Advertisements
Advertisements
प्रश्न
Find the least number which when divides 35, 56 and 91 leaves the same remainder 7 in each case.
उत्तर
Least number which can be divided by 35, 56 and 91 is LCM of 35, 56 and 91.
Prime factorization of 35, 56 and 91 is:
35 = 5 × 7
56 = 23 × 7
91 = 7 × 13
LCM = product of greatest power of each prime factor involved in the numbers = 23 × 5 × 7 × 13 = 3640
Least number which can be divided by 35, 56 and 91 is 3640.
Least number which when divided by 35, 56 and 91 leaves the same remainder 7 is 3640 + 7 = 3647.
Thus, the required number is 3647.
APPEARS IN
संबंधित प्रश्न
Prove that the square of any positive integer is of the form 3m or, 3m + 1 but not of the form 3m +2.
Define HOE of two positive integers and find the HCF of the following pair of numbers:
475 and 495
Find the largest number which exactly divides 280 and 1245 leaving remainders 4 and 3, respectively.
Using prime factorization, find the HCF and LCM of 144, 198 In case verify that HCF × LCM = product of given numbers.
Is it possible to have two numbers whose HCF if 25 and LCM is 520?
Find the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively.
If a = 23 ✕ 3, b = 2 ✕ 3 ✕ 5, c = 3n ✕ 5 and LCM (a, b, c) = 23 ✕ 32 ✕ 5, then n =
If n is a natural number, then 92n − 42n is always divisible by ______.
The remainder when the square of any prime number greater than 3 is divided by 6, is
For any positive integer n, prove that n3 – n is divisible by 6.
