Number System - Questions & Discussions

MBA CAT 2022
2230
51^1375/7
Now,
(49+2)^1375/7
We know that 7^2=49. Thereby we can eliminate 49 as it is a factor of 7.
So now our problem becomes
2^1375/7
=>*2/7
Now,
the above statement may be re-written as
(7+1)^458*2/7
Hence, the remainder becomes 2.


It's 2
As ,on every power of 6, 7 Will leave remainder 1
So at the end 51/7,leave remainder 2
2231
N is a set of all the natural nos less than 500 which can be written as sum of two or more consecutive natural nos .Find d max no of elements possible in N?
a) 250
b) 492
c) 493
d) none of these


Is it 492 ?
V jst hav to find all numbers which can be written only in terms of two
2232

When is [x - 41 equal to 4 - x?;)

2233

When is x - 4 equal to 4 - x ?

2234

Is (1.000001)^1000000 > 2 ?


Plz help me hhw to solve such pblms.

2235

1 and 8 are the first two natural numbers for which 1+2+3+......+n is a perfect square.Which number is a 4th such number?

2236

Sorry for being Puys.I have opened a new thread of Verbal,which deals with odd word/pair out,antonym-synonym type questions.All are welcome :

The thread is http://www.pagalguy.com/discussions/odd-word-pair-out-and-words-relationship-for-mba-2011-25069070/2821305

2238

Here comes my big one 😃 puys are requested to provide the solution and innovative ideas are most welcomed 😃 thanx in advance friends :)

1) the expression 333^555+555^333 is divisible by
a) 2 b) 3 c) 37 d)11
2) Define a number K such that it is the sum of squares of first M natural numbers(i.e. K=1^2+2^2+...+M^2), where M a) 10 b)11 c)12 d)none
3) M is a two didgit number 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
4) find the 28383rd term of the series 123456789101112...
a) 3 b)4 c)9 d)7
5) If you form a subset of integers chosen from between 1 to 3000, such that no two integers add up to a multiple of nine, what can be the maximum number of elements in the subset?
a) 1970 b)1971 c)1972 d)none
6) A triangular number is defined as a number which has the property of being expressed as a sum of consecutive natural numbers starting with 1. how many triangular numbers less than 1000, have the property that they are difference of squares of two consecutive natural numbers?
a) 20 b)21 c)22 d)23
7) The remainder when number 123456789101112...484950 is divided by 16 is
a) 3 b)4 c)5 d)6
What is the remainder when (1!)^3+(2!)^3+...(1152!)^3 is divided by 1152?
a)125 b)225 c)325 d)205
9) What is the remainder when 2(8!)-21(6!) divides 14(7!)+14(13!)?
a) 1 b)7! c)8! d)9!
10) How many integer values are there such that 4x+7y=3, while x a) 144 b)141 c)143 d)142
11) How many co-digit numbers less than or equal to 50, have the product of factorials of their digits less than or equal to the sum of the factorials of their digits?
a) 17 b)16 c)15 d)none
12) N=202x20002x200000002x20000000000000...2(31 zeroes), the sum of digits in this multiplication will be?
a)112 b)160 c)144 d) cant be detrmined.
13) Find the remainder when (50^56)^52 is divided by 11?
a) 7 b)5 c)9 d)10

2239

what is remainder when 777777.....50times divides by 74?

i got the answer as 7,(777777........48 times) remainder is
but the answer is 3 . whats the OA

2240

1). How many integer values of x and y satisfy the expression 4x+7y=3 where
x(a). 284
(b). 285
(c). 286
(d). none of these

2). Let Sm denote the sum of the squares of the first m natural numbers.
For how many values of m(a). 50
(b). 25
(c). 36
(d). 24

3). For the above question, for how many values will the sum of cubes of the first
m natural numbers be a multiple of 5 (if m(a). 20
(b). 21
(c). 22
(d). None of these

Please share the approach also.

2241
Im finding some difficulty in solving these kinds of questions. The questions below are from Arun Sharma but im not too convinced even after seeing the solution. I'll be glad if anyone can help me out with the concept.

Find the remainder when 32^33^34 is divided by 11 ?
a). 5
b). 4
c). 10
d). 1

Find the remainder when 30^72^87 is divided by 11.
a). 5
b). 9
c). 6
d). 3

Please share the approach also.
2242

Are the answers
d)1
b)9???
If i m ryt, i ll explain....... or else i ll bang my head for answers :banghead:;)

2243
Im finding some difficulty in solving these kinds of questions. The questions below are from Arun Sharma but im not too convinced even after seeing the solution. I'll be glad if anyone can help me out with the concept.

