CAT Past Year Questions: HCF & LCM - General Test Preparation for CUET

Both the given sequences are A.P.

The common difference for the first sequence, d1 = 4

The common difference for the second sequence, d2 = 5

The first common term in the two given A.P. is 19.

The common terms will also be in arithmetic progression with common difference

LCM (d1, d2) = LCM (4, 5) = 20

Let there be ‘n’ terms in this sequence, then the last term would be ≤ 415

Hence for A.P. of common terms

a + (n – 1) d ≤ 415

⇒ 19 + (n – 1) × 20 ≤ 415

⇒ (n – 1) × 20 ≤ 396 

⇒ (n – 1) = [396 / 20] where [396 / 20] is the greatest integer

⇒ (n –1) = 19

∴ n = 20

Here, 120 ≤ n ≤ 240.

120 = 23 (3)(5) and 240 = 24 (3)(5)

So, the prime factors involved in 120 and 240 are the same.

So, number of co-primes of 240 lying between 120 and 240 = φ(240) – φ(120).

1000 = 23 × 53 and 2000 = 24 × 53

Since LCM (c, a) and LCM (b, c) is 24 × 53 and LCM (a, b) = 23 × 53, so the factor 24 must be present in c.

Hence c = 24 × 5x, where x ranges from 0 to 3 Therefore, there are four possible values of C.

Since, HCF of (a, b) = K × 53, it means

a = 2y × 53

b = 2z × 53

x = 0 to 3, y = 0, then z = 3 → 4 cases.

x = 0 to 3, y = 1, then z = 3 → 4 cases.

x = 0 to 3, y = 2, then z = 3 → 4 cases.

x = 0 to 3, y = 3, then z = 3 → 4 cases.

x = 0 to 3, y = 3, then z = 2 → 4 cases.

x = 0 to 3, y = 3, then z = 1 → 4 cases.

x = 0 to 3, y = 3, then z = 0 → 4 cases.

Hence, total cases = 28.

Let the two numbers be x and y according to question,

x + y = 600 and x/y = 7/8

∵ x + y = 600 ...(i)

x/y = 7/8 ⇒ 8x - 7y = 0 .....(ii) 

From equation (i) and (ii)

x = 280 and y = 320

∴ LCM of 280 and 320 = 2240

1080 = 23 × 33 × 51

(where N = apbqcr

∴ No. of factors (p + 1) (q + 1) (r + 1)

∴ No. of factors of 1080 (3 + 1) (3 + 1) (1 + 1) = 4 × 4 × 2 = 32

1800 = 23 × 32 × 52

(where N =  apbqcr

∴ No. of factor (p + 1) (q + 1) (r + 1)

∴ Number of factor of 1800 = (3 + 1)(2 + 1)(2 +1)

= 4 × 3 × 3 = 36

∴  HCF of 1080 and 1800 = 23 × 32 × 5

where N = ap, bq, cr

No. of factors HCl = (p + 1)(q + 1)(r +1)

∴ No. of factors HCF of two numbers = (3 + 1) (2 + 1) (1 + 1)

= 4 × 3 × 2 = 24

So, the required number of divisors

= (32 + 36) – 2 × 24 = 20

LCM of 6, 4 and 3 = 12

Multiply by 12 of each number in power

⇒ 214, 39, 58

So, ascending order is

58 > 39 > 214  or 52/3 > 33/4 >27/6

The general term is of the form

(X + n(n + 2)) / (n + 2)

n(n + 2) is always divisible by (n + 2).

So we can say that n(n + 2) ± 1 would never be divisible by (n + 2).

If we put X = –1, the numerator and denominator of all the terms would be co-prime.