HCF and LCM – Railway Maths Part IV

Here we have to find the HCF of (155 – 5), (260 – 10), (315 – 15)
So the HCF of 150, 250, 300 is = 50,
i.e. the greatest number that will divide 155, 260, 315 and leave the remainder 5, 10 and 15 respectively is 50.

The smallest 5 digit number = 10000
Number nearest to 10000, divisible by 7 = 9996
So, the smallest 5 digit number which will leave a remainder of 6 when divided by 7 is
9996 + 6 = 10002

Let the two numbers are a and b.
As the sum of the two numbers is 60 and their HCF and LCM are 12 and 72 respectively,
⇒ a + b = 60 and ab = 12 × 72
We have to find the sum of the reciprocal of the two numbers,
⇒ 1/a + 1/b = (a + b)/ab = 60/(12 × 72) = 5/72

Let the two numbers are x and y,
i.e. x + y = 1500 … (i)
and xy = 16379 (LCM)

(x – y)² = (x + y)² – 4xy
x – y = 1478 … (ii)

Adding (i) and (ii)
(x + y) + (x – y) = 1500 + 1478 = 2978
2x = 2978, x = 1489, y = 11

Short Tricks:-

Given that,
x + y = 1500 and (L.C.M) xy = 16379
In this type of question, make a factor of L.C.M = 16379 = 11 × 1489
Clearly, condition satisfy,
x + y = 11 + 1489 = 1500

The smallest four digit number = 1000
So, nearest perfect number to it = 961,
So, the smallest number which should be subtracted from 1000 to make it a perfect square is
(1000 – 961) = 39

Greatest 3 digit number = 999
Remainder left when greatest number divided by 24 = 15
Greatest 3 digit number divisible by 24 =
999 – 15 = 984
Similarly, smallest 4 digit number = 1000
Remainder left when smallest number divisible by 24 = 16
Smallest number divisible by 24 = 1000 + (24 – 16) = 1008
Hence, the sum of the greatest 3 digit number and the smallest 4 digit number that have 24 as their HCF =
(984 + 1008) = 1992

The sum of the LCM and HCF of two numbers is 369 and the difference between the LCM and HCF of two numbers is 351. So,
LCM = (369 + 351)/2 = 360 &
HCF = (369 – 351)/2 = 9
⇒ (product of numbers) a × b = LCM × HCF
⇒ 72 × b = 360 × 9 ⇒ b = 45
i.e. the second number is 45.

The least number that which is divisible by 144, 108 and 72 is its LCM = 432
So, the least number that when divided by 144, 108 and 72 leaves the remainder 3 in each case is (432 + 3) = 435

The LCM of 5, 6, 7 is the least number which is divisible by 5, 6, 7 = 210.
And the multiples of the LCM = 210 are 420, 630, 840 which lies between the range (250 - 1000).
So, only 3 numbers from 250 to 1000 are exactly divisible by 5, 6 and 7.

Let the numbers be 26a and 26b.
Now we know, LCM × HCF = product of two numbers
26a × 26b = 156 × 26 ⇒ ab = 6
Again, the difference between the two numbers is also 26
26(a – b) = 26 ⇒ (a – b) = 1
So, ab = 6 & (a – b) = 1, which is only possible if a = 3 and b = 2.
Then the two numbers are,
a = 26 × 3, b = 26 × 2,
∴ the sum of the numbers = 26 × (3 + 2) = 130