=> 1(ans), 16!^1777 is even with more than 1 zero (16! has 3 zeros, the same raised to the power 1777 has >= 1777 * 3 zeros)
123

Originally Posted by BIJAY PRASAD View Post

2.define a no: k such thtit is the sum of the squires of the first m natural nos where m

the numbers of the form 8n - 1 ,8n satisfies the condition

so number of values = 6*2 = 12


Naga,

Honestly i don't understand the question as well as your answer. Kindly elaborate..
124
Originally Posted by BIJAY PRASAD View Post

2.define a no: k such thtit is the sum of the squires of the first m natural nos where m


Naga,

Honestly i don't understand the question as well as your answer. Kindly elaborate..



question : for how many value of m
answer : the number of form 8n , 8n - 1 satisfy the condition

if u still cant understand pm me
125
question : for how many value of m


dude... i think the question says that we have to find a general form for all those numbers which can be expressed as the sum of first m natural numbers where m
corrections are invited!!
126
Originally Posted by BIJAY PRASAD View Post

2.define a no: k such thtit is the sum of the squires of the first m natural nos where m


Naga,

Honestly i don't understand the question as well as your answer. Kindly elaborate..


let me do it this way
k=1^2+2^2+....+m^2

so k =M(M+1)(2M+1)/6
so accrdin to question K should be divisible by 4
so M(M+1)(2M+1) should be divisible by 24

this is only possible when M(M+1)(2M+1) has 2^3 and 3 terms in it

so 1st m=7 ,2nd m=8...
this will be true for all the multiples(edit: of the form) of 8n-1 and 8n
thus 12 values less than 55
127
Sum of infinite series S=a/(1-r).
What is the restriction for 'r' here, Will r
r11gupta Says
Yes..the summation of an infinite GP is possible only if r

a little bit of correction...the condition is r
Cheers!!
128
The no. is 1056. Here's how I did it:

Let x^2 = n + 100
and y^2 = n + 169, where n is the number which we have to find

Now, y^2 - x^2 = 69
(y-x)(y+x) = 69

Now, 69 can be got either by 1*69 or 3*23.

Using 1*69, let's assume (y+x) = 69 and (y-x) = 1. Then, we can get x = 34 and y = 35

Thus, n + 100 = 34^2 = 1156
n + 169 = 35^2 = 1225

Either way, we can find out that n = 1056 :)

let me do it this way
k=1^2+2^2+....+m^2

so k =M(M+1)(2M+1)/6
so accrdin to question K should be divisible by 4
so M(M+1)(2M+1) should be divisible by 24

this is only possible when M(M+1)(2M+1) has 2^3 and 3 terms in it

so 1st m=7 ,2nd m=8...
this will be true for all the multiples(edit: of the form) of 8n-1 and 8n
thus 12 values less than 55




Avinav/kinscool,

Thanks a lot for explaining.

Guys,

This thread is opened for all of us and not only to seek the answer of a problem but also to help each other.Only posting answer will not gonna be of much help.Most the time the answers and answer options are given.What is important,is sharing the way/method each of us do it. This will help us to reach solution in various way,expanding our way of thinking.

So please do post the method while posting your answer.



Thanks,
129

Avinav,

Thanks a lot for explaining.

Guys,

This thread is opened for all of us not only to seek the answer of a problem but to help each other.Only posting answer will not gonna be of much help.Most the time the answers and answer options are given.What is important,is sharing the way each of us did it. This will help us to reach solution in various way,expanding our way of thinking.

So please do post the method while posting your answer.



Thanks,


This should be my words fuzzyLogic....Thanks guys...i m new to forum...its a great initiative....thanks a ton...

130
some questions from arun sharma...

1.There is a natural no tht become equal to the squire of a natural no when 100 is added to it and squire of another no when 169 is added to it. find the no.


