28383rd term of series 12345678910111213...??
plz explain the procedure..
28383rd term of series 12345678910111213...??
plz explain the procedure..
last digit 3
9+2*90+3*900+4*6373 +9 +2 more digits
=3
tricky number system question::
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Q-If n is a multiple of 5 and n=(p^2) X q where p and q are prime numbers, which of the following must be a multiple of 25?
a. p^2
b. q^2
c. pq
d. p^2.q^2
e. p^3.q
(Please read p^2 as square of p)
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Substitution by integers does not help , since it leaves you with two choices - b and d.
answer as in OG :: (d).
why not (b)?
see..it's easy....if n is a multipls of 5 then either P is 5 or Q is 5.....so when P=5 then P^2=25 and q can be any other prime number.....so N= 25*q
also when q=5 then N=P^2*5
now chk fr the options you'll get only D as the correct one
tricky number system question::
--------------------------------------------------------------------
Q-If n is a multiple of 5 and n=(p^2) X q where p and q are prime numbers, which of the following must be a multiple of 25?
a. p^2
b. q^2
c. pq
d. p^2.q^2
e. p^3.q
(Please read p^2 as square of p)
------------------------------------------------------------------
Substitution by integers does not help , since it leaves you with two choices - b and d.
answer as in OG :: (d).
why not (b)?
At least one of p,q or both can be 5 so that n is a multiple of 5. If only one of p,q is 5, then options (a) and (b) obviously fail and for the same reason options (c) and (e) fail.
p^2 q^2 must be a multiple of 25 even if one of p,q is 5
tricky number system question::
--------------------------------------------------------------------
Q-If n is a multiple of 5 and n=(p^2) X q where p and q are prime numbers, which of the following must be a multiple of 25?
a. p^2
b. q^2
c. pq
d. p^2.q^2
e. p^3.q
(Please read p^2 as square of p)
------------------------------------------------------------------
Substitution by integers does not help , since it leaves you with two choices - b and d.
answer as in OG :: (d).
why not (b)?
i think the question is generalised, so either p can be 5 or q can be 5. we are not sure which of them is 5. thus, if we choose option b then we discard the possibility of p to be 5, hence a no. that must be a multiple of 25 can be p^2*q^2.
correct me if i m wrong..
how many 3 digit nos. are there , the cubes of which end with "56" ?
1. 30
2. 27
3. 18
4. 20
how many 3 digit nos. are there , the cubes of which end with "56" ?
1. 30
2. 27
3. 18
4. 20
my take:
option c)..
how many 3 digit nos. are there , the cubes of which end with "56" ?
1. 30
2. 27
3. 18
4. 20
option c)18
unit digit = 6
yx6 * yx6 * yx6 = ___(78x+21)6
78x+21 should give unit digit as 5
x can be 3&8
_36, _86
total 18 numbers
how many 3 digit nos. are there , the cubes of which end with "56" ?
1. 30
2. 27
3. 18
4. 20
Tens digit 3*a*b +1 . 2 nos within a space of 100 numbers satisfies this.
18 is the ans
1. Find the first non zero integer from right in 24!
a. 4 b.2 c. 8 d none
2. In an Ap nth term =123123.....3000 digits , commmon diff =333. Find the sum of all 3 digit terms of this AP.
a 1672 b 1935 c 1835 d none
-- wrong post--- sorry--
1. Find the first non zero integer from right in 24!
a. 4 b.2 c. 8 d none
2. In an Ap nth term =123123.....3000 digits , commmon diff =333. Find the sum of all 3 digit terms of this AP.
a 1672 b 1935 c 1835 d none
1. its 6 use the formula 4^k (2k!) to find the non zero last digit of 20!then multiply by 21,22,23,24
1. Find the first non zero integer from right in 24!
a. 4 b.2 c. 8 d none
2. In an Ap nth term =123123.....3000 digits , commmon diff =333. Find the sum of all 3 digit terms of this AP.
a 1672 b 1935 c 1835 d none
1) 6
2) none (1368 )
1) 6
2) none (1368 )
Can you please tell the process how to solved the second question
1. Find the first non zero integer from right in 24!
a. 4 b.2 c. 8 d none
my take : none 6
2. In an Ap nth term =123123.....3000 digits , commmon diff =333. Find the sum of all 3 digit terms of this AP.
a 1672 b 1935 c 1835 d none
my take : none
My Answers:-
1)6 (Last none zero digit of 10! is 8.so,20!=8*8=64 here last digit is 4 and at last 4! now last none zero digit is 4.so, 4*4 =16,hence last none zero digit is 6.
2)none of these.
ashish13dec SaysCan you please tell the process how to solved the second question
divide 123123.....3000 digits by 333
u get remainder 123
now AP is 123 ,456 , 789
sum = 1368
My Answers:-
1)6 (Last none zero digit of 10! is 8.so,20!=8*8=64 here last digit is 4 and at last 4! now last none zero digit is 4.so, 4*4 =16,hence last none zero digit is 6.
2)none of these.
How did you arrive none zero digit of 10! to be 8..... can you explain it further
My Answers:-
How did you arrive none zero digit of 10! to be 8..... can you explain it further
1*2*3*4*5*6*7*8*9*10= 3628800
Here last none zero digit is 8
24! can be in the form of 10! + 10! + 4!
here we can't add last digit instead of adding we multiple because it is about factorial so, 8*8*4 and after this we can find the value of last digit is 6.