Numbers high level Exercise 2 - 4

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19. P is the product of all the factors of 15552. If P = ${12^N} \times M$, where M is not a multiple of 12, then find the value of M. [M and N are positive Integers]
a. ${3^{42}}$
b. ${3^{46}}$
c. ${3^{48}}$
d. ${3^{52}}$

Answer: A

Explanation:
15552 = ${2^6} \times {3^5}$
Product of all the factors of 15552 = ${\left( {{2^6} \times {3^5}} \right)^{\displaystyle\frac{{(6 + 1).(5 + 1)}}{2}}} = {2^{126}} \times {3^{105}}$
$ \Rightarrow $${\left( {{2^2}} \right)^{63}} \times {3^{63}} \times {3^{42}}$ = ${12^{63}} \times {3^{42}}$
So M = ${3^{42}}$

20. Let M be the set of all the distinct factors of the number N=${6^5} \times {5^2} \times 10$,Which are perfect squares. Find the product of the elements contained in the set M.
a. \({5^{20}}\)
b. \({5^{22}}\)
c. \({5^{24}}\)
d. \({5^{26}}\)

Answer: C

Explanation:
N = ${6^5} \times {5^2} \times 10$ = ${2^6} \times {3^5} \times {5^3}$
Even powers of 2 available: ${2^0},{2^2},{2^4},{2^6}$
Even powers of 3 available: ${3^0},{3^2},{3^4}$
Even powers of 5 available: ${5^0},{5^2}$
Therefore number of factors of the number N that are perfect squares = 4 x 3 x 2 = 24
Product of the elements contained in M
= ${2^{\left( {2 + 4 + 6} \right) \times 6}} \times {3^{\left( {2 + 4} \right) \times 8}} \times {5^{2 \times 12}} = {2^{72}} \times {3^{48}} \times {5^{24}}$

21. In a hostel there are 1000 students in 1000 rooms. One day the hostel warden asked the student living in room 1 to close all the doors of the 1000 rooms. Then he asked the person living in room 2 to go to the rooms which are multiples of his room number 2 and open them. After he ordered the 3rd student to reverse the condition of the doors which are multiples of his room number 3. If He ordered all the 1000 students like the same, Finally how many doors of those 1000 rooms are in open condition?
a. 31
b. 168
c. 169
d. 969

Answer: D

Explanation:
We understand that a door is in open or in close condition depends on how many people visited the room.
If a door is visited by odd number of persons it is in close condition, and is visited by even number of persons it is in open condition.
The number of people who visit a certain door is the number of factors of that number. Let us say room no: 24 is visited by 1, 2, 3, 4, 6, 8, 12, 24 which are all factors of 24. Since the number of factors are even this door is in open condition.
we know that the factors of a number N=${a^p}.{b^q}.{c^r}...$ are (p+1).(q+1).(r+1)...
From the above formula the product is even if any of p, q, r... are odd, but the product is odd when all of p, q, r are even numbers.
If p, q, r ... are all even numbers then N=${a^p}.{b^q}.{c^r}...$ is a perfect square.
So for all the perfect squares below 1000 the doors are in closed condition.
There are 31 perfect squares below 1000 so total doors which are in open condition are (1000 - 31) = 969

22. What is the product of all factors of the number N = ${6^4} \times {10^2}$ which are divisible by 5?
a. ${2^{210}} \times {3^{140}} \times {5^{105}}$
b. ${2^{206}} \times {3^{130}} \times {5^{105}}$
c. ${2^{225}} \times {3^{180}} \times {5^{125}}$
d. ${2^{215}} \times {3^{140}} \times {5^{125}}$

Answer: A

Explanation:
N = ${6^4} \times {10^2}$ = ${2^6} \times {3^4} \times {5^2}$
Total product of the factors = ${\left( {{2^6} \times {3^4} \times {5^2}} \right)^{\displaystyle\frac{{\left( {6 + 1} \right).\left( {4 + 1} \right).\left( {2 + 1} \right)}}{2}}}$ = ${\left( {{2^6} \times {3^4} \times {5^2}} \right)^{\displaystyle\frac{{105}}{2}}}$
So total product of the factors N which are not multiples of 5 =
${\left( {{2^6} \times {3^4}} \right)^{\displaystyle\frac{{\left( {6 + 1} \right).\left( {4 + 1} \right)}}{2}}}$ = ${\left( {{2^6} \times {3^4}} \right)^{\displaystyle\frac{{35}}{2}}}$
So, total product of the factors of N which are multiples of 5 = $\displaystyle\frac{{{{\left( {{2^6} \times {3^4} \times {5^2}} \right)}^{\displaystyle\frac{{105}}{2}}}}}{{{{\left( {{2^6} \times {3^4}} \right)}^{\displaystyle\frac{{35}}{2}}}}}$ = ${2^{210}} \times {3^{140}} \times {5^{105}}$

23. Let N = ${2^3} \times {3^{17}} \times {5^6} \times {7^4}$ and M = ${2^{12}} \times {3^5} \times {5^4} \times {7^8}$. P is total number of even factors of N such that they are not factors of M. Q is the total number of even factors of M such that they are not factors of N. Then 2P -Q = ?
a. 40
b. 126
c. 69
d. 195

Answer: A

Explanation:
N = ${2^3} \times {3^{17}} \times {5^6} \times {7^4}$
M = ${2^{12}} \times {3^5} \times {5^4} \times {7^8}$

So by the above diagram, we can see that P is the number of even factors of N which are not factors of M and Q is number of even factors of M which are not factors of N.
Now to calculate common even factors of N and M, we take HCF of N and M which is ${2^3} \times {3^{5}} \times {5^4} \times {7^4}$ ($\because$ take minimum powers of both numbers).
Even factors of above number = 3 × (5 + 1) × (4 + 1) × (4 + 1) = 450 ( $\because$ We have to include atleast one multiple of 2 to make it even number.
Even factors of P = 3 × 18 × 7 × 5 = 1890
Even factors of Q = 12 × 6 × 5 × 9 = 3240

So 2P - Q = 90

1234

Number System: BasicsNumber System: HCF and LCMNumber System: Factors and CoprimesNumber System: Divisbility Rules Number System: Power of a number in a Factorial Number System: Units digit of an expression Number System: Last two digits of an expressionNumber System: Base SystemNumber System: Last non zero digit of a factorial (LNZ)Number System: Last two non zero digits of a factorialNumber System: High level Exercise