Find the remainder when 32^33^34 is divided by 11 ?
a). 5
b). 4
c). 10
d). 1



definitely (d) for dis.. 32/11 gives -1. (-1)^33 would give same (-1). then (-1)^34 would give 1. plz tell me if i am wrong.
2244
what is remainder when 777777.....50times divides by 74?

i got the answer as 7,(777777........48 times) remainder is
but the answer is 3 . whats the OA


The answer would be 3. Here is how:

7777...50 times / 74 = 7(111...50 times)/37*2
111 is divisible by 3. So, we can write the whole thing as:

7(11100..47 zeroes)/2*37 + ... + 7 * 11100/2*37 + 7*11/2*37
All the terms except the last would be divisible by 74.

The last term would be 77/74. So, remainder = 3.
2245


2). Let Sm denote the sum of the squares of the first m natural numbers.
For how many values of m(a). 50
(b). 25
(c). 36
(d). 24

3). For the above question, for how many values will the sum of cubes of the first
m natural numbers be a multiple of 5 (if m(a). 20
(b). 21
(c). 22
(d). None of these


for (2) sum of squares is n(n+1)(2n+1)/6, so to be divisible by 4 it would be n(n+1)(2n+1)/24... so with hit and trial for first 9 digits(1-9) we find that for n=7 and 8 the whole sum is divisible. then check for digits 10-20, we will find that 15 and 16 satisfy.. so we get an AP here.. as 7,15,23,... and 8,16,24,... here one may also confuse for last term in case of series with 7.. we know the last term for 8 would be 96, so for 7 it would be 95. simple!! so in total we get 12+12=24 terms. hope i am correct..

for (3), same sum of cubes is ^2, so to be divisible by 5, we have /20.. again with a quick scan wid digits 1-9 we get 5,9.. then from 10-20 we get 15,19.. thus we find 10 terms.. we will also get 20,40.. also u shud check odrs like 24,44.. so a total of 14 terms.. hope i am correct..
2246
1). How many integer values of x and y satisfy the expression 4x+7y=3 where
x(a). 284
(b). 285
(c). 286
(d). none of these

2). Let Sm denote the sum of the squares of the first m natural numbers.
For how many values of m(a). 50
(b). 25
(c). 36
(d). 24

3). For the above question, for how many values will the sum of cubes of the first
m natural numbers be a multiple of 5 (if m(a). 20
(b). 21
(c). 22
(d). None of these

Please share the approach also.


1). How many integer values of x and y satisfy the expression 4x+7y=3 where
x|(a). 284
(b). 285
(c). 286
(d). none of these

Approach:
4x + 7y = 3 -- x = (3-7y)/4

Taking y as positive:

If y=1, x=-1 | if y = 5, x = -8 If y = 9 , x = -15 and so on..
So, we see that y increases by 4 whereas x decreases by 7.
So, the number of possible terms would be limited by x.
Number of terms = (1000-1)/7 + 1 = 143

Taking y as -ve:

If y=-3, x = 6 (For values of y greater than -3, x is not an integer)
If y = -7, x=13 and so on.
Here also, the y will decrease by 4 whereas x will increase by 7. S, the limit will again be put by x.

Number of terms = (1000-6)/7 + 1 = 143.
But, since x
Hence total = 285

2). Let Sm denote the sum of the squares of the first m natural numbers.
For how many values of m(a). 50
(b). 25
(c). 36
(d). 24
Approach:

Sum of squares of m natural nos. = m(m+1)(2m+1)/6.
For this to be divisible by 4, m(m+1)(2m+1) has to be divisible by 24.
Now, 24 = 2^3 * 3

Since 2m+1 is always odd, so only one of m or m+1 has to be a factor of 8.
If m+1 is taken to be a factor of 8, then
m(m+1)(2m+1) = (8k-1)(8k)(16k-1)

Whatever be the values of k, one of these would be divisible by 3.

Similarly, if m is taken to be a factor of 8, then
m(m+1)(2m +1) = 8k(8k+1)(16k+1)

Whatever be the values of k, one of these would be divisible by 3.
So, the values m can take are: 7,8,15,16...
So, number of values = ((100-/8 +1) *2 = 24.

3). For the above question, for how many values will the sum of cubes of the first
m natural numbers be a multiple of 5 (if m(a). 20
(b). 21
(c). 22
(d). None of these
Approach:
Sum of cubes - ^2
m^2(m+1)^2 has to be div by 20.

20 = 4 * 5.
So, either m or m+1 has to be a multiple of 5.
Since if one of m or m+1 is odd, the other will be even, and hence its square will be divisible by 4.

