find the smallest number which,when divided by 12, 48, 60, leaves remainder is every case
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3
Answer:
7
- Every positive number is divisible by 1
- All even numbers(ending with 0,2,4,6,8) are divisible by 2. 12/2 is 6, 48/2 is 24 and 60/2 is 30
- 12/3=4, 48/3=16, 60/3=20. Thus all numbers are divisible by 3 with 0 as remainder
- 12/4= 3, 48/4= 12, 60/4= 15. 0 is remainder for all
- 12/5= 2.4 , 48/5= 9.6, 60/5= 12. though 12 and 48 leave a remainder when divided by 5, 60 is divisible by 5 without remainder.
- 12/6=2, 48/6=8, 60/6=10. 0 is remainder for all.
- 12 (rem is 5), 48 (rem is 6), 60 (rem is 4) Thus 7 is the answer.
PLS MARK BRAINLIEST
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