Number System Practice Gym by Anubhav Sehgal (NMIMS, Mumbai)  Part 10

How many positive divisors of 100^10 end in exactly two zeroes?
100^10 = 2^20 * 5^20
Taking exactly two trailing zeroes is equivalent to dividing the number by 10^2 = 100.
We get, 2^18 * 5^18
Now, since we have already taken out the two trailing zeroes, we must ensure no more zeroes (or tens) are considered else there would be more than 2 zeroes while we need EXACTLY two trailing/ending zeroes.
So, all factors of this number such that no 2,5 occur together will be:
{2^0, 2^1, …. , 2^18} = 19 factors
{5^0, 5^1, …. , 5^18} = 19 factors.
But note here that 2^0 = 5^0 = 1 has been counted twice.
Hence we remove that to get our final answer as 19 + 19  1 = 37.For what smallest positive integer N, will the number 3! + 5! + 7! + .. + (2N + 1)! be a perfect square ?
3! (1 + 4 * 5 + 4 * 5 * 6 * 7 + .... ) = 3! * odd.
Power of 2 is 1.
So no possible N as for a perfect square power of each distinct prime in the factorization of the number must be even.Find the sum of all numbers which give same remainder as the quotient when divided by 21.
N = 21q + q => N = 22q
Sum = 22 (0 + 1 + 2 + ... + 21)
= 22 * 21 * 22/2
= 4620 + 462
= 5082.Find the highest N such that 11^N divides (97! + 98! + 99!).
97! * (1 + 98 + 99 * 98) = 97! * (99 * 99)
Power of 11 in 97! = [97/11] + [97/121] + … = 8
Power of 11 inside bracket
= 2.
Highest power = 10.Given that 1002004008016032 has a prime factor P which is greater than 250000. Find P.
Note the pattern in the number. Last 7 digits are 8 times the first 7 digits.
Write the number as 1002004 * 10^9 + 8 * 1002004 = 1002004[10^9 + 8]
=> 1002004 is a factor of the given number.
1002004 = 2 * 2 * 250501
=> P = 250501.Find if the number 1000C500 is divisible by 7.
1000C500 = 1000! / 500!^2
Power of 7 in 1000! = 142 + 20 + 2 = 164
Power of 7 in denominator = Twice of Power of 7 in 500! = 2 * (71 + 10 + 1) = 2 * 82 = 164.
Power of 7 cancels out in numerator and denominator leaving no residual 7s.
Hence the number is not divisible by 7.There are two numbers between 1000 and 3000 that cannot be written as sum of two or more consecutive numbers. Find their sum.
We know that the number of ways in which N can be written as sum of two or more consecutive natural numbers is given by (No of odd factors) – 1.
Since we need such numbers which cannot be written as sum of two or more consecutive numbers we need to have the number of odd factors as only one. Also, 1 is a factor of every number and is odd. Hence our required number must be 2^k so that there are no other odd factors and we can have no ways to represent it as sum of two or more consecutive numbers.
Number of the form 2^k between 1000 and 3000 = 1024 and 2048.
Their sum = 3072.A base 7 threedigit number has its digits reversed when written in base 9. Find the decimal representation of the number.
Let the base 7 three digit number be abc.
Then the number in base 9 is cba.
Using the standard conversion we have, a * 49 + b * 7 + c = c * 81 + b * 9 + a
48a = 2b + 80c
24a = b + 40c
a = 1. Not possible.
a = 2 => b, c = 8, 1. But we can use only digits 0 to 6 as abc are digits of a number in base 7.
a = 3 => Not possible.
a = 4 => Not possible
a = 5 => b, c = 0, 3. Accepted. Number in decimal form = 49 * 5 + 0*7 + 3 = 248.
a = 6 => Not possible.
Hence our answer is distinct and equal to 248You are holding a book, which has 500 pages (not including cover). If you tear off all the pages starting from the one labelled 'page 311' until the one labelled 'page 458', this book will have 350 pages left. Now, if you tear off all the pages from the one labelled 'page 216' until the one labelled 'page 242', how many pages are left?
(Note: One paper has 2 pages, one page on one side and one page on the other)This is a question that tests your practical acumen as well along with mathematical skills.
When we removed page numbers 311 to 458, we should have lost (458 – 311 + 1) = 148 pages.
But since it says that 350 pages were left. It means that 150 pages were lost. What does this mean?
This means that the page numbered as 311(odd) was on the left side, making us lose 310 as well and similarly page numbered as 458 (even) made us lose 459 as well.
Now if we remove page 216 to page 242, we will lose (242 – 216 + 1) = 27 pages. Don’t make the same mistake twice. Since page 216 is even numbered, 217 is also lost but that has already been accounted for. But, tearing off page 242 would make us lose 243 as well. Hence in total we lose 28 more pages and hence are left with 350 – 28 = 322 pagesTwo different twodigit natural numbers are written beside each other such that the larger number is written on the left. When the absolute difference of the two numbers is subtracted from the fourdigit number so formed, the number obtained is 5481. What is the sum of the two twodigit numbers?
abcd  (ab cd ) = 5481
1000a +100b + 10c + d  10a + 10cb+d = 5481
990a + 99b + 20c + 2d = 5481
Clearly a=5 as b, c and d all are digits and a cd).
And sum = 55 + 18 = 73