Number System Practice Gym by Anubhav Sehgal (NMIMS, Mumbai)  Part 11

A natural number x leaves a remainder 1 when divided by p. The resultant quotient, when divided by q, leaves a remainder 2. The resultant quotient, when divided by r, leaves a remainder 3 and the quotient, thus obtained, is exactly divisible by 5. If p, q and r are all natural numbers, which of the following is the least possible value of x?
It is a case of successive division. In such a case the number is written as:
x= (p(q(r(5k) + 3) + 2) + 1.
As the number needs to be smallest, we must take as small values as possible.
Now, remainder with p is 1, so p >1. So, p = 2.
Remainder with q = 2, so, q > 2, so q = 3.
Remainder with r = 3, so, r > 3, so, r = 4.
And, k = 0. x = 2(3(4(0) + 4) + 2) + 1 = 2(3(4) + 2) + 1 = 2(14) + 1 = 29.One of the smaller sides of a right angled triangle is (2^2) * (3^3) * (4^4) * (5^5) * (6^6) * (7^7). It is known that other two sides are integers. How many triangles of this type are possible?
b^2 = (2^4) * (3^6) * (4^8 ) * (5^10) * (6^12) * (7^14)
= (2^32) * (3^18 ) * (5^10) * (7^14)
=> (a + c)(a  c) = (2^32) * (3^18 ) * (5^10) * (7^14)
Since both (a + c) and (a  c) are even, say 2k and 2n, then
k * n = (2^30) * (3^18 ) * (5^10) * (7^14)Now, k is greater than n, so we just have to write (2^30) * (3^18 ) * (5^10) * (7^14) as product of two numbers where k is the greater one and n is smaller one, then we can get values for a and c
So, answer will be (31 * 19 * 11 * 15  1)/2 = 48592In a test consisting of 15 questions, 3 marks are awarded for a correct answer, 1 mark is deducted for an incorrect answer and no mark is awarded for an unattempted question. If a student attempts at least one question in the paper, what is the number of distinct scores that he can get?
Basically if no of marks are +n for correct and 1 for incorrect, then
Number of score which are not possible is given by (n  1) + (n  2) + ... + 1 = n * (n  1)/2 = C (n, 2)
Assume a +6 and 1 marking scheme.
When one gets 30 correct; he gets 180.
And with 29 correct, he gets 174.
So, for a score of more than174, one needs 30 correct.
And with 30 correct i.e. all correct, you can have 0 wrong.
So, 30 correct, 1 wrong or 2 wrong or 3 wrong or 4 wrong or 5 wrong is not possible.
With 29 correct, 1 wrong is possible. But 2, 3 , 4 or 5 is not.
With 28 correct, 3, 4, 5 wrong is not possible.
So, basically, when we have +6 and 1: Number of invalid scores are: 1 + 2 + 3 + 4 + 5 = 15When we have +5 and 1:
Number of invalid scores are: 1 + 2 + 3 + 4 = 10.When we have +4 and 1:
Number of invalid scores are: 1+2+3 = 6.Here, 15 questions are there with +3 and 1 marking scheme.
Highest marks possible = 15 * 3 = 45
Lowest possible marks = 15 * 1 = 15
Total marks = 45 – (15) + 1 = 61 possible scores.
Out of which, C (3, 2) = 3 are not possible.
So, number of valid scores = 61 – C (3, 2) = 58.How many integers ‘n’ are there between 0 and 10^99, such that the unit digit of n^3 is 1?
Each number from 1 to 10 results in distinct unit digit for cube.
So, each digit in cube appears 1/10 times.
So, from 1 to 10^99 it will appear 10^98 timesHow many subsets of the set {1, 2, 3, 4 ... 30} have the property that the sum of the elements of a subset is less than or equal to 232?
Sum of all elements of set S = 465.
For any subset having sum of elements x, there will be one complimentary set which is having sum of elements equal to (465 – x).Example let there be a set {1, 2, 3, 4}.
Then subset which is having sum =3 will be {1, 2} {3}: (2)
and subset having sum (1+2+3+4) 3 =7 will also be 2: {3,4} {1,2,4}So, number of subsets having sum of elements x = Number of subsets having sum of elements (465  x)
=> No of subsets having sum 1 = No of subsets having sum 464
Also, No of subsets having sum 2 = No of subsets having sum 463
….
No of subsets having sum 232 = No of subsets having sum 233
Adding them we will get,
No of subsets having sum less than or equal to 232
= No of subsets having sum more than 233
= 2^30/2
= 2^29When a number N (N>500) is successively divided by 3,5,7, it leaves remainder 2,3,6 respectively. Find the remainder when the smallest such N is divided by 35?
N = 3(5(7k + 6) + 3) + 2 = 105k + 83.
N = 83, 188, 293, 398, 503, etc.
Our required N = 503.
503 mod 35 = 13.If the first 99 natural numbers are written side by side to form a new number 12345678 …. 9899, then how many minimum numbers are to be removed so that, the new number is completely divisible by 11?
Sum at odd places = 1+3+5+7+9+9(1+2+...+9) = 25+9(45) = 430
Sum at even places = 2+4+6+8+10(1+2+.....+9) = 30+10(45) = 470
Sum => even = 470 and odd => 430
(Sum of digits at odd position from right)  (Sum of digits at even positions from right) = 40
Even  Odd= 40
Remove 99 > diff =40
Remove 98> diff =41
Remove 97> diff =43
Remove 96> diff =46
Remove 95> diff =50
Remove 94> diff =55We should remove 6 numbers: {99, 98, 97, 96, 95, 94}.
N = 4^7 × 5^9 + 2^4 × 7 + 3 × 5^3 + 2^6 × 5^8. How many distinct digits are there in the number N?
= 32 * 10^9 + 84 + 375 + 25 * 10^6
= 32025000459.
So, 6 distinct digits.S1 = {2, 4, 6, 8, …, 800} S2 = {3, 6, 9, 12, ....., 900} If S3 = S1 ∪ S2, then what will be the 105th element of S3 if all its elements are arranged in increasing order?
S1 ∪ S2 is a set comprising of all numbers which are multiples of 2 or 3 (or both).
Now, we know that for a number N, there are N(1  1/2)(1  1/3) = N/3 numbers which are coprime to 2 and 3 and less than N.So, there are 2N/3 numbers less than or equal to N which are multiples of 2 or 3 (or both).
=> 2N/3 = 106.
=> N = 159 is 106th such number.
=> 158 should be the 105th number.N is the smallest number such that N/2 is a perfect square and N/3 is a perfect cube. Find the number of factors of N.
N/2 = x^2
N/3 = y^3
N^2 = 6.x^2.y^3
N = xy.root(6y)
For N to be a natural number,6y must be a perfect square => y = 6 => x = 2 * 3^2 and N = 2^3 * 3^4
=> Number of factors = 4 * 5 = 20