Number System

[gpta_varun]

how many 3 digit numbers ARE THERE which when divided by 5 and 7 leave remainder 2 and 4 respectively ???




-varun

[warrior]

Originally Posted by gpta_varun

how many 3 digit numbers ARE THERE which when divided by 5 and 7 leave remainder 2 and 4 respectively ???




-varun

no fo form 35k+32

its will start from 102 and end till 977 total there are 26

[Varun Khullar]

SB bina jeena bahut muskhil

7^37/11 remainder is
7^23^29^37 divided by 17 remainder is

[sakthi.abdullah]

Originally Posted by deepak

lemme try this..
a 2-digit no. havin a 3-digit no as its square..so the 2digit no is<32.. in other words A is 1,2 or 3.
but as B^2 has its last digit as A ..therefore A cant be 2 or 3.Hence A=1..and this leads us to B=9..so 19^2=361...
And X=3..

*phew*

nice...expln

[vikram_k51]

Bhai log pls help me wid these 2:

1.L.C.M(1,2,3,4...,200)/L.C.M(105,106,107...200)=k
Then k=?
1.205 2.10103 3.10403 4.10302


2.1^2!+2^3!+3^4!....+152^153!+153^154! has unit's digit:

1.)1 2.)9 3.)5 4.)7

Thnx

[saurabh_sist]

1.L.C.M(1,2,3,4...,200)/L.C.M(105,106,107...200)=k
Then k=?
1.205 2.10103 3.10403 4.10302

for finding L.C.M(1,2,3,4...,200) and L.C.M(105,106,107...200),
we need prime factorisation...
in both L.C.M(1,2,3,4...,200) and L.C.M(105,106,107...200),
highest power of 2 is 7...128
highest power of 3 is 4...81
..
..
..
similarily we will have different powers for different prime numbers
except 101 and 103 which will occur only in numerator
so ans is 101*103=10403

[saurabh_sist]

1^2!+2^3!+3^4!....+152^153!+153^154! has unit's digit:

1.)1 2.)9 3.)5 4.)7

1^2!+2^3!
+11^p+21^p..........15 terms
+12^p+22^p..........15 terms
+3^p+13^p+23^p....16 terms
+4^p+14^p+24^p....15 terms
+5^p+15^p+25^p....15 terms
+6^p+16^p+26^p....15 terms
+7^p+17^p+27^p....15 terms
+8^p+18^p+28^p....15 terms
+9^p+19^p+29^p....15 terms
+10^p+20^p....15 terms where p is multiple of 4 ( used since last digit repeats for every fourth power)

=1+4+15(for power of 11,21,..)+15*6(for 2^4)+16*1(for 3^4) +15*6(for 4^4)+15*5+15*6+15*1+15*6+15*9+0
=5+5+0+6+0+5+0+5+0+5+0
=1

[sakthi.abdullah]

Originally Posted by warrior

no fo form 35k+32

its will start from 102 and end till 977 total there are 26

35k+32 is ok..
but how tht 26 comes??
can u explain on thut

[sakthi.abdullah]

Originally Posted by Dan

can u pls explain 1000D - 10 D =a1a2 funda?

i did it this way:
any number divided by 11 gives reminder of the form 0.a1a2a1a2..
so if div by 110 we get 0.0a1a2a1a2...and the only choice thts a multiple of 110 is 1980

Dan

note=0.0a1a2a1a2..
10x=0.a1a2a1a2...
100000x=a1a2.a1a2a1a2....
so 99990x=a1a2
x=a1a2/99990

is it not

[gpta_varun]

Some important questions

1) There are 8436 steel balls , each of which have radius 1 , stacked in a pile such that 1 ball is at top , then 3 balls in second layer , then 6 in third , then 10 and so on. Number of layers ?? ....How do we cater to there progressions , where common diff itsel is in AP ?????

2) consider a graph with 12 points . It is possible to reach any point from any other thru a sequence of edges . Then the range of edges 'e' ...

a) 11<=e<=662 b) 10<=e<=66 c) 11<=e<=65 d) 0<=e<=11