HCF and LCM

Find the 4-digit smallest number which when divided by 12, 15, 25, 30 leaves no remainder:
Explanation :
LCM of 12, 15, 25 and 30 is 300
least number of 4-digit divided by 300 is 1200

Greatest number of 4-digit's is 9999.
L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
Required number (9999 - 399) = 9600.

Details

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., (37, 111). ∴ Greater number = 111.

The greatest number which when divides 564 and 467 leaves the remainder as 4 and 7.
564 - 4 = 560
467 - 7 = 460
HCF (560, 460) = 20
So, the greatest number is 20.

We know, Divisor, d₁ = 3, d₂ = 5, d₃ = 7
Remainders, r₁ = 2, r₂ = 3, r₃ = 4
Complete remainder = d₁ d₂ r₃ + d₁ r₂ + r₁
= (3×5×4) + (3×3) + 2
= 71.

Divisor×Quotient + Remainder = Number
6×7 + 5 = 47
5×47 + 3 = 238
4×238 + 2 = 954
∴ required number is 954.

Here, the HCF = 33
Then, let the two numbers be 33a and 33b,
Respectively, where a and b are coprime.
Then, 33a + 33b = 528
a + b = 16
Hence, values of (a, b) which are coprime to each other can be
(1, 15), (3, 13), (5, 11) and (7, 9).
Hence, required number of pairs is 4.

L.C.M. of 5, 6, 4 and 3 = 60. On dividing 2497 by 60, the remainder is 37. Number to be added = (60 - 37) = 23.

LCM of 7, 9 and 12 = 84
In the given options, only 253 is the least number which when divided by 7, 9 and 12 leaves the same remainder 1 in each case, since 253 leaves the remainder 1, when 253 is divided by 84.