HCF & LCM - UPSC Free MCQ Practice Test with solutions

Let the numbers be 37a and 37b.

Then, 37a x 37b = 4107

 ab = 3.

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

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

 Greater number = 111.

Given numbers are 43, 91, and 183.
Subtract smallest number from both the highest numbers.
We have three cases:
183 > 43; 183 > 91 and 91 > 43
Therrefore,
183 – 43 = 140
183 – 91 = 92
91 – 43 = 48
Now, we have three new numbers: 140, 48 and 92.

HCF (140, 48 and 92) = 4
The highest number that divides 183, 91, and 43 and leaves the same remainder is 4.

N = greatest number that will divide 105, 4665 and 6905 leaving the same remainder in each case

⇒ N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)
⇒ N = H.C.F. of 3360, 2240 and 5600 = 1120
Sum of digits in N = (1 + 1 + 2 + 0) = 4

Greatest number of 4−digits is 9999.

Now, 15=3×5

25 = 5×5

40 = 2×2×2×5

and 75 = 3×5×5

L.C.M. of 15,25,40 and 75 is 2×2×2×3×5×5 = 600.

On dividing 9999 by 600, the remainder is 399.

Required number = (9999−399) = 9600.

Step 1: Interpret the problem

• Six bells toll at intervals of 2, 4, 6, 8, 10, and 12 seconds.
• They start together at time t = 0.
• We need to find how many times they toll together in 30 minutes.
• This is an LCM problem → They will toll together every LCM(2, 4, 6, 8, 10, 12) seconds.

Step 2: Find the LCM

• Prime factorization:
• 2 = 2
• 4 = 2²
• 6 = 2 × 3
• 8 = 2³
• 10 = 2 × 5
• 12 = 2² × 3
• Take highest powers:
• 2³ × 3 × 5 = 120
• So, LCM = 120 seconds = 2 minutes

Step 3: How many times in 30 minutes?

• They toll together every 2 minutes.
• Total duration = 30 minutes
• Number of times = (30 ÷ 2) + 1
• (+1 because they also toll together at the start, i.e., t = 0)
• = 15 + 1 = 16

Given numbers are 1.08 , 0.36 and 0.90
G.C.D. i.e. H.C.F of 108, 36 and 90 is 18
Therefore, H.C.F of given numbers = 0.18            

Let the numbers 13a and 13b.

Then, 13a x 13b = 2028

⇒ ab = 12.

Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1
Now, the co-primes with product 12 are (1, 12) and (3, 4).

⇒ The required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).

L.C.M. of 6, 9, 15 and 18 is 90.
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
∴ Required number = (90 x 4) + 4 = 364.