Number System Practice Gym by Anubhav Sehgal (NMIMS, Mumbai)  Part 15

Rahul had 2n marbles numbered from 1 through 2n. He removed n marbles that were numbered consecutively. The sum on the remaining marbles was 1615. Find the sum of all possible values of n.
Let the marble numbers removed be (a + 1), (a + 2)… (a + n).
Then, 1615 = 2n * (2n + 1)/2 – n * (2a + n + 1)/2
3230 = 2n * (2n + 1) – n * (2a + n + 1)
3230 = n * (4n + 2 – 2a – n – 1) = n * (3n – 2a + 1)
2 * 5 * 17 * 19 = n * (3n – 2a + 1)
n = 34 or 38 satisfies.
Sum = 72.Find the smallest integer n for which (2^2 – 1) * (3^2 – 1) … (n^2 – 1) is a perfect square.
(n^2 – 1) = (n + 1) * (n – 1)
The given expression can be written as (1 * 3)(2 * 4)(3 * 5) … (n – 1)(n + 1) should be a perfect square
2n * (n + 1)[3 * 4 * 5 * 6 … (n – 1)]^2 should be a perfect square
2n * (n + 1) should be a perfect squaren and (n + 1) are coprime numbers.
Hence they must be individually perfect squares for the whole 2n * (n +1) to be a perfect square.
This gives us two cases:
Case 1
2n and (n + 1) are perfect squares
Case 2
n and 2(n + 1) are perfect squaresOne of n and n + 1 is odd while other is even. Also, 2 * odd is never a perfect square since power of 2 would be 1 in that case.
Hence, odd number of the two should be perfect square in itself and twice of the other even number should also be a perfect square.
Therefore either of n or (n + 1) must be a perfect square. Try for n = odd^2 and odd^2 – 1 values. n = 8 satisfies as the smallest.
For how many integers p, (30 – p)/(p – 10) is a prime number?
(30 – p) = 20 – (p – 10)
Given expression becomes 20/(p – 10) – 1 which should be prime.
Also, (p – 10) is a factor of 20 for it to be an integer at first place.
Substituting values for (p – 10) = 1, 2, 4, 5, 10, 20 out of which only (p – 10) = 1 and 5 lead to a prime number for expression. So, for p = 11 and 15 the expression is a prime number.
2 values.All the digits of a 50 digit number are 4 except the nth digit. The number is divisible by 13 for some choice of that n. Find the number of possible values of n.
4 not divisible
44 not divisible
44n divisible at n = 2
444n divisible at n = 6
44444n at n = 4 but n must not be 4. Not possible.
444444 completely divisible.
The cycle shall repeat.So we have for every 6 digits 3 numbers that will be divisible by 13.
So, for 50 digits we will have 3 * [50/6] = 3 * 8 = 24 possible values of n where [.] is the GIF.How many four digit perfect squares ABCD are possible such that DCBA is also a perfect square and is also a factor of ABCD given that A not equal to D.
abcd = k * (dcba)
k must also be a perfect square => k = 4 or 9.Case 1
k = 4
d has to be 1 or 2, but d cannot be 1 as abcd = 4(dcba) which is always even. Also, d cannot be 2 else abcd will not be a perfect square[Ending digits of perfect square cannot be 2, 3, 7 and 8]. Case 2 k = 9 d has to be 1=> a = 9
So, 1bc9 is a perfect square Unit digit 9 and number in the range of 1000 implies it may be 33^2 or 37^2 or 43^2. Out of which only 33^2 is a multiple of 3(We have k = 9 so square must be a multiple of 3). Hence our answer is only 1 possible value = 33^2 = 1089 whose reverse = 9801 is also a perfect squareLCM of three numbers is 1080 and their HCF is 2. How many such triplets are possible?
1080 = 2^3 * 3^3 * 5 and HCF = 2
Let the numbers be 2a, 2b, 2c.
Taking out one 2 from the LCM, 2^2 * 3^3 * 5
Power of 2 : Each can take : 0, 1, 2
Number of ways = 3^3 .
But not all can take from 0, 1 else LCM won’t be as required.
So, 3^3  2^3. We also can’t have all 1, 2 else HCF changes
3^3  2^3  2^3.
We removed all having 2^1 twice
3^3  2*2^3 + 1 = 12.Power of 3
4^3  2 * 3^3 + 2^3 = 18
Power of 5
2^3  1^3  1^3 = 6
Ordered triplets = 12 * 18 * 6Now for unordering,
2a, 2b, 2c where a, b, c coprime.
