sorry for the mistake ... i found it to be divisible by 2 and 3 both and hence said all of the above as per the options ... however now on actually doing it, i find it to be divisible by 2 , 3 and 37, but not by 11 ... am i rite? how come 5. all of the above is an option?
BalearicInsane Sayssorry for the mistake ... i found it to be divisible by 2 and 3 both and hence said all of the above as per the options ... however now on actually doing it, i find it to be divisible by 2 , 3 and 37, but not by 11 ... am i rite? how come 5. all of the above is an option?
I think the question should be 333^555+555^333 is not divisible by, then the answer wouls be 4th option, i.e, 11.
chillfactor SaysI think the question should be 333^555+555^333 is not divisible by, then the answer wouls be 4th option, i.e, 11.
That might be the case becoz according to the book the answer is no 4 i.e 11
hi....a problem frm the numbers chapter...
333^555+555^333 is divisible by
1. 2
2. 3
3. 37
4. 11
5. All of the above
333/37 or 11 or 3 ------ R =0
555/37 or 11 or 3 ------- R=0
and 333+555=888(a comman factor of 333^555 + 555^333)/2 ---R=0
hance Ans 5
hi....a problem frm the numbers chapter...
333^555+555^333 is divisible by
1. 2
2. 3
3. 37
4. 11
5. All of the above
it is divisible by 2,3 and 37.
==>mod2=1^555+1^333mod 2=0
==>also 333 and 555 are divisible by 37 and 3..so whole expression also divided by no's.
@mr. kool
but for 11 it is...mod 11=1+3^116*5 mod 11=6
so it is not divisible by 11..
Exponent of 2 in 10200! = 10192
Exponent of 2^3 in 10200! = 3397
Exponent of 3 in 10200! = 5094
Exponent of 3^2 in 10200! = 2547
Exponent of 7 in 10200! = 1698
Thus Exponent of 504 in 10200! = min(3397,2547,1698 ) = 1698
In book it is mentioned as 7>2*3 therefore we just need to check straight away for 7 rather than checking for 2^3 and 3^2... my question is can you explain it
Find the 28383rd term of the series 1234567891011121314.................
Ans-9
Find the 28383rd term of the series 1234567891011121314.................
Ans-9
Answer can't be 9, it's 3. Check my post here:
http://www.pagalguy.com/forum/quantitative-questions-and-answers/23813-quant-by-arun-sharma-167.html#post1636356
Answer can't be 9, it's 3. Check my post here:
http://www.pagalguy.com/forum/quantitative-questions-and-answers/23813-quant-by-arun-sharma-167.html#post1636356
I too got 3 as the answer but the book says 9
Individualist SaysI too got 3 as the answer but the book says 9
You will find loads of incorrect answers in Arun Sharma.
1.Define a number K such that it is the sum of the squares of the first M natural numbers(K=1^2+2^2+3^2+.....+M^2) where Ma.10
b.11
c.12
d.None of these
Correct answer:c
2.How many integer values of x and y are there such that 4x+7y=3 while xa.144
b.141
c.143
d.142
Correct answer:c
3.A triangular number is one which has the property of being expressed as the sum of consecutive natural numbers starting with 1.How many triangular numbers less than 1000 have the property that they are the difference of squares of 2 consecutive natural numbers?
a.20
b.21
c.22
d.23
Correct answer:c
Please post the approach
1.Define a number K such that it is the sum of the squares of the first M natural numbers(K=1^2+2^2+3^2+.....+M^2) where Ma.10
b.11
c.12
d.None of these
Correct answer:c
2.How many integer values of x and y are there such that 4x+7y=3 while xa.144
b.141
c.143
d.142
Correct answer:c
3.A triangular number is one which has the property of being expressed as the sum of consecutive natural numbers starting with 1.How many triangular numbers less than 1000 have the property that they are the difference of squares of 2 consecutive natural numbers?
a.20
b.21
c.22
d.23
Correct answer:c
Please post the approach
1. K=1^2+2^2+3^2+.....+M^2 = 1/6 * m*(M+1)*(2m+1)
R(K/4)=0 if K have a factor 24.
i.e table of 4 till 52 bcoz Mhence n{4,8,12,16......48,52}=12
3. Triangular number
tn = 1+2+3+4...+ n-1 +n = 1/2*n*(n+1)
for n=1,2,3,4,5,6,7 .......till 44 ------------------A
tn= 1,3,6,10,15,21,28..till 990
For case2
a=natural no.
(a+1)^2 - a^2 = 2a+1 i.e odd number ---------------B
in set n= n{1,2,4,5.....43,44}=44
in set n number of odd number = 44/2 =22 -----------C
eq B and C
we say that
22 triangular numbers less than 1000 have the property that they are the difference of squares of 2 consecutive natural numbers.
i have a doubt in questiopn no 11 of page 43,lod II,PLZ HELP......
gpatra30 Saysi have a doubt in questiopn no 11 of page 43,lod II,PLZ HELP......
Always post the question here, so that others can also try to help.
Answer should be a) 3.
We need to count the no. of digits:
1-9 = 9
10-99 = 90*2 = 180
100-999 = 900*3 = 2700
1000-9999 = 9000*4 = 36000
Thus, the number would be a 4 digit no. No. of digits till 999 is 2889. We need the 28383rd digit, i.e. we need to find the 25494th digit starting from 1000 (28383-2889 = 25494).
Since, we know it'll be a 4 digit number, we can find the last number by dividing 25492/4 (not 25494, as we're trying to find the number now. Once we find the number, we can find the next two digits to get to the 25494th digit).
25492/4 = 6373rd number starting from 1000. It means the number is 1000+6373-1 = 7372. Till 7372, we have got 25492+2889 = 28381 digits. Next number will be 7373, we're interested in only the 2nd number as then we'll reach the 28383rd digit, and this number is 3.
Hope you understood :)
hey i have a doubt here. I got answer 3 as well but why do u add 1000 instead u shd add 4(as we are counting digits here not sum) so it becomes 6377...next digit to follow would br 6378
1) 8 orators A,B,C,D,E,F,G and H, In how many ways can the arrangements be made so that A always comes before B and B always comes before C.
A) 8!/3! B) 8!/6! C) 5!.3! D) 8!/(5!.3!)
2) A train is running between Patna and Howrah. Seven people enter the train somewhere between patna and howrah. It is given that nine stops are there in between
Patna and Howrah. In how many ways can the tickets be purchased if no restriction is there with respect to the number of tickets at any station.
A) 45C7 B) 63C7 C) 56C7 D) 52C7
Please post the approach also...
EagleMenace Sayshey i have a doubt here. I got answer 3 as well but why do u add 1000 instead u shd add 4(as we are counting digits here not sum) so it becomes 6377...next digit to follow would br 6378
I've added 1000 since we need to find the 6373rd number after 1000. You can't find it by adding 4, can you?
A number xy is multipied by another number ab and the results comes as pqr,where r=2y,q=2(x+y) and p=2x where x,ya.11
b.13
c.31
d.22
e.none
A number xy is multipied by another number ab and the results comes as pqr,where r=2y,q=2(x+y) and p=2x where x,ya.11
b.13
c.31
d.22
e.none
The key lies in this line:
"where r=2y,q=2(x+y) and p=2x"
Due to this, the answer has to be option (d), i.e. 22. You can check by using any arbitrary number as xy and ab as 22.
A number xy is multipied by another number ab and the results comes as pqr,where r=2y,q=2(x+y) and p=2x where x,ya.11
b.13
c.31
d.22
e.none