practice problems pending

Q. Determine the smallest prime that does not divide any five-digit number whose digits are in a strictly increasing order.
Solution:
2 doesn't work: 12346
3,5 don't work: 12345
7 doesn't work: 12348 (since 348-12 = 336 div by 7)
11 => abcde => a-b + c-d + e
a-b <= -1 c-d <= -1 5<=e <= 9
Max sum = 7
Total sum can never be <= 0 because e + a + c > b + d always. Why? Because e > d c>b.
So sum can never be 0 or multiple of 11.
11 is the answer.
For finding the min value, also see:
a + (c-b) + (e-d)
1     1           1
So min value is 3


Q. Suppose for some positive integers ( r ) and ( s ), the number ( 2^r ) is obtained by permuting the digits of the number ( 2^s ) in decimal expansion. Prove that ( r = s ).

Solution:
Since both the number have same number of digits('d') and same sum of digits, we use those properties.
Sum of digits is same => 
2^r = 2^s mod 9
Why?
Proof here.

WLOG r >= s
=> 2^(r-s) = 1 mod 9
Now powers of 2 cycle mod 9 with period = 6.
=> (r-s) = 6k , i.e. a multiple of 6.

Since 2^r and 2^s have same number of digits
=> 2^r < 10.2^s
=> 2^(r-s) < 10
=> r-s < 4
Only condition satisfying both these conditions is r-s = 0
H.P.

Q. Find the 5 digit number abcde s.t. 4*abcde = edcba
Solution:

abcde < 25000 else the mult by 4 gives a 6 digit number.
So a = 1,2
But the new number is even, so a = 2.
e = 8,9
But 4*e should give last digit 2. So e = 8.

So far:
4*2bcd8 = 8dcb2

Now look at b.
If mult by 4 gives a carry from b then first digit won't remain 8.
For the first digit to remain 8, b can't give carry.
=> b < 3 => b = 0,1,2

Now look at d.
In the original number e*4 = 8*4 = 32 so carry is 3.
4d + 3 should give last digit as b = 0,1,2.
Let's try all values of d from 0 to 9.
d = 2,7 give last digit as 1. So d = 2,7

If d = 2
Number becomes
4*21c28 = 82c12
Which doesn't work since LHS > 84000 RHS < 84000
So d = 7
4*21c78 = 87c12 , this is likely work.

Now:
84312 + 400c = 87012 + 100c
=> 300c = 87018 - 84312 =   2700
=> c = 9
Answer: 21978




Q. Find the smallest positive integer with the property that, if you move the first digit to the end, the new number is 1.5 times larger than the old one.

Solution:
Let the number's first digit be 'd'. And after removing 'd' let the rest of the number be 'x'.
Let the total digits be 'k'.
So the original number is:
d*10^(k-1) + x

And new number is:
10x + d

Given:
10x+d = 1.5*d*10^(k-1) + 1.5x
Multiply by 2:
20x + 2d = 3d.10^(k-1) + 3x
17x = d(3.10^(k-1) - 2)

'd' is a single digit number, not divisible by 17.
For 'x' to be an integer, the other factor has to be divisible by 17.

(3.10^(k-1) - 2) = 0 mod 17
3.10^(k-1) = 2 mod 17

Multiply by 6:
18.10^(k-1) = 12 mod 17
1.10^(k-1) = 12 mod 17
10^(k-1) = 12 mod 17

Now we need to find the smallest power of 10 which is 12 mod 17
That is 15.
k-1 = 15 => k = 16

For the original number to be smallest, let's try d = 1.
So we get x = (3*10^15 - 2)/17

Solving it you get:
x = 176470588235294
And original number = 1176470588235294
New number = 1764705882352941