What would be the smallest natural number which when divided either by 20 or…
2018
What would be the smallest natural number which when divided either by 20 or by 42 or by 76 leaves a remainder of 7 in each case?
- A.
3047
- B.
6047
- C.
7987
- D.
63847
Attempted by 71 students.
Show answer & explanation
Correct answer: C
Key idea: any number that leaves remainder 7 when divided by 20, 42, and 76 must be of the form 7 + m, where m is divisible by each of 20, 42, and 76. To get the smallest such number, take m as the least common multiple (LCM) of the three divisors.
Find prime factorizations: 20 = 2^2 * 5, 42 = 2 * 3 * 7, 76 = 2^2 * 19.
Take highest powers of all primes: LCM = 2^2 * 3 * 5 * 7 * 19 = 7980.
Smallest number = 7 + 7980 = 7987.
Check: 7987 − 7 = 7980, and 7980 is divisible by 20, 42, and 76, so 7987 leaves remainder 7 in each case. Therefore 7987 is the smallest such natural number.
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