So, the values m can take are: 4,5,9,10
So, total number of values = 10+9 = 19

P.S: Please correct me if I am wrong.
2247


Taking y as positive:

If y=1, x=-1 if y = 5, x = -8 If y = 9 , x = -15 and so on..
So, we see that y increases by 4 whereas x decreases by 7.
So, the number of possible terms would be limited by x.
Number of terms = (1000-1)/7 + 1 = 143


dude correct me if im wrong, wont it be,
number of terms=(1000-(-1))/7 + 1=144

so total terms would be 286.
2248
dude correct me if im wrong, wont it be,
number of terms=(1000-(-1))/7 + 1=144

so total terms would be 286.

Sirjee, I guess you are missing a point here..

the value of x would go from -1 to -1000.
So, I just neglected the - sign and proceeded.
2249

1) the expression 333^555+555^333 is divisible by
a) 2 b) 3 c) 37 d)11

Clearly divisible by 3, 37 and 2. So, its not divisible by 11
2) Define a number K such that it is the sum of squares of first M natural numbers(i.e. K=1^2+2^2+...+M^2), where M a) 10 b)11 c)12 d)none

K = n(n + 1)(2n + 1)/6
=> One of n and (n + 1) should be divisible by 8
=> n = 8k or 8k - 1
=> n can take 12 values (7, 8, 15, 16, ...., 47, 4
4) find the 28383rd term of the series 123456789101112...
a) 3 b)4 c)9 d)7

Check post #3499 of this page
http://www.pagalguy.com/discussions/quant-by-arun-sharma-25023813
5) If you form a subset of integers chosen from between 1 to 3000, such that no two integers add up to a multiple of nine, what can be the maximum number of elements in the subset?
a) 1970 b)1971 c)1972 d)none

We should have all numbers of form 9k + 1, 9k + 2, 9k + 3, 9k + 4 and one number of form 9k
=> 334 + 334 + 334 + 333 + 1 = 1336
6) A triangular number is defined as a number which has the property of being expressed as a sum of consecutive natural numbers starting with 1. how many triangular numbers less than 1000, have the property that they are difference of squares of two consecutive natural numbers?
a) 20 b)21 c)22 d)23

Triangular numbers are of form n(n + 1)/2.
Any number that can be written as difference of squares of two consecutive natural numbers are odd numbers
=> n(n + 1)/2 should be odd
=> n should be of form 4k - 2 or 4k - 3
Also, n(n + 1)/2 => n => 22 such numbers
7) The remainder when number 123456789101112...484950 is divided by 16 is
a) 3 b)4 c)5 d)6

Consider just last 4 digits and find the remainder, i.e., 4950/16. So remainder will be 6
What is the remainder when (1!)^3+(2!)^3+...(1152!)^3 is divided by 1152?
a)125 b)225 c)325 d)205

Check this page:-
http://www.pagalguy.com/discussions/official-quant-thread-for-cat-2010-part2-25055013
9) What is the remainder when 2(8!)-21(6!) divides 14(7!)+14(13!)?
a) 1 b)7! c)8! d)9!

2(8!)-21(6!)
= 2*8(7!)-3(7)(6!)
= 16(7!) - 3(7!)
= 13(7!)
14(7!)+14(13!)
= 14(7!)+14(13!)
= (1+13)(7!) + 14(13!)
= 7! + 13(7!) + 14(13!)
=> Remainder will be 7!
10) How many integer values are there such that 4x+7y=3, while x a) 144 b)141 c)143 d)142

Check this page:-
http://www.pagalguy.com/discussions/quant-by-arun-sharma-25023813
12) N=202x20002x200000002x20000000000000...2(31 zeroes), the sum of digits in this multiplication will be?
a)112 b)160 c)144 d) cant be detrmined.

N = 202x20002x200000002x20000000000000...2(31 zeroes)
= 2^5 * 101x10001x100000001x10000000000000...1(31 zeroes)
= 32(10^2 + 1)(10^4 + 1)(10^8 + 1)(10^16 + 1)(10^32 + 1)
= 32(10^62 + 10^60 + .... + 10^2 + 1)
= 3232323232....64 digits
=> Sum = 5*32 = 160
13) Find the remainder when (50^56)^52 is divided by 11?
a) 7 b)5 c)9 d)10

I think the question is 50^(56^52)
Euler's number of 11 is 10
=> We need to find the remainder when 56^52 is divided by 10, i.e., 6
=> Question reduces to 50^6 (mod 11) = 6^6 (mod 11) = 36^3 (mod 11) = 3^3 (mod 11) = 5 (mod 11)

=> remainder will be 5
2250

dis question is an important property so its important if we remember dis