HCF and LCM

The HCF of two numbers is 1/6th of the LCM of those two numbers. if one of the numbers is 15, find the other?

LCM and HCF

The product of 2 digit numbers is 2160 and their LCM is 180. Find the 2 numbers.

HCF

Find the HCF of 2923 and 3239

HCF and LCM,

Delegates from 5 different schools are coming to Srimati sulochanadevi singhania school. The number of delegates from Kendriya vidyalaya is 42, 60 from Poddars', 210 from Hiranandani, 90 from Jamnalal and 84 from Lodha. They are being put up at Satkar Residency Hotel. What is the minimum number of rooms that would be required to accommodate so that each room has the same number of occupants and occupants are all from the same school?

HCF and LCM

A number p is such that:
p = 5k1 + 4,
p = 6k1 + 5,
p = 7k1 + 6,
p = 8k1 + 7,
p = 9k1 + 8,

Find the smallest number that satisfy this condition

HCF and LCM

The product of 2 numbers and its HCF is 1080. How many such pair of numbers are possible which satisfy the above condition?

HCF and LCM

Three numbers are in the ratio of 3:6:9 and their LCM is 2700. What is the HCF?

LCM and HCF

Three numbers are in the ratio of 3:4:5 and there LCM is 2400. What is the HCF?

LCM and HCF

Three wheels can complete 60, 36 and 24 rpm respectively. There is a red spot on each wheel that touches the ground at time zero. After how much time, all these spots will simultaneously touch the ground again?

LCM and HCF

On Ashok Marg three consecutive traffic change after 36,42 and 72 secs respectively. If the lights are first switched on at 9:00 A.M. sharp, at what time will they change simultaneously?

IMO STD IX

HCF and LCM ----MAT 2014

question number 45 of MAT

The HCF of two numbers is “x” and the sum of the uncommon factors is 20 and the product of the uncommon factors is 99. If the sum of the two numbers is 260, then thee two possible numbers are......................................

HCF and LCM

the traffic lights at 3 different road crossings change after 48 sec, 72 sec and 108 sec . If they change simultaneously at 7 a.m after what time will they change again simultaneously ?

Questions (with answers) on HCF and LCM

1. Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
2. The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
3. Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:
4. The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:  
5. The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is: 
6. The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is
   
   answers

harder HCF --- tutorial

Find the HCF of the following pair of numbers

·         703, 851

·         553, 869

·         2117, 2263

·         2773, 3009

HCF and LCM

The HCF of two numbers is 23, and two of the factors of LCM are 13 and 14. Find the greater of the two numbers.

HCF

Find the HCF of

x⁴ + x² + 1 and x¹⁰ + x⁵ + 1

Division , HCF and LCM

Use Euclid’s division algorithm to find the HCF. Also find which of the following pairs of numbers are co-prime: 

NCERT Exemplary Question

(i)                 231, 396

(ii)               847, 2160

Tougher questions in LCM

Find the greatest number less than 900 which is exactly divisible by 8, 12 and 28?

Answer:
LCM of 8,12, 28 = 168
to find the multiple of 168 = 900/168 =  5 60/168.
So, greatest number less than 900 which is exactly divisible = 900 - 60 = 840.

What greatest number can be subtracted from 2470 so that the Remainder may be divisible by 42, 98 and 105?

Answer:
Remainder will be LCM of 42, 98, 105 = 1470
So, the greatest number that can be subtracted = 2470 - 1470 = 1000.

Find the least number that being increased by 8 is divisible by 21, 35 and 48?

Answer:
LCM of 21, 35, 48 = 1680.
SO, the least number that being increased by 8 = 1680 -8 = 1672.

Find the least number which when divided by 18, 24, 30 and 42 will leave a remainder of 1 in each case.

Answer:
LCM of 18, 24, 30, 42 = 2520 
So, the least number that would leave a remainder of 1 = 2520 + 1 = 2521.

HCF and LCM of fractions

To find the HCF and LCM of several fractions:

Step 1: First express the fractions in their lowest terms.

Step 2: HCF = HCF of Numerator / LCM of denominator

Step 3: LCM = LCM of Numerator/ HCF of denominator



 Question:

The circumference of the fore wheels and hind wheels of a carriage are 2²/₅ and 3 ³/₇ feet respectively. A chalk mark is made at the point of contact of the wheels with the ground at any given point. How far will the carriage have traveled so that their chalk marks may be again on the ground at the same time?

 

tougher problems on LCM

A gardener had a number of shrubs to plant in rows. At first he tried to plant 5 in each row, then 6, then 8, then 9 and then 12, but had always 1 left. On trying 13 he had none left. What is the smallest number of shrubs that he could have had?    

Answer:

           

The LCM of 5,6,8,9 and 12 is 360.

So, the required number is 360y + 1

Given that 360y + 1 does not leave a remainder with 13.

360 = 13 x 27 + 9

So, 360y + 1 = (13 x 27 + 9)y + 1

= 13x27y + 9y+1

Since there is no remainder, 9y + 1 should be equal to 13 or its multiple.

For, y = 1, 9y +1 = 10 ..... (x)

Y = 2, 9y + 1 = 19 ..............(x)

Y = 3, 9y +1 = 28 ...............(x)

Y = 4, 9y +1 = 37 ...............(x)

Y = 5, 9y +1 = 46 ...............(x)

Y = 6, 9y +1 = 55 ...............(x)

Y = 7, 9y +1 = 64 ...............(x)

Y = 8, 9y +1 = 73 ...............(x)

Y = 9, 9y +1 = 82 ...............(x)

Y = 10, 9y +1 = 91 ...............(√) divisible by 13

So, the number is 360y + 1 = 360x10 + 1 = 3601