Few for share and few for help!

4.is it possible????????

That is what we have to search!!

One more Doubt:

Please provide a simpler simplification of this

4. Dunno what the procedure should be, but 2178 is the only 4-digit number satisfying that...
I made a program in C++ to find out :P

4)

Start with some inequalities:

1 <= A <= 9 (since ABCD is a four-digit number, A can't be zero)
0 <= B <= 9
0 <= C <= 9
4 <= D <= 9 (D must be at least 4*A which is at least 4)

Then we have some equations:

4*D = 10*z + A (z is the carry to the tens column)
4*C + z = 10*y + B (y is the carry to the hundreds column)
4*B + y = 10*x + C (x is the carry to the thousands column)
4*A + x = D

There are more inequalities:

0 <= z <= 3
0 <= y <= 3
0 <= x <= 3 (because the multiplier is 4, the largest a carry can be
is 3)

Now we combine the first and last equations by multiplying the last by
4 and substituting for 4*D in the first equation:

4*(4*A + x) = 10*z + A or 4*x = 10*z - 15*A = 5*(2*z - 3*A)

Since 5 divides the number on the right, 5 must divide 4*x, and
so 5 divides x. But x is small, from 0 to 3, so x = 0 is forced. Thus
2*z - 3*A = 0, so 2*z = 3*A.

Similarly, 3 divides z, so put z = 3*w, so A = 2*w.
Then D = 8*w, and 4 <= D <= 9.

Clearly D = 8, so w = 1, A = 2, and z = 3.

Similarly, using the middle two equations, we can determine B, C, and
y. This gives you the answer you seek. You can finish the rest.

source: http://mathforum.org/library/drmath/view/56759.html

2nd >
multiplying 12! in the numerator of LHS given u terms in the form of ¹²C_k
i.e LHS is ¹²C₁+¹²C₃+¹²C₅12! = 102412!

so n = 10 , m = 12

now f(x+y) = f(x)f(y)

put y = 1 , so f(x+1) =f(x)

hence f(x+n)=f(x)

so f'(x+n)=f'(x)

hence f ' (integer ) = f'(0) = 10 = n

hence b is correct

2nd done, 4th done. Rest are there!!

I'm posting my solution for the first one. There are two : One uses Binomial and other was just a luck by chance solution. I hope the guys out here will figure out the Binomial one as it involves mere calculation, but the LUCK BY CHANCE one is really great.

Soln:

We observe √11 = 3.3166....
√1111 = 33.3316666....

and so on.

Hence, N number of 1's in √N gives N2 3's after decimal then 1 and then N 6's {See Above}

Hence when N = 1998, We have

√N = (N-1) .{ N2 3's }{ 1}{ N 6's}

i.e, Hence

√111...1111 {1998 times} = 3333.... 33 {1997 times} (.) 333..333 {998 times} 1 666....666 {1998 times}

Therefore, 1000th digit = 1
1001th digit = 6

4 >

we see 2499 is the last number for which abcd and 4 abcd both are 4 digit

so 1≤a ≤2 , 1≤b≤4 and 1≤c,d≤9

we have

4(1000a + 100b + 10c + d) = (1000d + 100c + 8b ) + 2b + a

If x+y = z , and any two are divisible by d,the rest is also divisible by d

so 4|2b + a or a is even so a = 2

Now 4| 2b + 2 or 2|b + 1 so b is odd

Also we can say that d is either 3 or 8 because 4*3 and 8*4 only gives unit digit as 2

as a = 2 , d ≥8 so d= 8

putting
a= 2 and d=8 in the above equation we get

13b + 1 = 2c

which gives b =1 and c = 7

so the number is 2178

Nice,,

Wow..!! Great dig!

The power of '9' :P

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