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What would be the smallest natural number which when divided either by 20 or by 42 or by 76 leaves a remainder ‘7’ in each case is_
- a)3047
- b)6047
- c)7987
- d)63847
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
What would be the smallest natural number which when divided either by...
We need a number that can be written as 20x + 7, 42y + 7, 76z + 7 for three integers x, y and z. We basically need to find least common multiple of 20 (2 * 2 *5), 42(2 * 3 * 7) and 76(2 * 2 * 19) which is 2 * 2 * 5 * 3 * 7 * 19 = 20 * 21 * 19 = 420 * 19 = 7980 So our number is 7980 + 7 = 7987.
Note - You can also verify options using virtual calculator.
Note - You can also verify options using virtual calculator.
Most Upvoted Answer
What would be the smallest natural number which when divided either by...
To find the smallest natural number that satisfies the given conditions, we need to find the least common multiple (LCM) of 20, 42, and 76.
First, let's find the LCM of 20 and 42:
Factorizing 20: 2^2 * 5
Factorizing 42: 2 * 3 * 7
To find the LCM, we take the highest power of each prime factor:
LCM(20, 42) = 2^2 * 3 * 5 * 7 = 420
Now, let's find the LCM of 420 and 76:
Factorizing 420: 2^2 * 3 * 5 * 7
Factorizing 76: 2^2 * 19
Again, we take the highest power of each prime factor:
LCM(420, 76) = 2^2 * 3 * 5 * 7 * 19 = 7980
Therefore, the smallest natural number that leaves a remainder when divided by 20, 42, or 76 is 7980.
First, let's find the LCM of 20 and 42:
Factorizing 20: 2^2 * 5
Factorizing 42: 2 * 3 * 7
To find the LCM, we take the highest power of each prime factor:
LCM(20, 42) = 2^2 * 3 * 5 * 7 = 420
Now, let's find the LCM of 420 and 76:
Factorizing 420: 2^2 * 3 * 5 * 7
Factorizing 76: 2^2 * 19
Again, we take the highest power of each prime factor:
LCM(420, 76) = 2^2 * 3 * 5 * 7 * 19 = 7980
Therefore, the smallest natural number that leaves a remainder when divided by 20, 42, or 76 is 7980.
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What would be the smallest natural number which when divided either by 20 or by 42 or by 76 leaves a remainder ‘7’ in each case is_a)3047b)6047c)7987d)63847Correct answer is option 'C'. Can you explain this answer?
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