HCF and LCM Questions (Solved)
Example 1: Find the greatest number which when divides 259 and 465 leaves remainders 4 and 6 respectively.
Sol: Here, the numbers 259 and 465 leave the remainders 4 and 6 respectively. So, the required number will be obtained by finding the H.C.F. of 259 – 4 = 255 and 465 – 6 = 459. The HCF of 255 and 459 is 51 which is the number.
Example 2: Find the least number which when divided by 6, 14, 18 and 22 leaves remainder 4 in each case.
Sol: The LCM of 6, 14, 18 and 22 is 1386. In order to get remainder 4 in each case, we will add 4 to the LCM. So, the number is 1386 + 4 = 1390.
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Example 3: Find the least number which when divided by 8, 12, 20 and 36 leaves remainders 6, 10, 18 and 34 respectively.
Sol: Here, the numbers are 8, 12, 20 and 36 and the respective remainders are 6, 10, 18 and 34. The difference between numbers and the respective remainders is equal to 2. So, first of all find the LCM of 8, 12, 20 and 36 which is 360. The required number is 360 – 2 = 358.
Example 4: Find the greatest number, which when divides 41, 71 and 91, leaves the same remainder in each case.
Sol: Take the difference between all the three pairs of numbers and their HCF will be the answer i.e. 91 – 41 = 50, 71 – 41 = 30 and 91 – 71 = 20. Now, the HCF of 50, 30 and 20 is 10.
Sometimes in such questions, the common remainder can also be asked. You can divide any of the numbers given by HCF (41 ¸ 10) and find the remainder to be equal to 1.
Example 5: Find the L.C.M of 2/5, 3/10 and 6/25
L.C.M. of 2/5, 3/10 and 6/25 = L.C.M of 2,3 and 6 / H.C.F of 5,10 and 25
L.C.M. of 2, 3 and 6 = 6 ; H.C.F. of 5, 10 and 25 = 5. Thus, the LCM of these three fractions will be 6/5.
Example 6: How often will five bells toll together in one hour if they start together and toll at intervals of 5, 6, 8, 12, 20 seconds, respectively?
Sol: The time after which the bells will ring together is the L.C.M. of 5, 6, 8, 12 and 20 seconds, i.e. 120 seconds. The number of times they will toll together in one hour = (3600 ¸ 120) = 30. Thus, they will toll together 30 times in an hour.
⇒ Sometimes the question is how many times they toll together in the first hour, in that case after finding the answer like above, you need to add 1 for the start together as well i.e. in the first hour it is 1 more than the usual number of times.
Example 7: Find the greatest number of four digits which when divided by 11, 21, 15, and 28 leaves 5, 15, 9 and 22 as remainders, respectively?
Sol: The LCM of 11, 21, 15 and 28 is 4620. The greatest 4 digit multiple of 4620 is 9240. Now the difference between the number and the respective remainder is same here. So to find the greatest 4 digit number which satisfies the given condition, subtract the common difference from 9240. Hence the required number is 9240 – 6 = 9234.
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