practice problems

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 = 2^s mod 9
=> 2^s(2^(r-s) - 1) = 0 mod 9
2^s and 9 are coprime
=> 2^(r-s) - 1 = 0 mod 9
=> 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 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

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