The smallest number, which when divided by 5, 10, 12 and 15 leaves remainder 2 in each case; but when divided by 7 leaves no remainder, is
Answers
Answer:
The number is 182
Step-by-step explanation:
To find the LCM of 5, 10, 12 and 15
5 = 5
10 = 2 x 5
12 = 2 x 2 x 3 = 2^2 x 3
15 = 3 x 5
LCM = 2^2 x 3 x 5 = 60
leaves remainder 2 in each case we take 60 + 2 = 62
But 62 is not divisible by 7
To take multiples of 60 and add 2 to the number,
next multiple is 90.
But 92 is not divisible by 7.
Then next multiple is 180.
182 is divisible by 7.
Therefore smallest number, which when divided by 5, 10, 12 and 15 leaves remainder 2 in each case; but when divided by 7 leaves no remainder, is 182
Answer:
182
Step-by-step explanation:
Given The smallest number, which when divided by 5, 10, 12 and 15 leaves remainder 2 in each case; but when divided by 7 leaves no remainder, is
First let us find LCM of 5, 10, 12 and 15. Factors of 5,10,12 and 15 are 5 x 2 x 3 x 2 = 60. Now required number is of the form 60 k + 2. We need to find the least value of k where it is exactly divisible by 7.
So 60 x 1 + 2 = 62 which is not divisible by 7
60 x 2 + 2 = 122 is not divisible by 7
60 x 3 + 2 = 182 is exactly divisible by 7
The required number is 182