## Problems Package ( Number Theory I)

** 1) If set A ={ 26, 29, 34, …., n^2+25,…}
set B = {13501, 13504, …., n^2+13500,….}
how many common elements do the two sets have? **

**2) What is the smallest positive integer with 6 positive odd integer divisors and 12 positive even integer divisors? **

**3) 4. Compute the number of squares between 4^9 and 9^4. **

**4) Given any positive integer n, Fidn the sum of all possible last digits of n^5-n? **

Advertisements

2.180

Consider 5,9 then, 3,15,45 are automatically the factors.1 is already a factor. Now coming to even, when we have, 2*all the odd divisors and 4*all the odd divisors gives us 12. 45*4 is the answer

Celebrating LifeSeptember 15, 2008 at 1:39 pm

4.n(n^4-1) just trial and error. only 2 and 0 are the values in the unit digit.

Celebrating LifeSeptember 15, 2008 at 1:41 pm

4.Small correction, only zero..So only one value.

Celebrating LifeSeptember 15, 2008 at 1:43 pm

3.Consider 3^8 and 2^18.they can be rewritten as 81^2 and 512^2.So totally 430 values..

Celebrating LifeSeptember 15, 2008 at 1:55 pm

all correct 🙂

outtimedSeptember 15, 2008 at 1:56 pm

lets describe the elements of set A as a^2+25 and set B as b^2+13500

Now, for both to be equal, a^2+25=b^2+13500.

(a+b)(a-b)=13475

Now the factors of 13475 are 1,5,7,5,7,11

here a+b and a-b should be multiples of 5.

We now need to select the value of a-b such that the quotient we get when we divide a-b by 13475 is more than a-b( as it should be a+b)

so the possible values of (a-b) are 1/5/7/11/25/35/49/55/77.

So, in all 9 solutions.

milindSeptember 15, 2008 at 5:09 pm

2)the number should have 6 odd integer divisors. Apart from 1, 3 is the lowest. Now if we use the next as 5, we see that the product increases to 15. while another 3 increases the product to just 9. So we can have odd divisors are 1/3/9/27/81/243. the number=243

similarly for even divisors, it is 2^11

So, the least number is 2^11*3^5

milindSeptember 15, 2008 at 5:14 pm

4) since the cyclicity of all numbers is 4, the units digit becomes 0.

So, 0

milindSeptember 15, 2008 at 5:27 pm

man.. what debacle in ques no. 2.

If only we could edit our messages here.

people would have been saved from my disastrous soln(if I may call that).

@bhaskar.. nice soln for the 3rd problem.

milindSeptember 15, 2008 at 6:04 pm

Implex bhai.. yeh questions ke jawaab sahi hain kya?(except 2nd)

And the previous question sets too.

milindSeptember 17, 2008 at 6:06 am

yupppp

outtimedSeptember 17, 2008 at 10:32 am

ans to 4. n^5-n = n(n-1)(n+1)(n2+1)

–> n(n-1)(n+1) are three consecutive no. so product will always be 6

–> now for (n^2+1) will always have (1, 5, 6,7)

–> 6 *(1, 5, 6, 7) will give (6, 2) as distinct lst digits.

so sum of possible last digits will be 8.

ankitSeptember 19, 2008 at 12:21 pm

ankit check again

outtimedSeptember 19, 2008 at 1:05 pm