1cant we take the no as abcd.
i mean M=abcd
then N=dcba
and, N= 4 M
so, 4000Xa + 400Xb + 40Xc +4Xd = 1000d+100c+10b+a
but then i am geting 4a=d and 4d=a.
so, a=d !
can this be an approach ?
find out a four digit number M such that N=4xM
and N has the following properties :
i) it is also a four digit no.
ii) the digits of N are in reverse order of those in M.
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8 Answers
62M<=2499
First digit of M is 1 or 2
so the last digit of N has to be 1 or 2 (ButN is a multiple of 4 so it has to be 2)
Now can you solve it?
first digit of M becomes 2
last digit is either between 3 or 8
Now can you try some more?
1
24sorry to disturb everyone...........
but sky are u giving ISI ?????
1let m be 1000w+100x +10y+z
z,w≠0
m<2500
so w=1or 2
since n is a multiple of 4 w=2 →z=8
m looks like 2 x y 8
since n=4m
writin dem in eq leads to
332z+20y=1333w+130x
sub w & z
eq bcomz
2y-1=13x ---1
since n is div by 4 last 2 digit must be div by 4
10y+2/4
y can b 1,3,5,7,9
4m 1
y belongs to I
satisfied by 7 only
d no. is 2178
11Palani and the others forgot that there exists a much simpler soln....
If number is abcd, then only possible value of a is 2....
Meaning d=8....(it can't be 9)....
Only possible values of b={3,1,9}....
Easy to check only 1 is consistent with corresponding c's.....
Done!