Question

# If r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q is r^{2}t^{4}s^{2}, then the number of ordered pair (p, q) is –

- 224
- 225
- 252
- 256

Hint:

### Natural numbers known as prime numbers can only be divided by one (1) and by the number itself. In other terms, prime numbers are positive integers greater than one that only have the number itself and the number's first digit as factors. Here we have given r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q is r2t4s2, then what is the number of ordered pair (p, q).

## The correct answer is: 225

### An ordered pair is made up of the ordinate and the abscissa of the x coordinate, with two values given in parenthesis in a certain sequence.

Now we have given the LCM as: r^{2}t^{4}s^{2}

Consider following cases:

Case 1: if p contains r^{2} then q will have r^{k}, for the value k=0,1.

So 2 ways.

Case 2: if q contains r^{2} then p will have r^{k}, for the value k=0,1.

So 2 ways.

Case 3:Both p and q contain r^{2}

So 1way.

So after this we can say that:

exponent of r=2+2+1 = 5 ways.

Similarly

exponent of t=4+4+1=9 ways.

exponent of s=2+2+1 = 5 ways.

So total ways will be:

5 x 5 x 9 = 225 ways.

Finding the smallest common multiple between any two or more numbers is done using the least common multiple (LCM) approach. A number that is a multiple of two or more other numbers is said to be a common multiple. Here we understood the concept of LCM and the pairs, so the total pairs can be 225.

### Related Questions to study

### A rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having edge length one unit then no. of rectangles which have odd unit length

Here we used the concept of number system and the rectangle, we can also solve it by permutation and combination. herefore, we get the number of rectangles possible with odd side length = m^{2}n^{2}.

### A rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having edge length one unit then no. of rectangles which have odd unit length

Here we used the concept of number system and the rectangle, we can also solve it by permutation and combination. herefore, we get the number of rectangles possible with odd side length = m^{2}n^{2}.

^{n}C_{r} + ^{2n}C_{r+1} + ^{n}C^{r+2} is equal to (2 r n)

^{n}C_{r} + ^{2n}C_{r+1} + ^{n}C^{r+2} is equal to (2 r n)

The coefficient of in is

The coefficient of in is