Find the smallest positive integer that leaves a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5.
59
Each remainder is exactly one less than its divisor (2 is 1 less than 3, 3 is 1 less than 4, 4 is 1 less than 5).
So the required number plus 1 must be divisible by the LCM of 3, 4 and 5: \(\text{LCM}(3,4,5) = 60\).
Smallest positive integer: \(60-1 = 59\).
Hence, the smallest positive integer satisfying all conditions is 59.
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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?
