CMAT Number Systems Questions

$$(A)$$: $$3^{3^{3^{3}}}$$ = $$3^{3^{27}}$$

$$(B)$$: $$3^{(33)^{3}}$$ = $$3^{(33)^{3}}$$

$$(C)$$: $$(3^3)^{33}$$ = $$3^{99}$$

$$(D)$$: $$3^{333}$$

So, the powers of 3 in (A), (B), (C) and (D) are: $$3^{27},33^3,99,333$$ respectively.

Now, we know, $$3^{27}>33^3>333>99$$

So, the powers of $$3$$ in $$(A)>(B)>(D)>(C)$$

So, correct answer is option $$(A)$$

We can rewrite the following as:

A) $$\sqrt[4]{3}$$ = $$3^{\dfrac{1}{4}}$$

B) $$\sqrt[3]{2}$$ = $$2^{\dfrac{1}{3}}$$

C) $$\sqrt[6]{5}$$ = $$5^{\dfrac{1}{6}}$$

D) $$\sqrt[6]{7}$$ = $$7^{\dfrac{1}{6}}$$

Now, we will make the denominator of all equal:

A)$$3^{\dfrac{3}{12}}$$

B) $$2^{\dfrac{4}{12}}$$

C) $$5^{\dfrac{2}{12}}$$

D) $$7^{\dfrac{2}{12}}$$

Simplifying further, we get, 

A)$$3^{\dfrac{3}{12}}$$ = $$\left(3^3\right)^{\dfrac{1}{12}}=\left(27\right)^{\dfrac{1}{12}}$$

B) $$2^{\dfrac{4}{12}}$$ = $$\left(2^4\right)^{\dfrac{1}{12}}=\left(16\right)^{\dfrac{1}{12}}$$

C) $$5^{\dfrac{2}{12}}$$ = $$\left(5^2\right)^{\dfrac{1}{12}}=\left(25\right)^{\dfrac{1}{12}}$$

D) $$7^{\dfrac{2}{12}}$$ = $$\left(7^2\right)^{\dfrac{1}{12}}=\left(49\right)^{\dfrac{1}{12}}$$

Arranging them in sequence: We get: B < C < A < D

If abc is positive, it can have two cases:

Case 1: when all three integers are postive

Case 2: when any two are negative and the third is positive (as the two negative will cancel while multiplying)

We need to find the option which must be true:

Option 1: $$\dfrac{a}{b} < 0$$

It is not necessary that one of a and b is negative and the other one is positive. It can be possible that but both a and b are negative and hence the fraction can be greater than 0. Therefore, this option is not necessarily true. 

Option 2: $$\dfrac{ab}{c} > 0$$

Here, if all the three integers are positive then the expression will definitely be greater than 0. On the other hand, if any two are negative the according to the expression, the negatives will cancel out and the expression will again be greater than 0. Hence, this option must be true. 

Option 3: $$bc < 0$$

Same as option 1, it is not necessary that one of  b and c is negative and the other one is positive for the expression to be true. There can also be the possibility when both are negative and the product can be greater than 0 in that case.

Option 4: $$a > bc$$

We don't know the numeric values so we can't say that a will be greater than the product of b and c necessarily. 

We are given the surds, $$\sqrt[3]{2}$$, $$\sqrt{3}$$, $$\sqrt[4]{5}$$ and $$\sqrt[6]{7}$$.

The values can be written as, $$2^{\frac{1}{3}},\ 3^{\frac{1}{2}},\ 5^{\frac{1}{4}},\ 7^{\frac{1}{6}}$$ which can be rewritten as,

$$2^{\frac{8}{24}},\ 3^{\frac{12}{24}},\ 5^{\frac{6}{24}},\ 7^{\frac{4}{24}}$$.

In the above values, comparing the values of $$2^8,\ 3^{12},\ 5^6,\ 7^4$$ is enough and they can be rewritten as,

$$\left(2^4\right)^2,\ \left(3^6\right)^2,\ \left(5^3\right)^2,\ \left(7^2\right)^2\ =\ $$ $$16^2,\ 729^2,\ 125^2,\ 49^2$$

We can see that the value of A < D < C < B.

Therefore, the correct answer is option D.

Let the two-digit number be xy.

(2x)(y/2) = yx

So, 2x = y or y/2=x

This means the digit in the unit's place is twice the digit in ten's place.

We are given that the HCF of two numbers is 21. Let the numbers be X, and Y. Then X and Y can be written as,

X = 21a and Y = 21b where a and b are coprimes.

We know that, 

product of the numbers =  LCM * HCF

X * Y = 3003 * 21

21a * 21b = 3003 * 21

a * b = 143

a * b = 11 * 13

There are two possibilities for (a,b) with a and b as coprimes, one being (1, 143) and the other being (11, 13). We cannot take (1, 143) because, in this case, one of the numbers would be 21, but we are given that both numbers are greater than 21. So, the numbers X and Y are 21 * 11 and 21 * 13, which are 231 and 273. So, the sum of the numbers = 231 + 273 = 504.

Therefore, the correct answer is option A.