The problem asks for the least number that leaves a remainder of 4 when divided by 5, 6, 8, 9, and 12.
Finding LCM of Divisors
To find this least number, we first need to find the Least Common Multiple (LCM) of the divisors: 5, 6, 8, 9, and 12.
First, find the prime factorization of each divisor:
- $5 = 5$
- $6 = 2 \times 3$
- $8 = 2^3$
- $9 = 3^2$
- $12 = 2^2 \times 3$
The LCM is found by taking the highest power of each prime factor present in the factorizations:
LCM = $2^3 \times 3^2 \times 5$
LCM = $8 \times 9 \times 5$
LCM = $72 \times 5$
LCM = $360$
Determining the Least Number
The question states that the number leaves a remainder of 4 in each case. To find the least such number, we add the remainder to the LCM.
Least Number = LCM + Remainder
Least Number = $360 + 4$
Least Number = $364$
Therefore, the least number that leaves a remainder of 4 when divided by 5, 6, 8, 9, and 12 is 364.
Similar Questions
- The H.C.F. and the L.C.M. of two numbers are 5 and 495, respectively. If one of the numbers is 55, find the other one.
- The HCF of two numbers is 16 and their LCM is 240. If one number is 48, find the other number.
- The sum of two numbers is 132 and their L.C.M. is 630. What are the two numbers?
- The LCM of the numbers 12.8 and 0.004 is:
- Find the least number which is divisible by first ten natural numbers.
- Find the HCF of ($3^{45} - 1$) and ($3^{35} - 1$).
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- HCF of $\frac{1}{3}, \frac{3}{4}, \frac{4}{5}$ and $\frac{5}{6}$ is:
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- If the LCM of $20x^3y^2$ and $10x^4y^4$ is $20x^4y^4$, find the HCF.
Important Questions from LCM and HCF
Six bells begin to toll together and toll, respectively, at intervals of 3, 4, 6, 7, 8 and 12 seconds. After how many seconds, will they toll together again?
A and B are two prime numbers such that A > B and their LCM is 209. The value of A 2 - B is:
Find the least number which when divided by 12, 18, 24 and 30 leaves 4 as remainder in each case, but when divided by 7 leaves no remainder.
Calculate the HCF of \(\frac{12}{5}\) , \(\frac{14}{15}\) and \(\frac{16}{17}\) .
Three numbers are in the proportion of 3 : 8 : 15 and their LCM is 8280. What is their HCF?
