Number System - Questions & Discussions

MBA CAT 2022
1148
X = (10000 10001 10002 10003) + 1 and X = Y2, where Y is a natural number.

What is the value of Y?
1149
1)P ={1, 2, 3, 6,12, 24, 48, 96}. How many distinct values of 'x' are possible, if 'x' is defined as the sum of
one or more elements of the set 'P'?
(1) 150 (2) 192 (3) 148 (4) 191

2)
The LCM of three positive integers X, Y and Z is 1192. Find the total number of ordered triplets
(X, Y and Z).
(1) 400 (2) 361 (3) 289 (4) 225

3)
V is a 56 digit number. All the digits except the 32nd from the right are the same. If V is divisible by 13,
then which of the following can never be the unit's digit of V?
(1) 4 (2) 7 (3) 1 (4) Both 4 and 7


please explain me above questions in very simple way...
1150
X = (10000 10001 10002 10003) + 1 and X = Y2, where Y is a natural number.

What is the value of Y?


X = (k)(K+1)(k+2)(k+3) + 1
X = (k^2 + 3k)(k^2 + 3k + 2) + 1
let K^2 + 3k = Z
X = Z(Z+2) + 1
Y^2 = X = (Z+1)^2
Y = (Z+1) = (K^2 + 3k + 1) => 10000^2 + 3*10000 + 1 => 100030001
1151
X = (10000 10001 10002 10003) + 1 and X = Y2, where Y is a natural number.
What is the value of Y?

Getting 100030001 i.e. None of these.........

As y^2 = (10000 * 10001 * 10002 * 10003) + 1 or
y^2 - 1 = 10000 * 10001 * 10002 * 10003 or

(y+1)(y-1) = 10000 * 10001 * 10002 * 10003

Pairing.........

(y+1)(y-1) = (10000 * 10003) * (10001 * 10002) or

(y+1)(y-1) = 100030000 * 100030002 i.e.

y = 100030001..........

FOR QSTN OF THIS TYPE USE THIS trick:-

firstly we shud know multiplying 4consecutive digits +1=perfect square

n*(n+1)*(n+2)*(n+3)=^2
so here n=10000
ans is 100030001. option 5 ans


Check these out.........
1152
1)P
={1, 2, 3, 6,12, 24, 48, 96}. How many distinct values of x are possible, if x is defined as the sum of

one or more elements of the set P?
(1) 150 (2) 192 (3) 148 (4) 191
please explain me above questions in very simple way...


Ans is (2) 192, check the set P, you can get all the digits from
1 ~ till the summation of all the numbers in the set, which is 192
1153
1)P
={1, 2, 3, 6,12, 24, 48, 96}. How many distinct values of x are possible, if x is defined as the sum of
one or more elements of the set P?

(1) 150 (2) 192 (3) 148 (4) 191



all the sum possible from 1 to 192 so answer is 192...
1154

What is the 12th digit from the left of (6007)3?

1155
madnikhil Says
What is the 12th digit from the left of (6007)3?

12th digit frm left is nothin but the units digit
7^3 = 3???
1156
12th digit frm left is nothin but the units digit
7^3 = 3???



yep..it will be the unit's digit ie 3..but how did you know that the cube of 6007 has exactly 12 digits?

6007^3=(6000+7)^3

then we will apply the formula..did you check with the no. of digits in 6000^3 or some other way to find out no. of digits in any no's cube?
1157

Please elucidate me with your approach..
Problem:
Inradius of a right-angled triangle is 32 units and sum of the other two sides other than the hypotenuse is 42 units. What is the circum radius?
(a)26,(b)32,(c)52,(d)43,(e)More than one.
Edited:typo

1158
Please elucidate me with your approach..
Problem:
Inradius of a right-angled triangle is 32 units and some of the other two sides other than the hypotenuse is 42 units. What is the circum radius?
(a)26,(b)32,(c)52,(d)43,(e)More than one.


