100+ HCF and LCM Questions for SBI PO, SO, Clerk Pdf - 1

Ans: D

L.C.M. of 10, 12, 15 and 18 = 540.

Dividing (99999 + 3769) by 540, the remainder is 88.

∴ Required number = 99999 - 88 = 99911.

Ans: C

99 = 1 × 3 × 3 × 11

101 = 1 × 101

176 = 1 × 2 × 2 × 2 × 11

182 = 1 × 2 × 7 × 13.

So, divisors of 99 are 1, 3, 9, 11, 33 and 99.

divisors of 101 are 1 and 101

divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176

divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.

Hence, 176 has the most number of divisors.

Ans: B

Let the numbers be x and 4x.

Then, x × 4x = 84 × 21

⇔ x² = $$({84 × 21} / {4})$$

⇔ x = 21.

Hence, large number = 4x = 84.

Ans: A

Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43)

                          = H.C.F. of 48, 92 and 140 = 4.

Ans: B

Let the numbers be 37a and 37b.

Then, 37a × 37b = 4107 ⇒ ab = 3.

Now, co-primes with product 3 are (1, 3)

So, the required numbers are (37 × 1, 37 × 3) i.e., (1, 111).

∴ Greater number = 111.