Find the least number that when divided by 18 ,24 and 56 leaves a remainder of 3 in each case
Answers
The least number which when divided by 18,24 and 56 leaves a remainder 3 in each case is 507.
Concept used:
- To find the least number which when divided by 18,24,and 56 leaves remainder 3 in each case, First we have to find the L.C.M. of 18,24,and 56 and then add 3 in the L.C.M. of these 3 numbers.
- LCM of two or more numbers = Product of the greatest power of each prime factor involved in the numbers, with highest power.
Given:
Number that when divided by 18 ,24 and 56 leaves a remainder of 3 in each case.
To find :
The least number
Solution:
Step 1: Find the L.C.M of 18 ,24 and 56 by prime factorization method:
18 = 2 × 3 × 3 = 2¹ × 3²
24 = 2 × 2 × 2 × 3 = 2³ × 3¹
56 = 2 × 2 × 2 × 7 = 2³ × 7¹
L.C.M = 2³ × 3² × 7¹
LCM = 2 × 2 × 2 × 3 × 3 × 7 = 504
L.C.M of 18 ,24 and 56 = 504
Step 2: Add the remainder to the LCM:
Add remainder 3 to the L.C.M of 18 ,24 and 56 = 504 + 3 = 507
Hence, the least number which when divided by 18,24 and 56 leaves a remainder 3 in each case is 507.
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Step-by-step explanation:
answer is
507
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