It is sure that number too be added must be added such that difference between two consecutive squares is 69...this is possible when......35^2 - 34^2 = 69

so number to be added is 34^2 - 100 = 1056....:shocked:
131

Consider all possible seven-digit binary numbers having four 1s and three 0s. What is the sum of all such numbers?

(a) 1470 (b) 1615 (c) 1740 (d) 1825 (e) 1910

132
Consider all possible seven-digit binary numbers having four 1s and three 0s. What is the sum of all such numbers?

(a) 1470 (b) 1615 (c) 1740 (d) 1825 (e) 1910


ans:sum will be 20*2^6 +10(2^6-1) = 1910
133
Consider all possible seven-digit binary numbers having four 1s and three 0s. What is the sum of all such numbers?

(a) 1470 (b) 1615 (c) 1740 (d) 1825 (e) 1910

option e)1910
7th position will always hold 1,==> 64*6!/(3!*3!) = 1280
for all other position
*(32+16+8+4+2+1) = 630
total = 1910
134
Consider all possible seven-digit binary numbers having four 1s and three 0s. What is the sum of all such numbers?

(a) 1470 (b) 1615 (c) 1740 (d) 1825 (e) 1910


CAses - when all four 1's are together....
i.e.

1) 1111000 - 120

2) 1110100, 1110010, 1110001 - 112*3 + 7

3) 1101100, 1101010, 1101001 - 104*3 + 7

4) 1100110, 1100101, 1100011 - 3*96 + 14


After this you will notice a pattern...whenever 3 1's come in first four places....number which we get from first four places is decreased by 8....and we add 7 to it....

whenever we have 2 1's in first 4 1's i.e case of 1010, 1100, 1001...we decrease the preceding number by 8 and add 14 to it....

I.e.
Sum = 120 + (112*3 + 7) + (104*3 + 7) + (96*3 + 14) + (88*3 + 7) + (80*3 + 14) + (72*3 + 14) + 71
= 1910......:
135

_ _ _ _ _ _ _

we'll convert binary to decimal to find the sum-

7 digit no -> first digit will be always 1...remaining 6 places wil be filled by 3 1's and 3 0's...
so 2^6 will be repeated 6!/(3!*3!) ways
sum=2^6 *(6!/3!*3!)

now 6th digit- no of times 1 will be there

1 1 _ _ _ _ _

remaining blanks will be filled by 2 1's and 3 0's -> 5!(2!*3!) ways
sum = 2^5 * 5!/(2!*3!)

similarly for rest of terms....

so total sum =
2^6 (6!/3!*3!) + 5!/(2!*3!) (2^5+2^4+2^3+2^2+2^1+2^0)

= 1910

my take is 1910

136
Consider all possible seven-digit binary numbers having four 1s and three 0s. What is the sum of all such numbers?

(a) 1470 (b) 1615 (c) 1740 (d) 1825 (e) 1910



7th digit should be 1

total number = 6! / 36 = 20

lowest number in decimal form = 71

highest number in decimal form = 120

so this series form some sequence some what equal to AP , though not AP

sum = 20/2 { 120 + 71 } = 191 * 10 = 1910

i think this shortest way to arrive at the answer

any comments welcome !
137

what is second digit (10s digit ) in 2^1001

Plz explain method of finding 10s digit for any given number..

138

u can write 2^1001 as(1024)^100*2..
24^any even power will give 76 as last 2 digits

so its 76*2=5 is the ten's digit

139
what is second digit (10s digit ) in 2^1001

Plz explain method of finding 10s digit for any given number..

2^1001 mod 100 will give the last two digits.
2^1001 mod 100 = 4*(2^999 mod 25).
Euler(25) = 20
=> 2^999 mod 25 = 2^19 mod 25 = 13.
=> 2^1001 mod 100 = 52.
Ten's digit is 5.
140
what is second digit (10s digit ) in 2^1001

Plz explain method of finding 10s digit for any given number..

2^1001mod100 = 2*1024^100mod100 = 2*24^100mod100 = 2*76^50mod100 = 2*76 =152
so tens digit is 5