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Find the least number which when divided by 12, leaves a remainder of 7, when divided by 15, leaves a remainder of 10 and when divided by 16, leaves a remainder of 11
- a)115
- b)235
- c)247
- d)475
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
Find the least number which when divided by 12, leaves a remainder of...
12 – 7 = 5, 15 – 10 = 5 and 16 – 11 = 5
Hence the desired number is 5 short for divisibility by 12, 15 and 16.
L.C.M. of 12, 15, 16 is 240
Hence the least number = 240 – 5 = 235
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Find the least number which when divided by 12, leaves a remainder of...
To find the least number that satisfies the given conditions, we can use the concept of the least common multiple (LCM) of the numbers 12, 15, and 16.
The LCM of three numbers is the smallest number that is divisible by all three numbers. In this case, we are looking for the smallest number that leaves remainders of 7, 10, and 11 when divided by 12, 15, and 16 respectively.
Let's solve this step by step:
Step 1: Find the LCM of 12, 15, and 16
The prime factorization of these numbers is as follows:
12 = 2^2 * 3
15 = 3 * 5
16 = 2^4
To find the LCM, we take the highest power of each prime factor that appears in any of the numbers:
LCM = 2^4 * 3 * 5 = 240
Step 2: Find the number that satisfies the given conditions
To find the least number, we need to find the smallest multiple of 240 that leaves remainders of 7, 10, and 11 when divided by 12, 15, and 16 respectively.
We can start by adding 240 to each of the remainders until we find a number that satisfies all the conditions:
Remainder of 7: Adding 240 to 7 gives us 247, which leaves a remainder of 7 when divided by 12.
Remainder of 10: Adding 240 to 10 gives us 250, which leaves a remainder of 10 when divided by 15.
Remainder of 11: Adding 240 to 11 gives us 251, which leaves a remainder of 11 when divided by 16.
Therefore, the least number that satisfies all the conditions is 247.
Hence, the correct answer is option B) 235.
The LCM of three numbers is the smallest number that is divisible by all three numbers. In this case, we are looking for the smallest number that leaves remainders of 7, 10, and 11 when divided by 12, 15, and 16 respectively.
Let's solve this step by step:
Step 1: Find the LCM of 12, 15, and 16
The prime factorization of these numbers is as follows:
12 = 2^2 * 3
15 = 3 * 5
16 = 2^4
To find the LCM, we take the highest power of each prime factor that appears in any of the numbers:
LCM = 2^4 * 3 * 5 = 240
Step 2: Find the number that satisfies the given conditions
To find the least number, we need to find the smallest multiple of 240 that leaves remainders of 7, 10, and 11 when divided by 12, 15, and 16 respectively.
We can start by adding 240 to each of the remainders until we find a number that satisfies all the conditions:
Remainder of 7: Adding 240 to 7 gives us 247, which leaves a remainder of 7 when divided by 12.
Remainder of 10: Adding 240 to 10 gives us 250, which leaves a remainder of 10 when divided by 15.
Remainder of 11: Adding 240 to 11 gives us 251, which leaves a remainder of 11 when divided by 16.
Therefore, the least number that satisfies all the conditions is 247.
Hence, the correct answer is option B) 235.
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Find the least number which when divided by 12, leaves a remainder of 7, when divided by 15, leaves a remainder of 10 and when divided by 16, leaves a remainder of 11a)115b)235c)247d)475Correct answer is option 'B'. Can you explain this answer?
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