Please verify the question for the part in blue.
1159
bishoo123 Says
Please verify the question for the part in blue.



it must be sum of the other two sides..
1160
what is the remainder when 123123123............(300 terms) is divided by 504 ?
1161
what is the remainder when 123123123............(300 terms) is divided by 504 ?

123123123............(300 terms) /8*7*9
using chinease remainder theorem:
123123123............(300 terms) mod 8=3
123123123............(300 terms) mod 7=0 (using the divisibility rule of 7)
123123123............(300 terms) mod 9=7

so 8x+3=7y=9z+6
1162
123123123............(300 terms) /8*7*11

using chinease remainder theorem:

123123123............(300 terms) mod 8=3
123123123............(300 terms) mod 7=0 (using the divisibility rule of 7)
123123123............(300 terms) mod 11=7

so 8x+3=7y=11z+7

ans 259


The answer cannot be 259......

123123123........(300 terms) is a multiple of 3 and so is 504.... Hence, the remainder necessarily has to be a multiple of 3.... 259 is not a multiple of 3..
1163
what is the remainder when 123123123............(300 terms) is divided by 504 ?


504 = 8*7*9
123123...300times mod8 = 3
123123...300times mod7 = 123*(10^297 + 10^294 + ...1)mod7
4(3^297 + 3^294 + 1)mod7 = 4(1(3^300 -1)/2 = 2(3^300 -1)mod7
e(7) = 6
hence 3^300mod7 = 1
hence 123123..300times mod 7 = 0
123123123....300times mod9 = 50(3+4+5)mod9 = 6

7x = 8y+3 = 9z+6
7x = 72y + 51
hence remainder = 483??
1164
123123123............(300 terms) /8*7*11

using chinease remainder theorem:

123123123............(300 terms) mod 8=3
123123123............(300 terms) mod 7=0 (using the divisibility rule of 7)
123123123............(300 terms) mod 11=7

so 8x+3=7y=11z+7

ans 259

simple mistake..chk the red part:)
1165
what is the remainder when 123123123............(300 terms) is divided by 504 ?

504 = 8*9*7.........

123123...(300 terms) mod 8 = 3
123123...(300 terms) mod 9 = 6
123123...(300 terms) mod 7 = 0

So
8a +3 = 9b + 6 = 7c

Solve we will have remainder as 483


And pls don't post the same question on two threads...........
1166
Please elucidate me with your approach..
Problem:
Inradius of a right-angled triangle is 32 units and sum of the other two sides other than the hypotenuse is 42 units. What is the circum radius?
(a)26,(b)32,(c)52,(d)43,(e)More than one.
Edited:typo

let the inradius be r and circumradius be R and the two non-hypotenuse sides be a and b
the given a+b=42 and r=32
now the hypotenuse will be 2R
also 1/2 ab = rs = 32(21 + R) ---(1)
also a^2+b^2 = 4R^2
so (a+b)^2 - 2ab = 4R^2 = 42^2 - 2ab
hence we get ab = 42^2/2 - 2R^2 = 21*42 - 2R^2
puttin this value in (1) we get
21*21 - R^2 = 32*21 + 32R
R^2 + 32R + 231 = 0
R = (-32 +- 10)/2 =-21,-11...both are not possible...therefore i can say tht data is inconsistent...correct me if i have done sumthin wrng here...
1167
504 = 8*7*9
123123...300times mod8 = 3
123123...300times mod7 = 123*(10^297 + 10^294 + ...1)mod7
4(3^297 + 3^294 + 1)mod7 = 4(1(3^300 -1)/2 = 2(3^300 -1)mod7
e(7) = 6
hence 3^300mod7 = 1
hence 123123..300times mod 7 = 0
123123123....300times mod9 = 50(3+4+5)mod9 = 6

7x = 8y+3 = 9z+6
7x = 72y + 51
hence remainder = 483??


Can someone please elaborate a bit on the bold part. pardon my ignorance.. 😃