One 2 taken out from LCM.
2^2 * 3^3 * 5
1,1, n
2^2 , 2^2 , 3^3 * 5
3^3, 3^3 , 2^2*5
5, 5 , 2^2 * 3^3
{3!/2! For permutation} * 2 {for x, x, y or y, y, x form} * 4 for the 4 cases
(1296  3 * 2 * 4)/6 + 2 * 4 = 220 unordered triplets.A number when divided by 15 leaves a remainder of 2 and when divided by 18 leaves a remainder of r. How many values can r take?
N = 15a + 2 = 18b + r
r = 3(5a – 6b) + 2
5a – 6b = 0, 1, 2, 3, 4, 5 giving corresponding values of r as 2, 5, 8, 11, 14, 17.
6 values.Maximum value of HCF of n^2 + 17 and (n + 1)^2 + 17 is?
Approach 1
Let (n^2 + 17) = h * m and (n + 1)^2 + 17 = h * n where h is the HCF and m, n are coprimes.Subtracting we get,
(2n + 1) = h * (n – m) = h * k (say)
n = (h * k – 1)/2Substituting back in any one of the original equations,
h^2 * k^2 + 1 – 2 * h * k + 68 = 4 * h * m
h(4m – 2k – h * k^2) = 69
h is a factor of 69.
Highest value of h = 69.How many numbers from 4 to 499 are neither divisible by 2 nor 4 nor 5?
For every p consecutive naturals, only one of them is divisible by p while rest (p – 1) are not divisible where p is some prime. This leads us to the Euler Totient function in such problems which gives us number of numbers not divisible by one or more primes.
4 to 499 = (499 – 4 + 1) = 496 numbers
We need to keep our number range a multiple of LCM (numbers present for check of divisibility)
LCM(2, 4, 5) = 20.
So extending our range for additional numbers, we have our new range = 1 – 500 with 500 numbers.Now, number of numbers not divisible by 2 nor 4 nor 5 = 500 * (1 – 1/2) * (1 – 1/5) = 200 as numbers not divisible by 2 will automatically be not divisible by 4.
Out of these, the number 1 and 3 would have been counted but do not belong to our actual range hence must be removed.
So, we have 200 – 2 = 198 required numbers.A census taker goes into a house and asks the owner, “How many children do you have and what are their ages?” Owner: I have 3 children whose product of ages is 36 and sum of ages is the house number right opposite to our house.
Census taker goes out to the opposite house but comes back to enquire again.
Census taker: I need more information, I still don’t know their ages.
Owner: I need to go, my youngest child is sleeping upstairs.
Census taker: Thank you, I have everything I need now.”
What is the opposite house’s number and what are the children’s individual ages?36
1 * 1 * 36 : Sum = 38
1 * 2 * 18 : Sum = 21
1 * 3 * 12 : Sum = 16
1 * 4 * 9 : Sum = 14
1 * 6 * 6 : Sum = 13
2 * 2 * 9 : Sum = 13
2 * 3 * 6 : Sum = 11
3 * 3 * 4 : Sum 10When he knows the product of ages he can come up with all these combinations. After having a look at house number, he still needs more information. This means that house number which is the sum of ages is not unique for the combinations he has. That leaves us with: 1 * 6 * 6 : Sum = 13 and 2 * 2 * 9 : Sum = 13
Now the owner says that his “youngest” child is sleeping upstairs. It means we cannot have 2, 2, 9 as our combination as there won’t be a “youngest” child then but two younger children. Hence the individual ages are 1, 6, 6 while the house number = 13.
Given that 34! = X95232799ab96041408476186096435cd000000 where X is the leading digit and a, b, c and d are missing digits from the actual value of 34! What is the sum of (a + b + c + d)?
a) 0
b) 1
c) 4
d) 6
e) none of thesecd can be found out with the help of last two nonzero digits of 34!
34 = 5(6) + 4.
Last two nonzero digits = 12^6 * 6! * 31 * 32 * 33 * 34 = 84 * 72 * 24 = 84 * 3 * 76 = 52 * 76 = 52.
Given the options you could directly mark none of these.Otherwise, let us see how to proceed. Choose any three primes in the range 1 34 whose divisibility test will involve a, b and X. Form equations and solve for the three variables.
You will get a = 0, b = 3 and X = 2.
(a + b + c + d) = 10.