find the smallest number which when divided by 35 , 56 and 105 leaves a remainder 6 in each case.
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Answered by
5
Answer:
Therefore smallest number which, when divided by 35,56 and 105, leaves a remainderof 5 in each case is 845. When a number n isdivided by 6 its remainder is 1 when thenumber is divided by 5 its remainder is 4 and when it is divided by 11 its remainder is 5.
Answered by
8
Answer:
846
Step-by-step explanation:
let the number be x.
A/q: dividend = divisor*quotient+remainder
1. x= 35a+6
2. x=56b+6
3. x=105c+6
As you can see above equations will give us
35a=56b=105c
So you need a number which should be divided by all or you need to find LCM of all three and then add 6 to it finally.
LCM is 840. Just add 6 to it and your answer is 846.
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